NUCLEAR CHEMISTRY

Contents

• Introduction
• Atom and atomic structure
• Atomic Mass and Atomic mass unit (amu)
• Forces that hold atoms together
• Nuclear Binding Energy and Mass Defect
• Nuclear Stability
• Radioactivity
• Radioactive Emissions
• Detection of Radiation
• Nuclear Transmutation
• Nuclear Fusion and Fission
• Half Life, Rate of Radioactive decay and Decay Rate Constant
• Worked Examples of Questions in Radioactivity
• Practice Questions on Radioactivity
• Application of Radioactivity
• Key Terms
• Recommended Videos
• Test Questions
• Discuss And Explain

Learning Objectives

By the end of this section, you should be able to:

• Describe a Nucleus.
• Describe Radioactivity and what leads to it.
• Describe Radioactive Emissions.
• Calculate Radioactive half life, rate of decay, and decay rate constant of any radioactive sample.

Introduction

Up until now, we have been discussing electrons, elements, and different kinds of chemical reactions, and these chemical reactions majorly involve electrons. How about we take a look at the nucleus of atoms. The nucleus of an atom does not play any major role in chemical bonding or chemical reaction, but it is a very wide and interesting topic to study. The nucleus of an atom can undergo changes, and can also be induced to undergo a change. The study of the nucleus of atoms, how it changes, and many other stuffs relating to it, is called Nuclear chemistry; I always think about nuke every time this topic is being discussed, maybe because it usually involves a massive amount of energy - like a nuke💥

Atom and atomic structure

Yes, before we delve into nuclear chemistry, let us take a quick revision on atom and atomic structure. 

Every matter has atoms. 

A matter is anything that has mass and take up space, anything at all, including yourself. 

The interesting thing is that a matter can be made up of pure or a mixture of pure substances in any state; talking about a human now, we are made up of several pure substances; take only one pure substance, then we will be talking about an element. 

Every element has a unique kind of atom, like I've once said in a previous topic, no two elements have the same atom; and like you know before now, an atom is the smallest unit of an element. 

An atom is mostly empty space, and is composed of two main regions: Nucleus, and the Electron Cloud.

A Nucleus is located at the centre of an atom, and it is veeery tiny, when I say veeery tiny, I mean very very very tiny. 

Found in a Nucleus are protons and neutrons; combination of both is referred to as Nucleon. 

Nucleons are enclosed within a Nucleus. 

Protons often carry positive charge, while neutrons have no charge. This makes a nucleus to be positively charged. 

However, despite the fact that a nucleus is very tiny, most of the mass of every atom is as a result of the mass of their nucleus.

When we talk about an Electron Cloud, an Electron Cloud takes up the most volume of an atom. 

Found in an Electron Cloud are electrons, which are spinning in orbitals. 

Electrons carry negative charge, which usually would cancel out the effect of the positive charge of a nucleus, making an atom to be neutral, or inert, or ground. 

When you compare the size of a nucleus to the size of an electron cloud, I once saw a publication that said it is like comparing the size of a stadium to one grain of sand at its center, and yes, that is how it is, nevertheless it will not be drawn that way, so you can get to picture a nucleus too.

Nucleus at the centre surrounded by electron cloud

PRO TIP: radius of an atom is usually measured in picometres (pm) which is 10^-12m, or Angstrom (Å) which is 10^-10m or 100pm.

I believe you got the point of my analysis, a nucleus is tiny, yet it contributes the most to the mass of an atom, while an electron cloud takes up the most volume of an atom.

It is also important to note that not all atoms of an element are necessarily the same. 

Several Atoms of an element can differ in the number of neutrons that they carry. 

Every atom of a particular element must have the same number of protons, that must be constant, but, the number of neutrons can vary in different atoms, and when we talk about an atom (of an element) that has a specific number of neutrons, and another atom (of that same element) that has a different number of neutrons, they are referred to as isotopes of the element. I believe you must have discussed that during your secondary school level.

Mass number of an atom is the sum of the number of its protons, and the number of its neutrons. 

This can also be called Nucleon number, since nucleon is a collection of all protons and neutrons in an atom.

However, due to the existence of isotopes, we don't just take the mass number of any atom as being the real mass number for an element.

The average mass number of all its atoms is usually accepted, which is the average of the number of neutrons carried by all its atoms + number of protons (can be of anyone, since it is constant). 

It is due to this that you would see mass numbers having decimals, whereas atomic numbers, which is proton number, usually are whole numbers.

PRO TIP: you can always get the proton number of an element from the periodic table. The number (on the periodic table) where an element is found, is its atomic number and its proton number. It is also its electron number, if the atom is inactive.

The mass number and atomic number for an atom are usually written as:

$${}_Z^{A}X$$

Where A = Mass number,

             Z = Atomic number,

             X = Atom of a specific element.

Example, Carbon– 12 has a mass number of 12, and atomic number of 6, thus carbon– 12 can be written as:

${}_6{}^{12}C$

Another example, radioactive Uranium has a mass number of 238, and an atomic number of 92, this can be written as:

${}_{92}{}^{238}U$

Remember that mass number = number of protons + number of neutrons,

Hence, number of neutrons will mathematically = mass number - number of protons.

The Nucleus of a specific atom, of a specific element is called a Nuclide.

Atomic Mass and Atomic mass unit (amu)

While discussing atoms and their mass numbers before now, you saw that atomic mass numbers are large numbers, and you may be thinking, uhm, can a hydrogen atom really weigh 1g despite being extremely small? if you were thinking kilograms, haha, that's way too much. Even grams is way too much.

Well, let me put it like this; the actual mass of a proton is 1.67252 × 10^-27 Kg. Can you see how very small that is? 

Also, the mass of a neutron is 1.67496 × 10^-27 Kg and the mass of an electron is 9.1095 × 10^-31 Kg. 

The mass of atoms of elements are also nearly as small as these. 

So, because these masses are too small to be used for calculations all the time, the atomic mass unit was introduced.

Atomic mass unit is a unit for measuring extremely small mass in chemistry, and was introduced by studying the common Carbon–12 atom, at the end of the day, we could approximate the masses of atoms (of every other element) without using their very small amounts in kilograms. 

In short, an atomic mass unit (commonly written as amu) is $\frac{1}{12}$ of the total mass of a Carbon– 12 atom, which is in ground or inactive state. 

In other words, one atomic mass unit is gotten by dividing the mass of one carbon–12 atom by 12. 

When a Carbon– 12 atom is in ground state, its total mass in kilograms is approximately 1.992 × 10^-26 Kg.

Therefore, one atomic mass unit = 

$\frac{1.99265\times {10}^{-26}}{12}$ = $1.66054\times {10}^{-27}Kg$

The mass of a proton, neutron, and electron can therefore be expressed in atomic mass units, calculated mathematically as:

1.66054 × 10^-27 Kg = 1 amu                       —(1)

1.67252 × 10^-27 Kg = a amu                       —(2)

1.67496 × 10^-27 Kg = b amu                       —(3)

9.1095 × 10^-31 Kg = c amu                          —(4)

Dividing equations (2), (3), (4) by equation (1), we get:

a = 1.0072

b = 1.0087

c = 0.000549

Therefore, mass of a proton = 1.0072 amu

                   mass of a neutron = 1.0087 amu

                   mass of an electron = 0.000549 amu

Finally, we can confidently express the mass of atoms of other elements in amu.

1 hydrogen atom has a mass of 1.00782 amu, 1 helium atom has a mass of 4.0017 amu, and as we go along the periodic table, the masses of elements are expressed primarily in amu, that is why you see them as large numbers.

How do we convert from amu, back to Kg?

I believe you should know how to do that by now-

You simply calculate:

   [mass (in amu)]   ×   [1.66054 × 10^-27] 

= mass (in Kg)

This explanation is given because at some point, we shall be dealing with conversion of amu to Kg, and Kg to amu in this topic.

Forces that hold atoms together

We primarily have two forces that hold atoms together

• Electromagnetic force (or Electrostatic force)
• Nuclear force (or Strong Force)

Electromagnetic force keep electrons near the nucleus. This is due to the electrostatic force of attraction between protons in the nucleus, and electrons in the electron cloud. 

Remember the law of electrostatic forces? opposite charges attract! and yes, that electromagnetic force holds the electrons close to a nucleus.

Now, if you also remember from the electrostatic law, same charges push away one another, which means protons are actively repelling one another, and this could make a nucleus to break apart, however, this does not happen due to Nuclear forces. 

Nuclear force binds protons and neutrons together, and it is so strong that electrostatic repulsions between protons cannot overcome it. 

Therefore, neutrons are called by many, a kind of nuclear glue, they have no charge so no repulsion or attraction is exerted on them.

Particle	Charge	Mass (amu)	Location	Function
Electron	-1	0	Electron cloud	Behaviour of element
Proton	+1	1	Nucleus	Identity of Element
Neutron	0	1	Nucleus	Stability of Nucleus

Nuclear Binding Energy and Mass Defect

When we take a look at a hydrogen atom, a hydrogen atom has only one proton in its nucleus, and no neutron is present. 

This proton exerts its mass on the atom, therefore, a hydrogen atom has a total mass of about 1.00782 amu.

However, when we consider a helium atom, helium has 2 protons, 2 neutrons in its nucleus. 

Adding the masses of these together, we have:

   2 × (1.0072)    +   2 × (1.0087)

= 2.0144             +        2.0174

= 4.0318 amu

This value does not match the experimental value for mass of helium which is 4.0017 amu.

Why?

Scientists give an explanation that some of the mass is lost as energy when protons and neutrons come together to form a nucleon. 

It is nearly the same energy that will be required to break the nuclear forces that hold the nucleons together. 

This Energy is known as Nuclear Binding energy.

In this sense, it can be said that helium lost a mass of (4.0318 – 4.0017 =) 0.0301 amu which has been converted to energy. 

This difference between the calculated mass of helium's proton + neutron, and the actual mass of helium atom is known as Mass Defect. 

Mass defect defines the mass that is lost as energy during nuclear binding.

Albert Einstein showed, in his theory of special relativity, that mass and energy are simply different manifestations of the same fundamental quantity.

Mass and energy are related by the famous formula:

$$E=m{c}^2$$

Where E = Energy (in joules)

             m = mass (in Kg)

             c = speed of light (in ms^-1)

This relationship makes us to believe that we can convert matter to energy and energy to matter.

With Einstein's formula, we can calculate the amount of energy that was evolved as a result of lost mass conversion or transformation, for a helium atom.

E = unknown

m = lost mass (or mass converted into energy) = 0.0301 amu

We cannot use amu in our calculation, remember that we need to change it into a standard unit, hence, we have:

[0.0301 amu]  ×  [1.66054 × 10^-27 Kg/amu]

= 4.998 × 10^-29 Kg

c = 3.0 × 10⁸ m/s

Solving:

E = (4.998 × 10^-29)Kg × [(3.0 × 10⁸)m/s]²

E = (4.998 × 10^-29)Kg × (9.0 × 10¹⁶)m²/s²

E = (4.998 × 10^-29 × 9.0 × 10¹⁶) Kgm²/s²

E = (4.998 × 9.0 × 10^-29 × 10¹⁶) Kgm²/s²

E = (44.982 × 10^{-29 + 16}) Kgm²/s²

E = (44.982 × 10^-13) Kgm²/s²

E = (4.4982 × 10^-12) Kgm²/s²

E = 4.4982 × 10^-12 J

Thus, 4.4982 × 10^-12 J is calculated as energy lost by a helium atom during nuclear binding, and it is approximately the same energy that will be required to break apart the nucleon of a helium atom into separate protons and neutrons.

I got a friend who laughed and said that this energy is too small, that means he could break apart the nucleus of any helium he sees?

Well, he actually got cleared, that, this energy is for one atom of Helium only, One mole of Helium can contain as much as more than 10²³ atoms, and he was told to calculate the total amount of energy required for one mole, and then he ended up like wow!

The energy released from a Nuclear reaction is far much more than energy released from a Chemical reaction.

Additionally, Binding energy of atoms are usually very small, so scientists prefer to describe it in terms of electron volt, which is equivalent to 1.6 × 10^-19 Joules. 

Hence, for Helium, it is

1.6 × 10^-19 J = 1eV        –(1)

4.4982 × 10^-12 J = unknown eV          –(2)

Cross multiplying, and dividing

unknown eV = $\frac{4.4982\times {10}^{-12}}{1.6\times {10}^{-19}}$

                       = $2.81\times {10}^7\text{eV}$

This can be re-written as 28.1 MeV. 

Pronounced as twenty eight point one mega electron volt, or twenty eight point one million electron volt. This is binding energy for a helium atom.

The binding energy per nucleon for an atom is:

$\frac{\text{its binding energy}}{\text{its mass number which same as the nucleon number}}$

The binding energy per nucleon for helium will then become:

 $\frac{28.1\text{MeV}}{4\text{nucleons (mass number)}}$

$=7.025\text{MeV per nucleon}$

Graph of Nuclear binding energy per nucleon against nucleon number

Nuclear Stability

A nucleus can either be stable or unstable.

What makes a nucleus stable?

This actually depends on a variety of factors and no single rule allows us to predict whether a nucleus will be stable or not, unless we observe it. There are some observations that have been made to help us make predictions.

A popular one is Neutron to proton ratio.

Hydrogen has no neutron, so neutron to proton ratio is 0 : 1

Helium has two neutrons and two protons, thus neutron to proton ratio is 1 : 1

Lithium has four neutrons and three protons, thus neutron to proton ratio is 1.33 : 1

Strong nuclear forces exists between these nucleons. The more protons that are packed together, the more neutrons are needed to bind the nucleus together.

Elements upto number 20 have almost equal number of protons and neutrons. Heavier elements with higher atomic numbers have more neutrons to protons, and this keeps increasing along the periodic table. The neutrons help to glue the protons together in the nucleons. 

It then turns out to be observed that elements with much more protons and/or neutrons tend to start to become unstable, as they try to become stable, they end up emitting particles and rays, making them to be called radioactive. 

Beyond atomic-number of 83, it becomes impossible for the nuclear force to keep up with the proton - proton repulsion in the nucleus, so nucleus of atoms cease to be stable. 

This means that a nucleus with 84 protons or more will be radioactive regardless of how many neutrons it has (because of proton - proton repulsion).

Another good way of comparing the relative stabilities of different nuclei (nuclei is the plural of nucleus) is to look at their binding energies in terms of the binding energy per nucleon, that is, the nuclide's binding energy divided by the number of nucleons. 

A higher binding energy per nucleon means more stability, = the nucleons are more tightly bound in the nucleus. 

Carefully take a look at the graph illustrated above, you will find out that nuclides around mass number 60 have the highest binding energies per nucleon, meaning that they are most stable.

Iron nuclide with mass number 56 is known to have the most stable nucleon.

There are several important conclusions that we can draw from the graph above. 

First, it shows that certain nuclides are more stable than we might have expected them to be if the slope of the graph were smooth. The stabilities of ${}_2{}^4\mathrm{He}$ ${}_6{}^{12}C$ ${}_8{}^{16}O$ and ${}_{10}{}^{20}\mathrm{Ne}$ are all high. 

This is explained by the fact that they each have an even number of protons and an even number of neutrons. In short, paired nucleons (like paired electrons) are more stable than unpaired ones. 

We also have a set of numbers dubbed the magic numbers, nucleons that possess these magic numbers seem to be stable. These are predictions, and predictions do not always go pretty as planned, therefore you will see lots of exceptions, all these will be fully understood by Chemistry students in higher classes.

When a nucleus is unstable, it tries to achieve stability by emitting particles and rays from its nucleus, therefore transforming into another nucleus that tends to be stable, this is termed as radioactivity.

Radioactivity

Every element has atleast one isotope which is unstable. Unstable atoms try their best to achieve stability, hence they begin to spontaneously transform into another nuclei. 

How do they do this? 

They transform into another nuclei by emitting high energy particles and rays. This is known as radioactivity or radioactive decay. 

Atoms that exhibit radioactive decay are called radioactive atoms or radioisotopes. A Nucleus that undergoes radioactive decay is called a radionuclide.

A certain level of radiation naturally occur around us, even in our bodies.

It should be noted that radioactivity is not affected by temperature, pressure, chemical change or physical state.

Although radioactivity is spontaneous, it normally occur slowly, but the rates of occurrence vary widely.

When a radioactive atom emits rays and particles to become stable, it is said to undergo nuclear transformation, since its nucleus is actively changing form.

A radioactive atom can achieve stability by emitting one kind of particle and/or ray, while some others need to emit particles and rays more than once, the systematic changes that a radioactive atom undergo, transforming into several nuclei until it finally achieves stability is termed radioactive series. 

In other words, Some nuclei will not gain stability by one emission, and may undergo many until they reach stability. For example, Uranium-238 decays to Thorium-234 which decays to Protactinium-234 which decays… until eventually, lead-206 is obtained. The decay series is called a radioactive series.

What kind of particles or rays do radioactive atoms emit to achieve stability?

Radioactive Emissions

Alpha particles (α-particles)

When we talk about an alpha particle, alpha particle is given off when a nucleus has too many protons, because too many protons cause excessive repulsion. 

In an attempt to reduce the repulsion between the protons, the nucleus has to eventually remove some of its protons, and it does this by removing two protons, which will be accompanied by two neutrons. Thus the nucleus is said to decay by emitting an alpha particle, and this is termed as an alpha decay. 

How does an alpha particle look like?

Many call it a helium nucleus, because it is composed of two protons and two neutrons like a typical helium nucleus.

Two protons + two neutrons = nucleon number of 4

Alpha particle: 

$${}_2^4\text{He}$$

A radionuclide that emits an alpha particle has lost two protons and two neutrons.

This means that as it lost two protons, its atomic number will be reduced by two, and this makes it to become an atom of another element.

And also, the mass number will be reduced by four.

It is therefore shown in an equation as:

${}_Z{}^{A}X$        →         ${}_{Z-2}{}^{A-4}Y$     +        ${}_2{}^4\mathrm{He}$

For example:

Uranium–238 loses an alpha particle to become Thorium– 234 can be written as:

${}_{92}{}^{238}U$      →       ${}_{90}{}^{234}\mathrm{Th}$      +       ${}_2{}^4\mathrm{He}$

You should see that this is almost just like the regular chemical equations, no variable is lost. The sum of the masses (of Th and He) on the RHS is the same as the mass on the LHS. Also, the sum of the atomic numbers on the RHS is the same as the LHS.

This equation can also be written as:

${}_{92}{}^{238}U$      →       ${}_{90}{}^{234}\mathrm{Th}$      +       ${}_2{}^4\alpha$

Alpha Decay 

What are the properties of an Alpha particle?

The first you should know, is that an alpha particle is positively charged. An alpha particle has no electron, and because proton is positively charged and neutron has no charge, the net charge in an alpha particle results in positive.

An alpha particle is relatively heavy. The mass of two protons and two neutrons combined makes an alpha particle about 4 amu, which is heavy when compared to other particles.

Due to the positive charge and relatively heavy mass of an alpha particle, it can be slightly attracted (or deflected) towards a negative plate in an electrostatic field. Again this is due to electrostatic force of attraction and repulsion, an alpha particle gets repelled by the positive plate in an electrostatic field, while the negative plate welcomes it. This is slight though.

An alpha particle has 5 - 10% the speed of light. I believe you should be able to approximate its speed (in m/s) by yourself, right now.

Due to the fact that an alpha particle has only a positive charge, it has great affinity for electrons, and when it collides with other particles, it tends to remove their electrons, making itself a stable helium atom, while the particle which has lost an electron will become ionised. This is one of the facts that explains why an alpha particle is quickly destroyed. It ionises the particles in air and becomes a helium atom which may considerably be pointless, however, high amounts of alpha particles can travel further since they will need to collide with, and ionise more particles. 

An alpha particle can be absorbed by a thin sheet of paper and exposure to high amounts can only cause irritation to the skin.

A Reddit user once asked that what is the shape and colour of an alpha particle? Well, I've never seen one before, I only know it as a particle and very very very microscopic, if perhaps you know the answer to that you can let us know via the comments section. Please remember not to give an irrelevant comment because it will get removed in the next instant.

It is noteworthy that only elements with atomic number of more than 83 tend to achieve stability by emitting alpha particles.

Beta Particles (β-particles)

Talking about a beta particle, this is usually interesting; when a nucleus has too many neutrons than is needed, the 'too many neutrons' cause instability, and what happens thereafter, is that a neutron will split into a proton and an electron, and an antimatter particle called antineutrino. The proton is kept, the antineutrino has hardly any effect, while the electron is emitted at high speeds. The emitted electron is called a beta particle. They are fast moving streams of particles that are electron-like, and yes, very fast because they have a very much lesser mass, although slightly higher than of electrons.

How does a beta particle look like?

Many call it a lost electron from the splitting of a neutron. It is different from losing an electron from the electron cloud. A beta particle can be said to be composed of just one electron, no proton, no neutron, hence it has a negative charge.

A beta particle is shown as:

$${}_{-1}^0\beta$$

A radionuclide that emits a beta particle has no changes in its mass number, but its atomic number is increased by one. 

There's no net change in its mass number due to the fact that despite a neutron is split and lost, a proton is absorbed, and as a result, the nucleus transforms into an atom of another element, yet still having the same mass number. 

Atoms of different elements that have the same mass numbers are referred to as Isobars. 

Emission of beta Particles lead to different Isobars. 

Examples are ${}_{55}{}^{140}\mathrm{Cs}$     ${}_{56}{}^{140}\mathrm{Ba}$     ${}_{57}{}^{140}\mathrm{La}$     ${}_{58}{}^{140}\mathrm{Ce}$

Equations showing the emission of a beta particle are written as:

$${}_Z^A\text{X}\to {}_{Z+1}^A\text{Y}+{}_{-1}^0\beta$$
$${}_{49}^{116}\text{In}\to {}_{50}^{116}\text{Sn}+{}_{-1}^0\beta$$
$${}_{82}^{207}\text{Pb}\to {}_{83}^{207}\text{Bi}+{}_{-1}^0\beta$$
$${}_6^{14}\text{C}\to {}_7^{14}\text{N}+{}_{-1}^0\beta$$

It can be seen that addition of the atomic numbers on the RHS must be the same as the one on the LHS.

Beta Decay

Beta particle emission is the most common form of beta decay.

Beta decay occurs anywhere along the periodic table. Beta decay has other non-common, complex types which I believe full discussion is not for your level yet:

We have Positron emission, which is characterised as another type of beta particle emission, but here, instead of having a negative charge, it has a positive charge, and is called a positron. Positron emission occurs when there are more protons than neutrons, proton therefore splits into a neutron and a positron, leading to a reduction in the proton number, hence atomic number drops, but mass number remain unchanged.

Second is Electron Capture, this also occurs when there are more protons than neutrons. Here, an electron is captured from the electron cloud (predominantly the k shell) which counteracts the charge of a proton, making it transform into a neutron, and an electron neutrino. Chemistry students will get detailed explanation in higher classes.

What are the properties of a beta particle?

A beta particle has a negative charge.

A beta particle has a mass that is close to that of an electron.

Due to the negative charge and relatively small mass of a beta particle, it is strongly deflected towards the positive plate of an electrostatic field; I believe you know why by now- electrostatic forces.

A beta particle, due to its small mass and strong emission, has 90% the speed of light, thus it is always moving very fast and has higher penetration power than alpha particles.

Beta particles are smaller and fast moving which means that they collide less easily with other particles. When they eventually collide with another particle, they transfer energy to the particle, this can make electrons in the particle to become excited or become lost, making them ionised. As a result, beta particles are also ionising, but not very much like alpha particles.

Beta particles can pass through paper, but will be stopped by a thin sheet of metal.

Gamma Rays ($\gamma$ - rays)

When a nucleus has just emitted an alpha or a beta particle to attain stability, the newly formed nucleus is usually at a very high energy, this high energy nucleus has to drop to a lower energy state, and by so doing, it emits a high energy photon, known as gamma rays.

Unlike alpha and beta particles, gamma ray is not a particle, but a high energy electromagnetic radiation, often occurs as a way to release energy during alpha or beta decay.

Gamma radiation

A gamma ray has no charge or mass so it moves very fast and with a high penetrating power.

An atom that emits gamma rays only drops to a lower energy state, no neutron or proton is lost.

Gamma rays can penetrate deeply through paper, thin sheet of metal, but is absorbed by a very thick lead block, not very completely though. 

When a person is exposed to gamma rays, it can penetrate the body tissues and reach the cells, too much uncontrolled exposure can cause mutation of a cell's DNA, turning it into a cancerous cell which may grow and attack the others. A systematic and controlled exposure can also be used to alter the DNA of a cancerous cell.

Detection of Radiation

So, I was watching a Transformers movie many years back, and then men from one sector 7 came into the Witwicky guy's house and placed some instrument close to his body, the instrument started giving a loud beep indicating that he had been touched by the alien transformers, and then I could remember, it was one of the parts that I would reverse back to and think that "how is this possible?" The movie was filled with a lot of movie tricks, like a lot lot, but when I got to understand radiation, I got to understand why the instruments were able to give a loud beeping sound.

Unless radiation changes into another form of energy, it cannot be felt, and neither can it be seen but it can be detected by instruments.

A popular one is Geiger Müller Counter which majorly exploits the conductivity of ions of inert gases, like argon ions to make its speaker give a click sound for each particle. High energy particles remove electrons from argon to ionize argon, argon ions which can conduct electricity will then make the connected speaker to give sound.

Geiger Müller Counter | photo credit: Wikimedia commons

Others are scintillation counter (which uses a substance that gives off light when struck by high energy particles) and Thermoluminiscent dosimeters (in which electrons that are excited by high energy particles are trapped in crystals, and then the crystals are heated to make the electron return to ground state, therefore emitting light). There are several others that you will see as you reach higher classes.

Nuclear Transmutation

The nucleus of a particular element can be made to transform into a nucleus of another element. This can result from radioactive decay of the parent nuclide or by artificial methods. 

The conversion of one nucleus of a particular atom, into the nucleus (of an atom) of another element is termed as Nuclear Transmutation. 

An example is when a carbon nuclide changes into a nitrogen nuclide, or when a uranium nuclide changes into a thorium nuclide.

Like said earlier, nuclear transmutation can occur by natural methods or by artificial methods.

When we say natural methods of nuclear transmutation, we mean a type of transmutation that occurs without human intervention. It is usually spontaneous. 

Many scientists believe that the universe was created by a sudden and tremendous explosion, and then one thing started becoming the other, where one atom changed into another atom, it is explained as due to natural transmutation. This never stops, and is even going on right now around you.

Artificial transmutation, on the other hand, is an induced type of transmutation where a particular nuclide is forced to change into another one, usually intentionally done by human. 

The first known artificial transmutation was carried out by Sir Rutherford in his Lab many years ago, when he converted Nitrogen– 14 to Oxygen— 17, by bombarding Nitrogen– 14 with alpha particles from Radium. 

What happens in artificial transmutation is that a nuclide is bombarded with particles to form a new nuclide. The particles that are used for bombardment can be a neutron, proton, or another nuclei. 

When using neutrons, it can be directly used to bombard nuclei since a neutron has no charge, but using a proton or another nuclei will require acceleration, because the particles must be moving fast to overcome the electrostatic repulsion between the charged particle and the charged nucleus. This is where particle accelerators come in. 

Known examples of particle accelerators are Cyclotron and The Large Hadron Collider. A cyclotron accelerates charged particles using a high frequency alternating voltage, and a magnetic field. The Cyclotron then produces beam of particles that is then used to shoot at other nuclei, to create and study isotopes.

Cyclotron | photo credit: Wikimedia commons

Large Hadron Collider | source: internet

Elements that have been made by artificial transmutation include Natural Elements like Technetium, Astatine, Francium and Promethium. Others are known as Transuranium Elements, they are the elements beyond atomic number 92, and have not been discovered in nature. All of these elements are radioactive.

Nuclear Fusion and Fission

When we take a look at the terms fusion and fission, fusion simply means joining together, while fission means dividing.

We can then look at Nuclear fusion as the joining together of small nuclei to give a larger nucleus, while Nuclear fission is the splitting of a heavy nucleus to give smaller nuclei.

Both of these nuclear reactions give off a lot of energy. Fusion is known to give more energy, and it is theoretically believed to occur in the sun, but it is usually not feasible to carry out in real life, therefore, fission is mostly being used.

Nuclear Fusion

Nuclear Fission 

A nuclear fission is usually kind of induced, because no nucleus will decide to make itself to split just like that. How then do we make fission happen? We usually do this by bombarding neutrons to a large nucleus; when the large nucleus finally absorbs a neutron, it becomes unstable, and then splits into smaller nuclei, and some other particles are given off.

One thing that is usually noticed, is that, when one radionuclide undergoes nuclear fission and produces two new nuclei, and we make another of that same radionuclide to undergo another nuclear fission, it is not everytime that same two products are given off, and the particles given off do vary too.

But, when Uranium- 235 is bombarded with a neutron to make it undergo nuclear fission, uranium-235 produces two smaller nuclei and 3 or 2 neutrons. And its property of always continually giving out neutrons as particles is exploited in Nuclear chain reactions.

In a Nuclear chain reaction, neutrons are bombarded to a uranium- 235 nucleus, causing it to transform into uranium- 236, thus becoming unstable. Therefore it splits into two nuclei and usually 3 or 2 neutrons are given off. These neutrons that are given off are absorbed by another uranium atoms, leading to another fission that will produce two nuclei and 3 or 2 neutrons; the neutron products can then be absorbed by another uranium atoms, causing another fission. This is primarily known as a chain reaction, and the mass needed in order to sustain a chain reaction is called the critical mass. If the mass is too small, instead of the neutrons to hit another uranium-235 atom, the neutrons escape, and this is known as subcritical mass, the supercritical mass is the opposite and is the good one, where the neutrons cannot escape, but have to hit another uranium- 235 atom, and the chain keeps going on like that.

Nuclear Chain Reaction

This chain reaction is usually used in nuclear power plants which generate energy that is used by electric generators to generate electricity for your homes and hostels. However, when accidents happen, the damage can be devastating, like the Chernobyl disaster in Ukraine which is still suffered till today. There's a reason Uranium- 235 isotope is being used and not the others like uranium– 234 or uranium– 238, which may not support chain reaction, and there are reasons Boron or cadmium rods are used in the Nuclear chain reactor, these usually act as neutron absorbers and control rods, chemistry students will see extensive details on these in higher classes.

Half Life, Rate of Radioactive decay and Decay Rate Constant

When a radioactive element undergoes radioactive decay, it happens with respect to time. Some radionuclides can decay very fast while some others would decay very slowly. The time taken for a given sample of a radioactive nuclei to decay and only half of its original is left, is termed as half life. In other words, half life is a measure of the time taken for half of the total atoms in a given radioactive element to decay.

In actuality, half life of a radioactive element is independent of any sample size. This means that no matter how large or how small a sample is, the half life of that nuclei still remains the same, so the nuclei will continue to decay by half everytime it reaches its half life.

An uproar was stirred on the internet quite sometime ago, when people proposed that a radioactive element would never decay to absolute zero, some argued and said no, while many others said decay will happen asymptotically (=scientific term for endless, till infinity) and then finally stop when nobody knows or can calculate.

Well, discussion on that is not for your level yet, but for now, you should note that as an element decays by half, the rate of its decomposition continues to decrease. This rate of decay is directly affected by the active number of atoms.

In other words, the rate at which a nucleus emits radiation (and decay) does not depend on temperature, surface area or any catalyst; but it depends on how many radioactive atoms are present in a sample.

And, since the number of radioactive atoms present in a sample continue to decrease, rate of decay will also continue to decrease, hence, there is a direct proportionality relationship between rate of radioactive decay and the number of atoms in any given sample.

The rate of a radioactive decay is a first order kinetic process, which I believe you must have seen in CHM 101, this property makes us to be able to get a decay rate constant.

Decay rate constant (λ) is as a result of the direct proportionality of rate of radioactive decay, to number of atoms that are disintegrating

R  $\alpha$  N

R = λN

Where R is the rate of decay,

            N is the number of atoms that are disintegrating at a particular given time,

            λ is the constant of proportionality, and is known as the decay rate constant.

By mathematical methods, we got to know that the half life of a radioactive element is related to its decay rate constant by:

$$t_{1/2}=\frac{0.693}{\lambda }$$

Where,

t_1/2 = half life

λ = decay rate constant.

This makes us conclude that decay rate constant is dependent on half life, and half life is dependent on decay rate constant. Both of these do not depend on number of atoms or sample size.

We can then proceed to know how to calculate the size of a radioactive nuclei that will remain after a particular time.

This is given as: 

$$ln\left(\frac{N{}_o}{N{}_t}\right)=\lambda T$$

Where N_o = Original number of atoms,

             N_t = Number of atoms left after time T,

             λ = decay rate constant,

              T = time spent.

When we know half life but do not know the decay rate constant, we substitute λ with $\frac{0.693}{t_{1/2}}$ and the equation becomes:

$$ln\left(\frac{N{}_o}{N{}_t}\right)=\frac{0.693}{t{}_{1/2}}\times T$$

A very long time ago during my secondary school days, I found this formula for half life kind of complex to use because my then calculator couldn't solve equations involving Natural Logarithm (ln), so I had to look for a way out, and after a thorough brain rack and several trials and errors, I was able to come up with a formula that works perfectly for half life.

$$m=M\times \left(\frac{1}{2}\right){}^{{}^{\frac{T}{t}}}$$

Where m = mass left after time T,

            M = original mass,

            T = time spent,

            t = half life.

I used a small m for mass left because it is usually smaller than the original mass which is M.

By then, I'd thought I had created something new which only I knew how to use. Not until I entered FUNAAB and discovered I wasn't actually the only one using it. It was painful, but by then I have already used it to pass several exams, so it was all cool.

My actual formula was looking kind of scattered, it was actually $m=\frac{M}{2{}^{{}^{\frac{T}{t}}}}$ but it's still the same with the other formulas, just that never use it in any written exam, the lecturer may fail it because it is unconventional.

What we are going to be basically doing in half life and decay rate constant are calculations.

You should note that the unit for rate of radioactive decay is number of disintegrations per second (dis/s) and in standard units, it is called Becquerel (Bq).

Another unit for rate of decay is Curie (Ci). 1 Curie is equal to 3.7 × 10¹⁰ disintegrations per second.

You will understand all these as we move into the calculations aspect.

Calculations on Half Life, Rate of decay, and Decay Rate Constant

Example 1:

After 32 days, 5 milligrams of an 80 milligram sample of a radioactive isotope remains unchanged. What is the half life (in days) of this element?

Solution

I always tell students to always write out every given parameters in a question before proceeding to solve, it gives you a very clear view of what you are dealing with.

$T$ = 32 days

Mass Left $\left(N{}_t\right)$ = 5mg

Original Mass $\left(N{}_o\right)$ = 80mg

$t_{1/2}$ = ?

Solving:

Recall,

$ln\left(\frac{N{}_o}{N{}_t}\right)=\lambda T$

Replacing $\lambda$ with $\frac{0.693}{t_{1/2}}$

We therefore have:

$ln\left(\frac{N{}_o}{N{}_t}\right)=\frac{0.693}{t_{1/2}}\times T$

Substituting known values:

$ln\left(\frac{80}{5}\right)=\frac{0.693}{t_{1/2}}\times 32$

$ln(16)=\frac{0.693\times 32}{t_{1/2}}$

$2.773=\frac{22.176}{t_{1/2}}$

Cross Multiplying;

$2.773\xcancel{=}\frac{22.176}{t_{1/2}}$

$2.773\times t_{1/2}=22.176$

Dividing both sides by $2.773$;

$\frac{2.773\times t_{1/2}}{2.773}=\frac{22.176}{2.773}$

$\frac{\cancel{2.773}\times t_{1/2}}{\cancel{2.773}}=\frac{22.176}{2.773}$

$t_{1/2}=\frac{22.176}{2.773}$

$t_{1/2}=7.997$

$t_{1/2}=8$

It is calculated that the element has a half life of 8 days.

Using the Unconventional Short Formula

$T$ = 32 days

$m$ = 5mg

$M$ = 80mg

$t$ = ?

Solving:

Recall;

$m=M\times \left(\frac{1}{2}\right){}^{{}^{\frac{T}{t}}}$

Substituting known values;

$5=80\times \left(\frac{1}{2}\right){}^{{}^{\frac{32}{t}}}$

$5=80\times \frac{1{}^{{}^{\frac{32}{t}}}}{2{}^{{}^{\frac{32}{t}}}}$

$1$ raised to power anything is still $1$, we then have;

$5=80\times \frac{1}{2{}^{{}^{\frac{32}{t}}}}$

$5=\frac{80\times 1}{2{}^{{}^{\frac{32}{t}}}}$

Cross Multiplying;

$5\xcancel{=}\frac{80\times 1}{2{}^{{}^{\frac{32}{t}}}}$

$5\times 2{}^{{}^{\frac{32}{t}}}=80$

Dividing both sides by 5;

$\frac{5\times 2{}^{{}^{\frac{32}{t}}}}{5}=\frac{80}{5}$

$\frac{\cancel{5}\times 2{}^{{}^{\frac{32}{t}}}}{\cancel{5}}=\frac{80}{5}$

$2{}^{{}^{\frac{32}{t}}}=16$

Expressing $16$ in power of $2$, is $2^4$;

$2{}^{{}^{\frac{32}{t}}}=2^4$

By exponential relation rule;

$\cancel{2}{}^{{}^{\frac{32}{t}}}={\cancel{2}}^4$

$\frac{32}{t}=4$

Cross multiplying;

$\frac{32}{t}\xcancel{=}4$

$32=4t$

Dividing both sides by 4;

$\frac{32}{4}=\frac{4t}{4}$

$\frac{32}{4}=\frac{\cancel{4}t}{\cancel{4}}$

Rearranging;

$t=\frac{32}{4}$

t = $8$

Half life is 8 days

Example 2:

The half life of ${}_{53}^{131}\text{I}$ is 8 days. How much ${}_{53}^{131}\text{I}$ from a 32g sample remains after 5 half-lives?

Solution

$t_{1/2}$ = 8 days

Original Mass $\left(N{}_o\right)$ = 32g

$T$ = 5$t_{1/2}$ = $5\times 8=40\text{days}$

Mass Left $\left(N{}_t\right)$ = ?

From;

$ln\left(\frac{N{}_o}{N{}_t}\right)=\frac{0.693}{t_{1/2}}\times T$

Substituting known values;

$ln\left(\frac{32}{N{}_t}\right)=\frac{0.693}{8}\times 40$

$ln\left(\frac{32}{N{}_t}\right)=\frac{0.693\times 40}{8}$

$ln\left(\frac{32}{N{}_t}\right)=\frac{0.693\times {\cancel{40}}^5}{{\cancel{8}}_1}$

$ln\left(\frac{32}{N{}_t}\right)=0.693\times 5$

$ln\left(\frac{32}{N{}_t}\right)=3.465$

At this stage, we have to change $ln$ to its log form, which is $Lo{g}_e$

$Lo{g}_{{}_e}\left(\frac{32}{N{}_t}\right)=3.465$

Changing from Logarithm to index form;

$\frac{32}{N{}_t}=e^{3.465}$

$\frac{32}{N{}_t}=31.976$

Cross multiplying;

$\frac{32}{N{}_t}\xcancel{=}31.976$

$N{}_t\times 31.976=32$

Dividing both sides by $31.976$

$N{}_t=\frac{32}{31.976}$

$N{}_t\approx 1g$

It is calculated that 1g of ${}_{53}^{131}\text{I}$ will remain from 32g after 5 half lives.

Using the Unconventional short formula

$t$ = 8 days

$M$ = 32g

$T$ = 5$t$

$m$ = ?

$m=\frac{M}{2{}^{{}^{\frac{T}{t}}}}$

Substituting known values;

$m=\frac{32}{2{}^{{}^{\frac{5t}{t}}}}$

$m=\frac{32}{2{}^{{}^{\frac{5\cancel{t}}{\cancel{t}}}}}$

$m=\frac{32}{2^5}$

$m=\frac{32}{32}$

$m=1g$

Example 3:

Cobalt-60 is used in radiation therapy. It has a half life time of 5.26 years.
    (a)Calculate the rate constant (in s^-1) for radioactive decay.
    (b)What fraction of a certain sample will remain after 12 years?

Solution

(a)
Half life of Cobalt-60 ($t_{1/2}$) = 5.26 years

Decay rate constant (λ) = ?

Decay rate constant is given by $\lambda =\frac{0.693}{t_{1/2}}$

Since we are asked to calculate the decay rate constant in per seconds, we need to convert the half life in years to seconds, how do we do that?

365 days make one year, so 1 year will have (365) days

24 hours make one day, One year will therefore have (365 × 24) hours

60 minutes make one hour, so 1 year will have (365 × 24 × 60) minutes

60 seconds make one hour, so 1 year will have (365 × 24 × 60 × 60) seconds

This becomes 1 year = 31536000 seconds

5.26 years will then become 5.26 × 31536000 = 165879360 seconds

We can then go ahead to calculate the decay rate constant by;

$\lambda =\frac{0.693}{165879360}$

$\lambda =4.18\times {10}^{-9}{\text{s}}^{-1}$

Decay rate constant in per seconds for Cobalt- 60 is $4.18\times {10}^{-9}{\text{s}}^{-1}$

(b)
Time ($T$) = 12 years

Fraction that will be left = $\frac{\text{size of leftover after time T}}{\text{size of original sample}}$ = $\frac{N{}_t}{N{}_o}$ = $\frac{1}{x}$

Recall,

$ln\left(\frac{N{}_o}{N{}_t}\right)=\frac{0.693}{t_{1/2}}\times T$

$ln\left(\frac{N{}_o}{N{}_t}\right)=\frac{0.693}{5.26}\times 12$

$ln\left(\frac{N{}_o}{N{}_t}\right)=\frac{0.693\times 12}{5.26}$

$ln\left(\frac{N{}_o}{N{}_t}\right)=\frac{8.316}{5.26}$

$ln\left(\frac{N{}_o}{N{}_t}\right)=1.58$

Changing from $ln$ to $Log$;

$Lo{g}_{{}_e}\left(\frac{N{}_o}{N{}_t}\right)=1.58$

Changing from Log form to index form

$\frac{N{}_o}{N{}_t}=e^{1.58}$

$\frac{N{}_o}{N{}_t}=4.86$

We are trying to get $\frac{N{}_t}{N{}_o}$ not $\frac{N{}_o}{N{}_t}$, so we take the inverse of both sides, which is simply changing numerator to denominator, and denominator to numerator, we have:

$\frac{N{}_t}{N{}_o}=\frac{1}{4.86}$

$\frac{N{}_t}{N{}_o}=\frac{100}{486}$

$\frac{N{}_t}{N{}_o}=\frac{50}{243}$

The fraction that will be left of a sample of Cobalt- 60 after 12 years is $\frac{50}{243}$

Using the unconventional short formula

Calculation for (a) still remain the same method, (b) can follow alternate method:

Time ($T$) = 12 years

Half life ($t$) = 5.26 years

Fraction left after time ($T$) = $\frac{m}{M}$

Solving:

$m=\frac{M}{2{}^{{}^{\frac{T}{t}}}}$

$m=\frac{M}{2{}^{{}^{\frac{12}{5.26}}}}$

$m=\frac{M}{2^{2.28}}$

Cross multiplying;

$m\xcancel{=}\frac{M}{2^{2.28}}$

$m\times 2^{2.28}=M$

$m\times 4.86=M$

Dividing both sides by M;

$\frac{m\times 4.86}{M}=\frac{M}{M}$

$\frac{m\times 4.86}{M}=\frac{\cancel{M}}{\cancel{M}}$

$\frac{m\times 4.86}{M}=1$

Dividing both sides by 4.86;

$\frac{m\times 4.86}{M\times 4.86}=\frac{1}{4.86}$

$\frac{m\times \cancel{4.86}}{M\times \cancel{4.86}}=\frac{1}{4.86}$

$\frac{m}{M}=\frac{1}{4.86}$

$\frac{m}{M}=\frac{100}{486}$

$\frac{m}{M}=\frac{50}{243}$

The fraction that will be left of a sample of Cobalt- 60 after 12 years is $\frac{50}{243}$

Example 4:

If $\frac{1}{8}$ of an original sample of Krypton- 74 remain unchanged after 34.5 minutes, what is the half life of Krypton- 74?

Solution

$T$ = 34.5 mins

$N{}_t=\frac{1}{8}N{}_o$

$t_{1/2}$ = ?

Solving:

$ln\left(\frac{N{}_o}{N{}_t}\right)=\frac{0.693}{t_{1/2}}\times T$

$ln\left(\frac{N{}_o}{\frac{1}{8}N{}_o}\right)=\frac{0.693}{t_{1/2}}\times 34.5$

$ln\left(N{}_o÷\frac{N{}_o}{8}\right)=\frac{0.693\times 34.5}{t_{1/2}}$

$ln\left(N{}_o\times \frac{8}{N{}_o}\right)=\frac{0.693\times 34.5}{t_{1/2}}$

$ln\left(\cancel{N{}_o}\times \frac{8}{\cancel{N{}_o}}\right)=\frac{23.9085}{t_{1/2}}$

$ln\left(8\right)=\frac{23.9085}{t_{1/2}}$

$2.079=\frac{23.9085}{t_{1/2}}$

$2.079\xcancel{=}\frac{23.9085}{t_{1/2}}$

$t_{1/2}\times 2.079=23.9085$

Dividing both sides by $2.079$

$\frac{t_{1/2}\times 2.079}{2.079}=\frac{23.9085}{2.079}$

$\frac{t_{1/2}\times \cancel{2.079}}{\cancel{2.079}}=\frac{23.9085}{2.079}$

$t_{1/2}=\frac{23.9085}{2.079}$

$t_{1/2}=11.5$

It is calculated that Krypton- 74 has a half life of 11.5mins

Using the unconventional short formula

$T$ = 34.5 mins

$m=\frac{1}{8}M$

$t$ = ?

Solving:

$m=\frac{M}{2{}^{{}^{\frac{T}{t}}}}$

$\frac{1}{8}M=\frac{M}{2{}^{{}^{\frac{34.5}{t}}}}$

$\frac{M}{8}\xcancel{=}\frac{M}{2{}^{{}^{\frac{34.5}{t}}}}$

$2{}^{{}^{\frac{34.5}{t}}}\times M=8\times M$

Dividing both sides by $M$;

$\frac{2{}^{{}^{\frac{34.5}{t}}}\times M}{M}=\frac{8M}{M}$

$\frac{2{}^{{}^{\frac{34.5}{t}}}\times \cancel{M}}{\cancel{M}}=\frac{8\cancel{M}}{\cancel{M}}$

$2{}^{{}^{\frac{34.5}{t}}}=8$

Exponentially, $8=2^3$;

$2{}^{{}^{\frac{34.5}{t}}}=2^3$

$\cancel{2}{}^{{}^{\frac{34.5}{t}}}={\cancel{2}}^3$

$\frac{34.5}{t}=3$

$\frac{34.5}{t}\xcancel{=}3$

$3t=34.5$

Dividing both sides by 3;

$\frac{3t}{3}=\frac{34.5}{3}$

$\frac{\cancel{3}t}{\cancel{3}}=\frac{34.5}{3}$

$t=\frac{34.5}{3}$

$t=11.5$

It is calculated that the half life of Krypton- 74 is 11.5 mins

Example 5:

What is the half life of Sodium 25 if 1.00 gram of a 16.00 gram sample of Sodium 25 remains unchanged after 237 seconds.

Solution

$t_{1/2}$ = ?

$N{}_t$ = 1 gram

$N{}_o$ = 16 grams

$T$ = 237s

Solving:

$ln\left(\frac{N{}_o}{N{}_t}\right)=\frac{0.693}{t_{1/2}}\times T$

$ln\left(\frac{16}{1}\right)=\frac{0.693}{t_{1/2}}\times 237$

$ln\left(16\right)=\frac{0.693\times 237}{t_{1/2}}$

$2.773=\frac{164.241}{t_{1/2}}$

$2.773\xcancel{=}\frac{164.241}{t_{1/2}}$

$2.773\times t_{1/2}=164.241$

Dividing both sides by $2.773$

$\frac{2.773\times t_{1/2}}{2.773}=\frac{164.241}{2.773}$

$\frac{\cancel{2.773}\times t_{1/2}}{\cancel{2.773}}=\frac{164.241}{2.773}$

$t_{1/2}=\frac{164.241}{2.773}$

$t_{1/2}=59.23$

Half life of Sodium 25 as given in the question is 59.23s

Example 6:

How many days are required for 200 grams of Radon-222 to decay to 50 grams?
(Half life of Radon-222 is 3.82 days).

Solution

$T$ = ?

$M$ = 200 grams

$m$ = 50 grams

$t_{1/2}$ = 3.82 days

Solving:

$m=\frac{M}{2{}^{{}^{\frac{T}{t}}}}$

$50=\frac{200}{2{}^{{}^{\frac{T}{3.82}}}}$

$50\xcancel{=}\frac{200}{2{}^{{}^{\frac{T}{3.82}}}}$

$2{}^{{}^{\frac{T}{3.82}}}\times 50=200$

Dividing both sides by $50$

$\frac{2{}^{{}^{\frac{T}{3.82}}}\times 50}{50}=\frac{200}{50}$

$\frac{2{}^{{}^{\frac{T}{3.82}}}\times \cancel{50}}{\cancel{50}}=\frac{{\cancel{200}}^4}{{\cancel{50}}_1}$

$2{}^{{}^{\frac{T}{3.82}}}=4$

$2{}^{{}^{\frac{T}{3.82}}}=2^2$

$\cancel{2}{}^{{}^{\frac{T}{3.82}}}={\cancel{2}}^2$

$\frac{T}{3.82}=2$

$\frac{T}{3.82}\xcancel{=}2$

$T=3.82\times 2$

$T=7.64$

It will take 7.64 days for 200g of Radon-222 to decay to 50g

Example 7:

Strontium- 90 is one of the products of fission of Uranium- 235. This strontium isotope is radioactive, with a half life of 28.1 years. Calculate how long (in years) it will take for 1.0g of the isotope to be reduced to 0.2g by decay.

Solution

$t_{1/2}$ = 28.1 years

$N{}_o$ = 1.0g

$N{}_t$ = 0.2g

$T$ = ?

Solving:

$ln\left(\frac{N{}_o}{N{}_t}\right)=\frac{0.693}{t_{1/2}}\times T$

$ln\left(\frac{1.0}{0.2}\right)=\frac{0.693}{28.1}\times T$

$ln\left(5\right)=0.02466\times T$

$1.6094=0.02466\times T$

Dividing both sides by $0.02466$;

$\frac{1.6094}{0.02466}=\frac{0.02466\times T}{0.2466}$

$\frac{1.6094}{0.02466}=\frac{\cancel{0.02466}\times T}{\cancel{0.2466}}$

$T=\frac{1.6094}{0.02466}$

$T=65.26$

It will take 65.26 years for 1.0g of a Strontium- 90 isotope to decay 0.2g.

Example 8:

Rubidium- 84 has a half life of 33 days. How many mg of a 10mg sample of this isotope would have disintegrated after 99 days?
(b) How many days are required for a 1mg sample of Rb- 84 to decay to 0.0625mg?

Solution

When you read the (a) part of this question well, you would see that it is a little bit different from what we have been solving before now. The question says that how many mg would have disintegrated , not how many mg will be left.
Therefore, sample that would have disintegrated is subtracting the sample that is left from the original.

Sample that would have disintegrated = $M-m$

The given parameters in the question are:

Original sample ($M$) = 10mg

Half life ($t$) = 33days

Time spent ($T$) = 99days

You can then calculate the leftover by using any formula you wish to use.

$m=\frac{M}{2{}^{{}^{\frac{T}{t}}}}$

$m=\frac{10}{2{}^{{}^{\frac{99}{33}}}}$

$m=\frac{10}{2^3}$

$m=\frac{10}{8}$

$m=\frac{{\cancel{10}}^5}{{\cancel{8}}_4}$

$m=\frac{5}{4}$

Mass that will be left after 99 days is $\frac{5}{4}$mg

Therefore, mass that would have disintegrated after 99 days is;

$10-\frac{5}{4}$

$\frac{40-5}{4}$

$\frac{35}{4}=8\frac{3}{4}$

Mass of 10mg of Rubidium- 84 that would have disintegrated after 99days is $8\frac{3}{4}$mg

(b)
$M$ = 1mg

$m$ = 0.0625mg

$T$ = ?

$t$ = 35 days

$m=\frac{M}{2{}^{{}^{\frac{T}{t}}}}$

$0.0625=\frac{1}{2{}^{{}^{\frac{T}{33}}}}$

$0.0625\xcancel{=}\frac{1}{2{}^{{}^{\frac{T}{33}}}}$

$2{}^{\frac{T}{33}}\times 0.0625=1$

Dividing both sides by $0.0625$;

$\frac{2^{\frac{T}{33}}\times 0.0625}{0.625}=\frac{1}{0.0625}$

$\frac{2^{\frac{T}{33}}\times \cancel{0.0625}}{\cancel{0.625}}=\frac{1}{0.0625}$

$2{}^{\frac{T}{33}}=\frac{1}{0.0625}$

$2{}^{\frac{T}{33}}=16$

$2{}^{\frac{T}{33}}=2^4$

$\cancel{2}{}^{\frac{T}{33}}={\cancel{2}}^4$

$\frac{T}{33}=4$

$\frac{T}{33}\xcancel{=}4$

$T=33\times 4$

$T=132$

It is calculated that 132 days would be required for 1mg of Rb- 84 to decay to 0.0625mg

Example 9:

A 100g of a radioactive substance has a half life of 10 years. How many grams are left after 30 years?

Solution

$N{}_o$ = 100g

$t_{1/2}$ = 10 years

$N{}_t$ = ?

$T$ = 30 years

Solving:

$ln\left(\frac{N{}_o}{N{}_t}\right)=\frac{0.693}{t_{1/2}}\times T$

$ln\left(\frac{100}{N{}_t}\right)=\frac{0.693}{10}\times 30$

$ln\left(\frac{100}{N{}_t}\right)=\frac{0.693\times 30}{10}$

$ln\left(\frac{100}{N{}_t}\right)=\frac{0.693\times {\cancel{30}}^3}{{\cancel{10}}_1}$

$ln\left(\frac{100}{N{}_t}\right)=0.693\times 3$

$ln\left(\frac{100}{N{}_t}\right)=2.079$

Changing from $ln$ to $Lo{g}_{{}_e}$;

$Lo{g}_{{}_e}\left(\frac{100}{N{}_t}\right)=2.079$

Changing from Log form to index form;

$\frac{100}{N{}_t}=e^{2.079}$

$\frac{100}{N{}_t}=8$

$\frac{100}{N{}_t}\xcancel{=}8$

$8\times N{}_t=100$

Dividing both sides by 8;

$\frac{8\times N{}_t}{8}=\frac{100}{8}$

$\frac{\cancel{8}\times N{}_t}{\cancel{8}}=\frac{100}{8}$

$N{}_t=\frac{100}{8}$

$N{}_t=\frac{{\cancel{100}}^{25}}{{\cancel{8}}_2}$

$N{}_t=\frac{25}{2}$

$N{}_t=12.5$

100g of the given radioactive substance will remain 12.5g after 30 years.

Example 10:

${}_{53}^{131}\text{I}$ has a half life of 8 days. How long will it take for a sample to decay so that only 10% of the original ${}_{53}^{131}\text{I}$ sample survives?

Solution

$t=8\text{ days }$

$m=10\mathrm{\%}\text{ of }M$

$m=\frac{10}{100}\times M$

$m=\frac{1}{10}\times M$

$m=\frac{M}{10}$

$T=$ ?

Solving:

$m=\frac{M}{2{}^{{}^{\frac{T}{t}}}}$

$\frac{M}{10}=\frac{M}{2{}^{{}^{\frac{T}{8}}}}$

There's a shortcut that I always use: Anytime there is a common variable on exact both sides of "=" they can always cancel out each other.

$\frac{a}{b}=\frac{a}{c}$ can become $\frac{\cancel{a}}{b}=\frac{\cancel{a}}{c}$

which becomes $\frac{1}{b}=\frac{1}{c}$

But in $\frac{a}{b}=\frac{c}{a}$ : One $a$ cannot cancel out the other $a$ because they are not on exactly same sides. Pls take note!

Coming back to our solution,

$\frac{M}{10}=\frac{M}{2{}^{{}^{\frac{T}{8}}}}$

Our equation becomes;

$\frac{\cancel{M}}{10}=\frac{\cancel{M}}{2{}^{{}^{\frac{T}{8}}}}$

$\frac{1}{10}=\frac{1}{2{}^{{}^{\frac{T}{8}}}}$

I'm only teaching you these shortcuts because as long as FUNAAB continues to use CBT for CHM 104, you will never have enough time for long calculations!

$\frac{1}{10}\xcancel{=}\frac{1}{2{}^{{}^{\frac{T}{8}}}}$

$2{}^{{}^{\frac{T}{8}}}=10$

But $10$ cannot be expressed in power of $2$, so we introduce Log;

$Log2{}^{{}^{\frac{T}{8}}}=Log10$

$\frac{T}{8}Log2=Log10$

Dividing both sides by Log 2;

$\frac{\frac{T}{8}Log2}{Log2}=\frac{Log10}{Log2}$

$\frac{\frac{T}{8}\cancel{Log2}}{\cancel{Log2}}=\frac{Log10}{Log2}$

$\frac{T}{8}=\frac{Log10}{Log2}$

$\frac{T}{8}=\frac{1}{0.3010}$

$\frac{T}{8}=3.32$

$\frac{T}{8}\xcancel{=}3.32$

$T=8\times 3.32$

$T=26.56$

It will take 26.56 days for 10% of a given sample of ${}_{53}^{131}\text{I}$ to survive.

Provided you're in a situation in which the options are expressed in days, down to seconds, you know you would have to convert your answer in decimal, to corresponding smaller units.

How do we convert 26.56 days to hours, minutes, and seconds?

We do that by:

26.56 days = 26 days, $\frac{56}{100}$ of one day.

This translates to;

26 days, $\frac{56}{100}$ × 24 hours

26 days, 13.44 hours

26 days, 13 hours, $\frac{44}{100}$ of one hour

26 days, 13 hours, $\frac{44}{100}$ × 60 minutes

26 days, 13 hours, 26.4 minutes

26 days, 13 hours, 26 minutes, $\frac{4}{10}$ of one minute

26 days, 13 hours, 26 minutes, $\frac{4}{10}$ × 60 seconds

26 days, 13 hours, 26 minutes, 24 seconds

Therefore, 26.56 days expressed in its smaller time units is 26 days, 13 hours, 26 minutes, 24 seconds.

Example 11:

${}^{99}\text{Tc}$ is used for brain imaging scan. The half life of ${}^{99}\text{Tc}$ is 6 hours. What percentage of a dose of ${}^{99}\text{Tc}$ is left after 24 hours?

Solution

$t_{1/2}$ = 6 hours

$T$ = 24 hours

$\mathrm{\%}$ left = $\frac{\text{amount left}}{\text{original amount}}\times 100$ = $\frac{N{}_t}{N{}_o}\times 100$

Solving;

$ln\left(\frac{N{}_o}{N{}_t}\right)=\frac{0.693}{t_{1/2}}\times T$

$ln\left(\frac{N{}_o}{N{}_t}\right)=\frac{0.693}{6}\times 24$

$ln\left(\frac{N{}_o}{N{}_t}\right)=\frac{0.693\times 24}{6}$

$ln\left(\frac{N{}_o}{N{}_t}\right)=\frac{0.693\times {\cancel{24}}^4}{{\cancel{6}}_1}$

$ln\left(\frac{N{}_o}{N{}_t}\right)=0.693\times 4$

$ln\left(\frac{N{}_o}{N{}_t}\right)=2.772$

Changing from $ln$ to $Lo{g}_{{}_e}$;

$Lo{g}_{{}_e}\left(\frac{N{}_o}{N{}_t}\right)=2.772$

Changing from Log form to Index form;

$\frac{N{}_o}{N{}_t}=e^{2.772}$

$\frac{N{}_o}{N{}_t}=16$

Finding the inverse of both sides;

$\frac{N{}_t}{N{}_o}=\frac{1}{16}$

$\mathrm{\%}$ left will therefore be;

$=\frac{1}{16}\times 100$

$=\frac{100}{16}$

$=6.25$

Therefore, $6.25\mathrm{\%}$ of a ${}^{99}\text{Tc}$ dose will remain after $24$ hours.

Example 12:

${}_6^{14}\text{C}$ has a half life of $5730$ years. Live Carbon- 14 has activity of $15.3$.
A shirt is claimed to be Jesus' but is found to have a Carbon- 14 activity of $14.0$. How old is the shirt? and can the claim be true?

Solution

This kind of question is usually posed during Carbon dating. Carbon dating measures the activity of carbon, and uses it to trace exact human history dates.

Carbon- 14 is synthesised in the atmosphere (and not from outer space like some people believe). It is synthesised by the transmutation of atmospheric Nitrogen- 14. Humans and Plants continue to maintain constant amount of Carbon- 14, until death. After death, carbon- 14 starts to decay. The half life of Carbon- 14 is 5730 years, and the activity of Carbon- 14 can be used to calculate the exact year a thing existed.

Now, Back to our Question;

This Question can be solved in so many ways, we'll still arrive at the same answer. But still, let us follow the normal way that we have been using.

The activity of a radioisotope is its Rate of Decay, and I believe you still remember that rate of radioactive decay is directly proportional to number of atoms. We will need it here.

$R=\lambda \times N$

We can get decay rate constant ($\lambda$) for Carbon- 14 by;

$\lambda =\frac{0.693}{t_{1/2}}$

$\lambda =\frac{0.693}{5730}$

$\lambda =0.0001209424$

Now, when Carbon- 14 has an activity of 15.3, the relative disintegrations of its number of atoms $N$ is;

$R=\lambda \times N$

$15.3=0.0001209424\times N$

Dividing both sides by $0.0001209424$;

$\frac{15.3}{0.0001209424}=\frac{0.0001209424\times N}{0.0001209424}$

$\frac{15.3}{0.0001209424}=\frac{\cancel{0.0001209424}\times N}{\cancel{0.0001209424}}$

$N=\frac{15.3}{0.0001209424}$

$N=126506.5$

Calculating $N$ for Carbon- 14 with activity $14.0$;

$R=\lambda \times N$

$14.0=0.0001209424\times N$

Dividing both sides by $0.0001209424$;

$\frac{14.0}{0.0001209424}=\frac{0.0001209424\times N}{0.0001209424}$

$\frac{14.0}{0.0001209424}=\frac{\cancel{0.0001209424}\times N}{\cancel{0.0001209424}}$

$N=115757.58$

If we make $N$ for live Carbon- 14 = $N{}_o$ and current $N$ for Carbon- 14 = $N{}_t$, we can calculate the time difference from live carbon to time of current decay rate.

We do that by:

$ln\left(\frac{N{}_o}{N{}_t}\right)=\lambda \times T$

$ln\left(\frac{126506.5}{115757.56}\right)=0.0001209424\times T$

$ln\left(1.092857\right)=0.0001209424\times T$

$0.0887954=0.0001209424\times T$

Dividing both sides by $0.0001209424$;

$\frac{0.0887954}{0.0001209424}=\frac{0.0001209424\times T}{0.0001209424}$

$\frac{0.0887954}{0.0001209424}=\frac{\cancel{0.0001209424}\times T}{\cancel{0.0001209424}}$

$T=734.196$

It is therefore calculated that the shirt is 734.2 years old.

Since Jesus existed more than 2000 years ago, the shirt's age is nowhere near this, so this claim is shown to be false.

A popular alternative to the above used method is to directly use the activities of the Carbon- 14 at the different times in place of their number of disintegrations.

This makes it that:

$ln\left(\frac{15.3}{14.0}\right)=0.0001209424\times T$

$ln\left(1.092857\right)=0.0001209424\times T$

$0.0887954=0.0001209424\times T$

Dividing both sides by $0.0001209424$;

$\frac{0.0887954}{0.0001209424}=\frac{0.0001209424\times T}{0.0001209424}$

$\frac{0.0887954}{0.0001209424}=\frac{\cancel{0.0001209424}\times T}{\cancel{0.0001209424}}$

$T=734.196$

It is therefore calculated again that the shirt is 734.2 years old.

Example 13:

${}^{{}^{108}}\text{Ag}$ has a half life of 2.4 minutes. Calculate the percentage of a sample of ${}^{{}^{108}}\text{Ag}$ which will have decayed after 7.2 minutes.

Solution

Now, this is another question that requires you to first understand it, before attempting to solve.

The simplest and most straightforward method is by:

Given a 100% sample of ${}^{{}^{108}}\text{Ag}$, after 2.4 mins which is its half life, only 50% of ${}^{{}^{108}}\text{Ag}$ will be left.

After another 2.4 mins, 25% of ${}^{{}^{108}}\text{Ag}$ will be left, which means total time spent is now 2.4 + 2.4 = 4.8 mins.

After another 2.4 mins, only 12.5% of ${}^{{}^{108}}\text{Ag}$ will be left, which means total time spent is now 4.8 + 2.4 = 7.2 mins.

Therefore, after 7.2 mins, only 12.5% of ${}^{{}^{108}}\text{Ag}$ is left. But we were told to find the percentage that will have decayed after 7.2 mins.

This makes it that, since 12.5% remained after 7.2 mins, amount that have decayed is 100% - 12.5% = 87.5%

This means that 87.5% of ${}^{{}^{108}}\text{Ag}$ will have decayed after 7.2 mins.

You should note that this method only worked because 7.2 is a direct multiple of 2.4.

If we assume that the time given in the question is 7.0 mins, this method would ultimately not work. This means the above simple method is selective.

Using our Formula method for solving:

Half life ($t$) = 2.4 minutes

Time spent ($T$) = 7.2 minutes

Original amount of sample = $M$

Leftover amount of sample = $m$

Amount that would have decayed = Original amount - Leftover amount = $M-m$

Percentage of amount of leftover = $\frac{m}{M}\times 100$

Percentage of amount that would decayed = $\frac{M-m}{M}\times 100$ = $100\mathrm{\%}-$ Percentage of amount of leftover

We can first get $\frac{m}{M}$ by;

$m=\frac{M}{2{}^{{}^{\frac{T}{t}}}}$

$m=\frac{M}{2{}^{{}^{\frac{7.2}{2.4}}}}$

$m=\frac{M}{2^3}$

$m=\frac{M}{8}$

$m\xcancel{=}\frac{M}{8}$

$8\times m=M$

Dividing both sides by m;

$\frac{8\times m}{m}=\frac{M}{m}$

$\frac{8\times \cancel{m}}{\cancel{m}}=\frac{M}{m}$

$\frac{M}{m}=8$

Since we are looking for $\frac{m}{M}$ not $\frac{M}{m}$, we take the inverse of both sides;

$\frac{m}{M}=\frac{1}{8}$

We can now calculate the percentage of leftover by;

$\frac{m}{M}\times 100$

$=\frac{1}{8}\times 100$

$=\frac{100}{8}$

$=12.5\mathrm{\%}$

It is calculated that 12.5% of a sample will be left after 7.2 mins

The percentage of sample that have decayed will therefore be;

$100-12.5$

$=87.5\mathrm{\%}$

This means that 87.5% of ${}^{{}^{108}}\text{Ag}$ will have decayed after 7.2 mins.

Example 14:

It is found that the activity of a sample of ${}^{{}^{212}}\text{Bi}$ falls to 25% of its original value after 121 minutes. Calculate the half life of ${}^{{}^{212}}\text{Bi}$.

Solution

It can be seen from Example 12 that activity of a radioisotope can be used in place of $N$ in $ln\left(\frac{N{}_o}{N{}_t}\right)=\lambda \times T$

$N{}_o$ = 100$\mathrm{\%}$

$N{}_t$ = 25$\mathrm{\%}$

$T$ = $121$ minutes

$t_{1/2}=?$

$ln\left(\frac{N{}_o}{N{}_t}\right)=\frac{0.693}{t_{1/2}}\times T$

$ln\left(\frac{100\mathrm{\%}}{25\mathrm{\%}}\right)=\frac{0.693}{t_{1/2}}\times 121$

$ln\left(4\right)=\frac{0.693\times 121}{t_{1/2}}$

$1.3863=\frac{83.853}{t_{1/2}}$

$1.3863\xcancel{=}\frac{83.853}{t_{1/2}}$

$t_{1/2}\times 1.3863=83.853$

$\frac{t_{1/2}\times 1.3863}{1.3863}=\frac{83.853}{1.3863}$

$\frac{t_{1/2}\times \cancel{1.3863}}{\cancel{1.3863}}=\frac{83.853}{1.3863}$

$t_{1/2}=\frac{83.853}{1.3863}$

$t_{1/2}=60.5$

The half life of ${}^{{}^{212}}\text{Bi}$ in the given question is calculated to be 60.5 minutes.

Example 15:

How long will it take a radioactive element with a decay rate constant of 4.0 × 10^-4 s^-1 to decay to 0.10g from 0.50g.

Solution

$T$ = ?

$\lambda$ = 4.0 × 10^-4 s^-1

$N{}_t$ = 0.10g

$N{}_o$ = 0.50g

Solving:

$ln\left(\frac{N{}_o}{N{}_t}\right)=\lambda \times T$

$ln\left(\frac{0.50}{0.10}\right)=4.0\times {10}^{-4}\times T$

$ln\left(5\right)=4.0\times {10}^{-4}\times T$

$1.609438=4.0\times {10}^{-4}\times T$

$\frac{1.609438}{4.0\times {10}^{-4}}=\frac{4.0\times {10}^{-4}\times T}{4.0\times {10}^{-4}}$

$\frac{1.609438}{4.0\times {10}^{-4}}=\frac{\cancel{4.0\times {10}^{-4}}\times T}{\cancel{4.0\times {10}^{-4}}}$

$T=\frac{1.609438}{4.0\times {10}^{-4}}$

$T=\frac{1.609438}{4.0}\times {10}^4$

$T=0.4\times {10}^4$

$T=4.0\times {10}^3$

It will take 4000 s for the given radioactive element to decay from 0.50g to 0.10g.

Example 16:

A radioactive element decays with a decay rate constant of 2.02 × 10^-4 mins^-1. How long will it take for 90g of the substances to decay to 10g?

Solution

$\lambda$ = 2.02 × 10^-4 mins^-1

$T$ = ?

$N{}_o$ = 90g

$N{}_t$ = 10g

$ln\left(\frac{N{}_o}{N{}_t}\right)=\lambda \times T$

$ln\left(\frac{90}{10}\right)=2.02\times {10}^{-4}\times T$

$ln\left(9\right)=2.02\times {10}^{-4}\times T$

$2.19723=2.02\times {10}^{-4}\times T$

$\frac{2.19723}{2.02\times {10}^{-4}}=\frac{2.02\times {10}^{-4}\times T}{2.02\times {10}^{-4}}$

$\frac{2.19723}{2.02\times {10}^{-4}}=\frac{\cancel{2.02\times {10}^{-4}}\times T}{\cancel{2.02\times {10}^{-4}}}$

$T=\frac{2.19723}{2.02\times {10}^{-4}}$

$T=\frac{2.19723}{2.02}\times {10}^4$

$T=1.0877\times {10}^4$

It will take 10877 s for the given radioactive element to decay from 90g to 10g.

Example 17:

Calculate the half life for an element that disintegrates with a decay rate constant of 1.24 × 10^-4 s^-1.

Solution

This kind of question should be easy peasy for you by now.

$\lambda$ = 1.24 × 10^-4 s^-1

$t_{1/2}$ = ?

$\lambda =\frac{0.693}{t_{1/2}}$

$1.24\times {10}^{-4}=\frac{0.693}{t_{1/2}}$

$1.24\times {10}^{-4}\xcancel{=}\frac{0.693}{t_{1/2}}$

$t_{1/2}\times 1.24\times {10}^{-4}=0.693$

$\frac{t_{1/2}\times 1.24\times {10}^{-4}}{1.24\times {10}^{-4}}=\frac{0.693}{1.24\times {10}^{-4}}$

$\frac{t_{1/2}\times \cancel{1.24\times {10}^{-4}}}{\cancel{1.24\times {10}^{-4}}}=\frac{0.693}{1.24\times {10}^{-4}}$

$t_{1/2}=\frac{0.693}{1.24\times {10}^{-4}}$

$t_{1/2}=\frac{0.693}{1.24}\times {10}^4$

$t_{1/2}=0.55887\times {10}^4$

$t_{1/2}=5.589\times {10}^3$

An element that disintegrates with a decay rate constant of 1.24 × 10^-4 s^-1 has a half life of 5589 s.

Example 18:

A radioactive element decays with a rate constant of 4.0 × 10^-4 s^-1. How long will it take for 120g of the substance to decay to 60%.

Solution

$\lambda$ = 4.0 × 10^-4 s^-1

$T$ = ?

$N{}_o$ = 120 g

$N{}_t$ = 60% of $N{}_o$

$N{}_t$ = $\frac{60}{100}\times 120$

$N{}_t$ = $\frac{6\cancel{0}}{{}_1\cancel{10}\cancel{0}}\times {\cancel{120}}^{12}$

$N{}_t$ = 72 g

Solving:

$ln\left(\frac{N{}_o}{N{}_t}\right)=\lambda \times T$

$ln\left(\frac{120}{72}\right)=4.0\times {10}^{-4}\times T$

$ln\left(1.6667\right)=4.0\times {10}^{-4}\times T$

$0.51083=4.0\times {10}^{-4}\times T$

$\frac{0.51083}{4.0\times {10}^{-4}}=\frac{4.0\times {10}^{-4}\times T}{4.0\times {10}^{-4}}$

$\frac{0.51083}{4.0\times {10}^{-4}}=\frac{\cancel{4.0\times {10}^{-4}}\times T}{\cancel{4.0\times {10}^{-4}}}$

$T=\frac{0.51083}{4.0\times {10}^{-4}}$

$T=\frac{0.51083}{4.0}\times {10}^4$

$T=0.1277\times {10}^4$

$T=1.277\times {10}^3$

A radioactive element with a decay rate constant of 4.0 × 10^-4 s^-1 will take 1277 s to decay to 60%.

Example 19:

The half life of ${}_7^{13}\text{N}$ is 10.1 mins. Starting with 1000 atoms of ${}_7^{13}\text{N}$, how many atoms will remain after one hour.

Solution

$t$ = 10.1 mins

$M$ = 1000 atoms

$m$ = ?

$T$ = 1 hour = 60 mins

Solving:

$m=\frac{M}{2{}^{{}^{\frac{T}{t}}}}$

$m=\frac{1000}{2{}^{{}^{\frac{60}{10.1}}}}$

$m=\frac{1000}{2^{5.941}}$

$m=\frac{1000}{61.435}$

$m=16.277$

Atoms of ${}_7^{13}\text{N}$ that will be left after 1 hour is 16.277 atoms.

Example 20:

${}_{92}^{238}\text{U}$ has a half life of 5.3 × 10⁷ years. What is the rate constant for the process of the decay?

Solution

Another easy peasy question for you.

$t_{1/2}$ = 5.3 × 10⁷ years

$\lambda$ = ?

$\lambda =\frac{0.693}{t_{1/2}}$

$\lambda =\frac{0.693}{5.3\times {10}^7}$

$\lambda =\frac{0.693}{5.3}\times {10}^{-7}$

$\lambda =0.13\times {10}^{-7}$

$\lambda =1.3\times {10}^{-8}$

It is calculated that ${}_{92}^{238}\text{U}$ has a decay rate constant of 1.3 × 10^-8 year^-1

Other Worked Examples

Question 1:

Given the nuclear equation: $${}_{92}^{235}\text{U}+{}_0^1\text{n}\to {}_{56}^{142}\text{Ba}+{}_{36}^{91}\text{Kr}+3_0^1\text{n}+\text{energy}$$ State the type of nuclear reaction represented by the equation.

Answer:

The Nuclear Reaction that is represented by the equation above is a Nuclear Fission Reaction.

Question 2:

Which of these options is best for determining the actual age of earth?

(a) ${}^{238}\text{U}$

(b) ${}^{90}\text{Sr}$

(c) ${}^{60}\text{Co}$

(d) ${}^{14}\text{C}$

Answer:

${}^{238}\text{U}$ is best for measuring the age of earth because of its very long half life.

Question 3:

Which of these equations is an example of artificial transmutation?

(a) ${}_4^9\text{Be}+{}_2^4\text{He}\to {}_6^{12}\text{C}+{}_0^1\text{n}$

(b) $\text{U}+3{\text{F}}_2\to {\text{UF}}_6$

(c) ${}_{92}^{238}\text{U}\to {}_{90}^{234}\text{Th}+{}_2^4\text{He}$

(d) $\text{Ca}+2{\text{H}}_2\text{O}\to {\text{Ca(OH)}}_2+{\text{H}}_2$

Answer:

Option (a) is an artificial transmutation. Option (b) is a chemical reaction, Option (c) is natural transmutation by alpha decay, Option (d) is a chemical reaction.

Question 4:

Alpha particles are emitted during the radioactive decay of...

(a) Carbon- 14

(b) Calcium- 37

(c) Neon- 19

(d) Radon- 222

Answer:

Option (d) Radon- 222 is the only option that can decay by emitting Alpha particles. Only heavy elements that are beyond atomic number 83 can exhibit alpha decay.

Question 5:

Which isotope is most commonly used in the radioactive dating of the remains of organic materials?

(a) ${}^{14}\text{C}$

(b) ${}^{16}\text{N}$

(c) ${}^{32}\text{P}$

(d) ${}^{37}\text{K}$

Answer:

Carbon- 14 is usually used in radioactive dating. The activity of Carbon- 14 in an organic matter can be used to trace back the date when the matter was active.

Question 6:

Which type of radioactive emission has a positive charge and weak penetrating power?

(a) Alpha particle

(b) Beta particle

(c) Gamma radiation

(d) Neutron

Answer:

Alpha particle is the only option that has a positive charge. It also has the weakest penetrating power.

Question 7:

What is the name of the process in which the nucleus of an atom of one element is changed into the nucleus of an atom of another element?

Answer:

Nuclear transmutation describes the changing of the nucleus of an atom of one element, into the nucleus of an atom of another element.

Question 8:

In the Reaction ${}_{93}^{239}\text{Np}\to {}_{94}^{239}\text{Pu}+\text{X}$, what does $\text{X}$ represent?

(a) A Proton

(b) A Neutron

(c) An Alpha particle

(d) A Beta particle

Answer:

The nuclear equation can be written as ${}_{93}^{239}\text{Np}\to {}_{94}^{239}\text{Pu}+{}_b^a\text{X}$

$239=239+a$

$93=94+b$

$a=0$

$b=-1$

$\text{X}$ is then ${}_{-1}^0\text{X}$ which is a Beta particle.

Question 9:

Alpha particles and beta particles differ in

(a) Mass only

(b) Charge only

(c) Both Mass and Charge

(d) Neither Mass nor Charge

Answer:

Alpha and Beta particles do not have the same mass nor do they have the same charge. Option (c) is correct.

Question 10:

Given the Nuclear reaction: $${}_{27}^{60}\text{Co}\to {}_{-1}^0\text{e}+{}_{28}^{60}\text{Ni}$$ The reaction is an example of...

(a) Fission

(b) Fusion

(c) Artificial transmutation

(d) Natural transmutation

Answer:

The nuclear reaction shown above is a beta decay, which is a type of natural transmutation.

Question 11:

Einstein showed that mass can be converted to energy. Which of these show the correct relationship between mass and energy?

(a) $m=E{c}^2$

(b) $E=m{c}^2$

(c) $E=$ $\frac{\text{m}}{{\text{c}}^2}$

(d) $m=$ $\frac{\text{E}}{{\text{c}}^2}$

Answer:

Einstein showed in his formula that mass is related to energy by E = mc²

Question 12:

The stability of an isotope is based on its...

(a) Number of Neutrons, only

(b) Number of Protons, only

(c) Ratio of Neutrons to Protons

(d) Ratio of Electrons to Protons

Answer:

Neutron to Proton ratio is an important factor that helps to see if an isotope will be stable or not.

Question 13:

Which statement best describes Gamma radiation?

(a) It has a mass of 1 and a charge of 1

(b) It has a mass of 0 and a charge of -1

(c) It has a mass of 0 and a charge of 0

(d) It has a mass of 4 and a charge of +2

Answer:

Option (c) is correct. Gamma ray has a mass of 0 and a charge of 0.

Question 14:

Which of these changes take place in a Nuclear fusion reaction?

(a) Matter is converted to Energy

(b) Energy is converted to Matter

(c) Ionic bonds are converted to Covalent bonds

(d) Covalent bonds are converted to Ionic bonds

Answer:

In a Nuclear Fusion reaction, mass is lost as energy, hence, matter is converted to energy.

Question 15:

Given the Nuclear Equation: $${}_{10}^{19}\text{Ne}\to \text{X}+{}_9^{19}\text{F}$$ What particle is represented by X?

(a) Alpha

(b) Beta

(c) Neutron

(d) Positron

Answer:

The above equation can be re-written as ${}_{10}^{19}\text{Ne}\to {}_b^a\text{X}+{}_9^{19}\text{F}$

$19=a+19$

$10=b+9$

$a=0$

$b=1$

$\text{X}$ can then be written as ${}_1^0\text{X}$ which is a Positron.

A positron is ${}_{+1}^0\text{X}$

Question 16:

Given the Nuclear Equation: $${}_{99}^{253}\text{Es}+\text{X}\to {}_0^1\text{n}+{}_{101}^{256}\text{Md}$$ Which particle is represented by $\text{X}$?

(a) ${}_2^4\text{He}$

(b) ${}_{-1}^0\text{e}$

(c) ${}_0^1\text{n}$

(d) ${}_{+1}^0\text{e}$

Answer:

The above nuclear equation can be re-written as ${}_{99}^{253}\text{Es}+{}_b^a\text{X}\to {}_0^1\text{n}+{}_{101}^{256}\text{Md}$

$253+a=1+256$

$99+b=0+101$

$a=4$

$b=2$

$\text{X}$ can then be written as ${}_2^4\text{X}$ which is a Helium nucleus

Question 17:

Which natural radioactive emission has a charge of +2.

(a) Alpha particle

(b) Beta particle

(c) Gamma radiation

(d) All

Answer:

An Alpha particle has two protons which carry one positive charge each, and two neutrons which have no charge, and these make the net charge on an Alpha particle +2.

Question 18:

What is the activity, in Curies, of a radioisotope that has a decay rate of 1.85 × 10⁷ disintegrations/s.

Answer:

3.7 × 10¹⁰ disintegrations/s = 1 Curie

1.85 × 10⁷ disintegrations/s = X Curie

Cross multiplying;

$1\times 1.85\times {10}^7=X\times 3.7\times {10}^{10}$

$1.85\times {10}^7=X\times 3.7\times {10}^{10}$

$\frac{1.85\times {10}^7}{3.7\times {10}^{10}}=\frac{X\times 3.7\times {10}^{10}}{3.7\times {10}^{10}}$

$\frac{1.85\times {10}^7}{3.7\times {10}^{10}}=\frac{X\times \cancel{3.7\times {10}^{10}}}{\cancel{3.7\times {10}^{10}}}$

$X=\frac{1.85\times {10}^7}{3.7\times {10}^{10}}$

$X=\frac{1.85}{3.7}\times {10}^{7-10}$

$X=0.5\times {10}^{-3}$

$X=5\times {10}^{-4}$

Unknown curie = 5 × 10^-4 Curie

Therefore, 1.85 × 10⁷ disintegrations/s = 5 × 10^-4 Curie.

Question 19:

If ${}_{74}^{184}\text{W}$ has an average weighed mass of 183.9510 amu. Calculate its mass defect during formation, its binding energy and binding energy per nucleon.
(P = 1.007825 amu, N = 1.008665 amu)

Answer:

Weighed mass of ${}_{74}^{184}\text{W}$ = 183.9510 amu

Calculated mass of ${}_{74}^{184}\text{W}$ = ?

${}_{74}^{184}\text{W}$ has 74 protons as we can see from its atomic number.

Its number of neutrons is gotten by Mass number - proton number

= 184 - 74

= 110 neutrons

Calculated mass will become:

mass of 74 protons = 74 × 1.007825 = 74.57905

mass of 110 neutrons = 110 × 1.008665 = 110.95315

Addition of mass of protons + mass of neutrons =

= 74.57905 + 110.95315

= 185.5322

Calculated mass for ${}_{74}^{184}\text{W}$ is 185.5322 amu

Mass defect = Calculated mass - weighed mass

Mass defect = 185.5322 - 183.9510

Mass defect = 1.5812

The Mass defect of ${}_{74}^{184}\text{W}$ is 1.5812 amu

Binding Energy can be calculated by using Einstein's formula;

Binding Energy = mc²

We cannot use Mass defect in amu, so we convert 1.5812 amu to Kg by;

1.5812 × 1.66054 × 10^-27

= 2.626 × 10^-27 Kg

We can then go ahead to calculate for binding energy;

(c = 3 × 10⁸ m/s)

E = 2.626 × 10^-27 × (3 × 10⁸)²

E = 2.626 × 10^-27 × 9 × 10¹⁶

E = 2.626 × 9 × 10^-27 × 10¹⁶

E = 23.631 × 10^-27+16

E = 23.631 × 10^-11

E = 2.3631 × 10^-10

Nuclear binding energy for one atom of ${}_{74}^{184}\text{W}$ is 2.3631 × 10^-10 J

Binding Energy per nucleon = $\frac{\text{Binding Energy}}{\text{Nucleon number}}$

Binding Energy per nucleon = $\frac{2.3631\times {10}^{-10}}{184}$

Binding Energy per nucleon = 1.283 × 10^-12 J/ nucleon

Question 20:

Estimates show that the total energy output of the sun is 5 × 10²⁶ J/s. What is the corresponding mass loss in Kg of the sun?

Answer:

E = 5 × 10²⁶ J/s

m = ?

From E = mc²

5 × 10²⁶ = m × (3 × 10⁸)²

5 × 10²⁶ = m × 9 × 10¹⁶

$\frac{5\times {10}^{26}}{9\times {10}^{16}}=\frac{m\times 9\times {10}^{16}}{9\times {10}^{16}}$

$\frac{5\times {10}^{26}}{9\times {10}^{16}}=\frac{m\times \cancel{9\times {10}^{16}}}{\cancel{9\times {10}^{16}}}$

$m=\frac{5\times {10}^{26}}{9\times {10}^{16}}$

$m=\frac{5}{9}\times {10}^{26-16}$

$m=0.55\times {10}^{10}$

$m=5.5\times {10}^9$

The calculated mass loss of the sun in Kg is 5.5 × 10⁹ Kg/s.

Question 21:

Identify X in the equation: ${}_{53}^{135}\text{I}\to {}_{54}^{135}\text{Xe}+\text{X}$.

Answer:

The above equation can be re-written as ${}_{53}^{135}\text{I}\to {}_{54}^{135}\text{Xe}+{}_b^a\text{X}$

$135=135+a$

$53=54+b$

$a=0$

$b=-1$

$\text{X}$ is therefore ${}_{-1}^0\text{X}$ which is a Beta particle.

Question 22:

Identify X in

${}_{19}^{40}\text{K}\to {}_{-1}^0\beta +\text{X}$.

Answer:

The above equation can be re-written as ${}_{19}^{40}\text{K}\to {}_{-1}^0\beta +{}_b^a\text{X}$

$40=0+a$

$19=-1+b$

$a=40$

$b=20$

$\text{X}$ is therefore ${}_{20}^{40}\text{X}$ which is ${}_{20}^{40}\text{Ca}$

Question 23:

Identify X in

${}_{27}^{59}\text{Co}+{}_0^1\text{n}\to {}_{25}^{56}\text{Mn}+\text{X}$.

Answer:

The above equation can be re-written as ${}_{27}^{59}\text{Co}+{}_0^1\text{n}\to {}_{25}^{56}\text{Mn}+{}_b^a\text{X}$

$59+1=56+a$

$27+0=25+b$

$a=4$

$b=2$

$\text{X}$ is therefore ${}_2^4\text{X}$ which is a Helium nucleus.

Question 24:

A piece of fossilised wood has 30% Carbon- 14 activity compared to that of a new wood. How old is the artifact?

Answer:

The half life ($t_{1/2}$) of Carbon- 14 is 5730 years.

$N{}_t$ = $30\mathrm{\%}N{}_o$

$N{}_t=\frac{30}{100}\times N{}_o$

$N{}_t=\frac{3N{}_o}{10}$

$T$ = ?

Recall,

$ln\left(\frac{N{}_o}{N{}_t}\right)=\frac{0.693}{t_{1/2}}\times T$

$ln\left(\frac{N{}_o}{\frac{3N{}_o}{10}}\right)=\frac{0.693}{5730}\times T$

$ln\left(N{}_o÷\frac{3N{}_o}{10}\right)=0.0001209424\times T$

$ln\left(N{}_o\times \frac{10}{3N{}_o}\right)=0.0001209424\times T$

$ln\left(\cancel{N{}_o}\times \frac{10}{3\cancel{N{}_o}}\right)=0.0001209424\times T$

$ln\left(\frac{10}{3}\right)=0.0001209424\times T$

$ln\left(3.333\right)=0.0001209424\times T$

$1.2039=0.0001209424\times T$

$\frac{1.2039}{0.0001209424}=\frac{0.0001209424\times T}{0.0001209424}$

$\frac{1.2039}{0.0001209424}=\frac{\cancel{0.0001209424}\times T}{\cancel{0.0001209424}}$

$T=\frac{1.2039}{0.0001209424}$

$T=9954.927$

The artifact is 9954.927 years old.

Question 25:

What begins Nuclear Fission?

(a) Radioactive materials fission by themselves

(b) A spark

(c) Nuclear Fission has always been occuring in the Sun

(d) Bombardment of high energy particles into a radioactive atom

(e) Particle accelerator

Answer:

Option (d) is correct.

Practice Questions

Question 1

Identify x and y in

${}_{92}^{238}\text{U}\to {}_y^x\text{K}+{}_{90}^{234}\text{Th}$

x = 0, y = 0

x = 5, y = 2

x = 4, y = 2

x = 2, y = 0

Question 2

Nucleons are composed of...

Neutron only

Neutron + Proton

Neutron + Electron

Proton only

Question 3

A beta particle is an electron that was emitted from...

Electron Cloud

Nucleus

A and B

None of the above options

Question 4

A Positron is a particle that has the same mass as an electron but opposite charge.

True

False

Ambiguous

I have no idea

Question 5

Nuclei with 2, 8, 20, 28, 50 or 82 protons, or 2, 8, 20, 28, 50, 82 or 126 neutrons, are nuclei that have the magic numbers and they tend to be unstable.

True

False

Ambiguous

I have no idea

Question 6

Neutrons must be accelerated before they can be used for bombardment.

True

False

Ambiguous

I have no idea

Question 7

If we start with 1.00 grams of Sr- 90, 0.953 grams will remain after 2 years. What is the half life of Sr- 90?
How much Sr- 90 will remain after 5 years?

28.8 years, 0.887 grams

2.88 years, 0.887 grams

28.8 years, 8.87 grams

None of the above options

Question 8

The half life of Cobalt- 60 is 5.3 years. How much of a 1.0 mg sample will remain after 15.9 years?

$\frac{1}{4}$ mg

$\frac{1}{8}$ mg

$\frac{1}{16}$ mg

$\frac{1}{32}$ mg

Question 9

How much energy is lost or gained when one atom of Cobalt- 60 undergoes Beta decay?

(mass of Co- 60 is 59.9338 amu)
(mass of Ni- 60 is 59.9308 amu)

4.4838 × 10^-10 J

4.4838 × 10^-11 J

4.4838 × 10^-12 J

4.4838 × 10^-13 J

Question 10

When a Deuterium and a Tritium nuclei are joined together to form Helium nucleus, neutron and energy, the reaction is termed...

Nuclear Fission

Nuclear Fusion

Beta Decay

Alpha Decay

Question 11

Which of these Nuclear Emission has the highest penetration power?

Alpha Particle

Beta Particle

Positron

Gamma Radiation

Question 12

A Nuclear Fusion and Fission reactions are similar because both reactions...

Form heavy nuclides from light nuclides

Form light nuclides from heavy nuclides

Release a Large amount of energy

Absorb a Large amount of energy

Question 13

The energy released from a Nuclear Reaction results primarily from the...

Breaking of Bonds between atoms

Formation of Bonds between atoms

Conversion of mass into energy

Conversion of energy into mass

Question 14

Which Nuclear Decay Emission consists of energy only?

Alpha Particle

Beta Particle

Gamma Radiation

Positron

Question 15

Which of these equations represents Nuclear Fusion?

${}_0^1\text{n}+{}_{92}^{235}\text{U}\to {}_{56}^{142}\text{Ba}+{}_{36}^{91}\text{Kr}+3_0^1\text{n}$

${}_{88}^{226}\text{Ra}\to {}_{86}^{222}\text{Rn}+{}_2^4\text{He}$

${}_3^6\text{Li}+{}_0^1\text{n}\to {}_1^3\text{H}+{}_2^4\text{He}$

${}_1^2\text{H}+{}_1^3\text{H}\to {}_2^4\text{He}+{}_0^1\text{n}$

Question 16

Which group of Nuclear Emission is listed in order of increasing charge?

Alpha particle, Beta particle, Gamma radiation

Gamma radiation, Alpha particle, Beta particle

Positron, Alpha particle, Neutron

Neutron, Positron, Alpha particle

Question 17

Atoms of one element are converted to atoms of another element through...

Fermentation

Oxidation

Transmutation

Reduction

Question 18

An atom of Potassium- 37 and an atom of Potassium- 42 differ in their total number of...

Protons

Electrons

Neutrons

Positrons

Question 19

If $\frac{1}{8}$ of an original sample of Krypton- 74 remains unchanged after 34.5 minutes, what is the half life of Krypton- 74?

11.5 mins

23.0 mins

46.0 mins

34.5 mins

Question 20

What is the half life of Na- 25 if 2 grams of a 32 gram sample remains unchanged after 500 seconds.

250 seconds

1000 seconds

125 seconds

50 seconds

Question 21

Cobalt- 60 has a half life of 5.26 years. Calculate its decay rate constant.

0.132 year^-1

1.132 year^-1

2.132 year^-1

0.0132 year^-1

Question 22

Carbon- 14 has a half life of 5730 years; estimate the age of a fossil which contains 12.5% of the Carbon- 14 content of a living tissue.

34380 years

17190 years

8595 years

4297.5 years

Question 23

Which of these is used for treating Thyroid Cancer.

K- 37

Na- 25

I- 131

Ca- 37

Question 24

The process in which a nucleus breaks down by emitting radiation is known as

Transmutation

Transformation

Fusion

Radioactive decay

Question 25

Iodine- 131 decays by beta decay to...

Hint: Use your knowledge of Iodine being a Halogen on the Periodic Table.

Iodine- 132

Tellurium- 131

Iodine- 130

Xenon- 131

Question 26

What radioactive particle is released in the nuclear equation below: $${}_{74}^{159}\text{W}\to {}_{72}^{155}\text{Hf}+?$$

Alpha Particle

Beta Particle

Positron

Gamma Ray

Question 27

When Aluminium- 27 is bombarded with a neutron, a gamma ray is emitted. What radioactive isotope is produced?

Silicon- 27

Silicon- 28

Aluminium- 28

Magnesium- 28

Question 28

The activity of a radioisotope is defined as:

The radiation absorbed by a gram of material

The biological effect of different kinds of radiation

The amount of radiation absorbed by a material

The number of disintegrations per second

Question 29

The unit(s) that can be used to describe activity is/are:

Gray (Gy)

Becquerel (Bq)

rad

Curie (Ci)

B and D

Question 30

Technetium- 99m is an ideal isotope for scanning organs because it has a half life of 6.0 hours and is a pure gamma emitter. Suppose that 80 mg were prepared in the Technetium generator this morning, how many mg of Technetium- 99m would remain after 18 hours?

15 mg

5 mg

10 mg

20 mg

Application of Radioactivity

Radioactivity has found endless lists of applications.

• Radioactivity has been used for for calculating human history dates, especially by exploiting the radioactivity of Carbon- 14.

• Radioactive nuclei are used in medical imaging technology. The radionuclides used here must have short half life.

• Gamma rays are used to kill rapidly dividing cancer cells.

• Radioactivity is used in radiotracing technology for medicine and studying photosynthesis in plants.

• Radioactivity is employed in nuclear bombs and nuclear reactors.

• Food irradiation is used to kill bacteria.

• Radiation is used to sterilise surgical equipments, or tools that need very high purity.

• Radiation has made study of some quantum properties possible.

• Radiation is also used for study of shapes of molecules, molecular structures and composition.

All these are just few out of the many applications of radiation, the list is endless.

We all emit radiation from inside of us, (K- 40), and we are always exposed to radiation everytime of our lives, mostly natural.

PRO TIP: rem is a unit that is used to estimate biological damage of radiation.

You should note that your reading should never stop here, there are many text-books that explain radioactivity in full, and with the knowledge you now have, you should be able to understand, review and beat any question in Basic Nuclear Chemistry.

Key Terms

Nucleus • Electron Cloud • Proton • Neutron • Nucleon • Nuclide • Electrostatic Force (or Electromagnetic Force) • Nuclear Force • Binding Energy • Mass Defect • Electron Volt • Radionuclide • Isotope • Radioactive decay • Alpha particle • Beta particle • Isobar • Positron • Electron capture • Gamma Ray • Radioactive series • Nuclear fusion • Nuclear Fission • Ionizing radiation • Chain Reaction • Subcritical mass • Supercritical mass • Nuclear Transmutation • Natural Transmutation • Artificial Transmutation • Bombardment • Half Life • Rate of Radioactive Decay • Decay Rate Constant

Recommended Videos

Test Questions

Discuss And Explain

1. Explain the term Nuclear Transmutation. Analyse the differences between Natural Transmutation and Artificial Transmutation.

2. Describe why the half life of a radioisotope is constant while rate of radioactive decay is not. Write an equation to show the relationship between rate of decay and half life.

3. Explain how the activity of Carbon- 14 is used for Radioactive dating.

4. Analyse the differences between Alpha particle, Beta particle, and Gamma radiation.

5. What do you understand by Nuclear Stability?

6. Differentiate isotopes from Isobars. Beta decay leads to formation of which of these?

7. Explain why elements beyond atomic number 83 tend to achieve stability by emitting alpha particle(s).

8. Explain what you understand by Binding Energy and Binding Energy per nucleon.

9. Explain why a Neutron need not be accelerated before it is used for bombardment.

10. Differentiate Nuclear Fusion and Nuclear Fission. Which of these produces more energy, and why do you think it does?

11. Write an equation to show the relationship between mass and energy. Calculate the energy released (in MeV) when 0.052 amu is lost during Nuclear fusion.

12. Describe a chain reaction. Describe what you understand by Critical mass.

13. Describe the Ionizing processes of radioactive emission. How is this exploited in instruments used for detecting radiation?

14. Explain the good and bad effects of radiation.

15. Explain why radioisotopes that are used for medical imaging have short half lives. What do you think would be the effect if a radioisotope with a longer half life was used?