Atomic Mass and Composition of Nucleus

To measure various objects, we use a gram or a kilogram as the

standards. However, the mass of an atom is really small as compared

to a kilogram. To give you a perspective, the mass of a carbon atom

12C, is 1.992647 × 10–26 kg. Imagine writing the mass of atoms in

kilograms – not very convenient is it? Hence, Atomic Mass is

expressed using different units. Let’s find more about the composition

of a nucleus in the section below.

Atomic Mass and Isotopes

One Atomic Mass unit (u) is defined as 1/12th of the mass of the

carbon atom (12C). Therefore,

1u = (mass of one 12C atom)/12 = (1.992647 × 10–26)/12 = 1.660539

10–27 kg … (1)

Now, the atomic mass of different elements expressed in atomic units

(u) is nearly equal to being integral multiples of the mass of a

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hydrogen atom. However, it is important to note that there are many

exceptions to this rule.

The next point is how do we measure atomic mass accurately? The

answer is – by using a mass spectrometer. An interesting observation

made by measuring atomic masses is that there are many atoms of the

same element which have different masses but exhibit the same

chemical properties. Such atoms are called ‘Isotopes’.

Further observation revealed that nearly all elements have a mixture of

many isotopes. However, the number of isotopes can vary, depending

on the element. Let’s look at some examples:

Example 1

Chlorine has two isotopes having masses 34.98u and 36.98u. These

are nearly the integral multiples of the atomic mass of a hydrogen

atom. Also, the relative abundance of these two isotopes is 75.4 % and

24.6% respectively. So, the average mass of chlorine is calculated by

finding the weighted average of the masses of these two isotopes;

which is,

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= {(75.4 x 34.98) + (24.6 x 36.98)}/100 = 35.47u = the atomic mass of

chlorine.

Example 2

[Source: Wikipedia]

Hydrogen, on the other hand, has three isotopes having masses 1.0078 u,

2.0141 u, and 3.0160 u. Of these, the lightest isotope has a relative

abundance of 99.985% and is called the Proton. Now, the mass of a proton is,

mp = 1.00727 u = 1.67262 10– 27 kg … (2)

If we take the mass of a hydrogen atom and subtract the mass of an

electron from it, we get

1.00783u – 0.00055u = 1.00728u = mp

https://www.toppr.com/guides/chemistry/the-p-block-elements/chlorine/

The isotope having mass 2.0141 u is deuterium and the one having

mass 3.0160 u is tritium. Important to note here that tritium nuclei are

unstable and do not occur naturally. They are produced artificially in

laboratories.

Important Note

A proton is stable and carries one unit of fundamental charge. It is the

positive charge in the nucleus. Post the Quantum theory, it was agreed

that the electrons lie outside the nucleus (earlier there was a lot of

debate about them being inside the nucleus). The number of electrons

is equal to the atomic number of the element – Z.

Hence, the total charge of the atomic electrons is –Ze. Since an atom

is electrically neutral, the nucleus carries a charge of +Ze. By this, we

can conclude that the number of protons is equal to the number of

electrons in an atom = the atomic number Z.

Discovery of Neutron

Deuterium and tritium are the isotopes of hydrogen. Hence, the must

contain one proton each. However, there is a clear difference in their

atomic masses. The ratio of the atomic mass of hydrogen, deuterium,

and tritium is 1:2:3. Hence, there has to be some more matter in these

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isotopes adding to their atomic masses. Also, this additional matter

needs to be electrically neutral since the protons and electrons are in

balance.

In this case, the additional matter is present in multiples of the mass of

the proton (deuterium has additional matter equal to the mass of one

proton and tritium has matter equal to the mass of two protons).

Hence, it is easy to conclude that the nuclei of atoms contain neutral

matter, in addition to protons, in multiples of a basic unit.

In 1932, James Chadwick observed the emission of a neutral radiation

when a Beryllium nucleus was bombarded with alpha particles. Now,

at that time, the only known neutral radiation was photons or

electromagnetic radiation. If the neutral radiation consisted of protons,

then

The energy of photons >> The energy available from the

bombardment of the Beryllium nuclei with alpha particles

Chadwick hypothesised that the neutral radiation consisted of a new

type of neutral particles – Neutrons. By further calculations, he proved

that the mass of a neutron is nearly equal to the mass of a proton.

Now, the mass of a neutron known to a high degree of accuracy is,

mn = 1.00866 u = 1.6749×10–27 kg ~ mp

Composition of Nucleus

A free neutron is unstable. It decays into a proton, an electron and an

antineutrino (an elementary particle) with a mean life of around 1000s.

However, it is stable inside the nucleus. Therefore, the composition of

a nucleus is described as follows:

A = Z + N

Where,

● Z – atomic number = number of protons

● N – neutron number = number of neutrons

● A – mass number = = total number of protons and neutrons

Protons or Neutrons are also called Nucleons. Hence, the mass

number (A) of an atom is the total number of nucleons in it. A typical

nuclide of an atom is AZX, where X is the chemical symbol of the

atom.

Examples

The nuclide of gold is denoted by 19779Au. Hence, we can conclude

that there are 197 nucleons in a nucleus of gold. Out of these 79 are

protons and 118 (197-79) are neutrons.

● Deuterium is denoted as 21H … it has one proton and one

neutron

● Tritium is denoted as 31H … it has one proton and two neutrons

Isobars and Isotones

● Isobars – Nuclides with same mass number A. For Eg. 31H and

32H

● Isotones – Nuclides with the same number of neutrons (N) but

different atomic numbers (Z). For e.g. 19880Hg and 19779Au.

Solved Examples for You

https://www.toppr.com/guides/chemistry/structure-of-atom/introduction-to-structure-of-atom/

Question: Two stable isotopes of lithium 63Li and 73Li have respective

abundances of 7.5% and 92.5%. These isotopes have masses 6.01512u

and 7.01600u, respectively. Find the atomic mass of lithium.

Answer: Mass of the first lithium isotope 63Li = m1 = 6.01512u

Mass of the second lithium isotope 73Li = m2 = 7.01600u

Abundance of 63Li = n1 = 7.5%

Abundance of 73Li = n2 = 92.5%

The atomic mass (m) of lithium is

m = (m1n1 + m2n2) / (n1 + n2)

= {(6.01512u x 7.5) + (7.01600u x 92.5)} / (7.5 + 92.5)

= 6.940934u

Hence, the atomic mass of lithium is 6.940934u.

https://www.toppr.com/guides/chemistry/atoms-and-molecules/atomic-mass/

Mass Energy and Nuclear Binding Energy

The famous Einstein’s theory of special relativity established the fact

that mass is another form of energy. Also, you can convert the

mass-energy into other forms of energy. This opened the doors to a

better understanding of nuclear masses and the interaction of nuclei

with each other. In this article, we will look at the binding energy of a

nucleus which is essential to understand the nuclear fission and fusion

processes.

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Structure of Atom 

 

Atomic mass and composition of nucleus 

 

Basics of Isotopes,Isotones and Isobars 

According to Einstein’s mass-energy equivalence relation, we know

that,

E = mc2 … (1)

where c is the velocity of light in vacuum and is ≅ 3 x 108 m/s. Also, E

is the energy equivalent of the mass, ‘m’. Experimental studies of

nuclear reactions between nucleons, nuclei, electrons and other more

recently discovered particles. According to the Law of Conservation

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of Energy, we know that in a reaction, the initial and final energy is

the same provided we take the mass-energy into consideration.

Nuclear Binding Energy

We know that the nucleus is made up of protons and neutrons. So,

logically, the mass of the nucleus = the sum of masses of the protons

and neutrons, right? Not really! The nuclear mass (M) is always less

than this sum. To e=understand this better, let’s look at an example,

168O has 8 protons and 8 neutrons. Now,

● Mass of 8 neutrons = 8 × 1.00866 u

● Mass of 8 protons = 8 × 1.00727 u

● Mass of 8 electrons = 8 × 0.00055 u

Therefore the expected mass of 168O nucleus = = 8 × 2.01593 u =

16.12744 u.

We know from mass spectroscopy experiments that the atomic mass

of 168O is 15.99493u. Subtracting the mass of 8 electrons from this,

we get the experimental mass of 168O nucleus = 15.99053u

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(15.99493u – [8 x 0.00055u]). Hence, we see that there is a difference

between the two numbers of 0.13691u (16.12744 u – 15.99053u).

In simple words, the mass of the 168O nucleus is less than the total

mass of its constituents by 0.13691u. This difference in mass of a

nucleus and its constituents is called the mass defect (ΔM) and is

given by

ΔM = [Zmp + (A – Z)mn] – M … (2)

So, what exactly does mass defect mean?

The mass of an oxygen nucleus < the sum of masses of its protons and

neutrons (in an unbounded state). Therefore, the equivalent energy of

the oxygen nucleus < the sum of the equivalent energies of its

constituents.

We can also say that if you want to break down an oxygen nucleus

into 8 protons and 8 neutrons, then you must provide the extra energy

(ΔMc2). The relation between this energy (Eb) to the mass defect

(ΔM) is derived from Einstein’s mass-energy equivalence relation

{equation (1)}. Therefore,

https://www.toppr.com/guides/+maths/relations-and-functions/relations/https://www.toppr.com/guides/maths/trigonometric-functions/trigonometric-equations/

Eb = ΔMc2 … (3)

In other words, if certain protons and neutrons are brought together to

form a nucleus of a certain charge and mass, then an energy Eb is

released in the process. This energy is called the Binding Energy of

the nucleus. So, if we want to separate a nucleus into protons and

neutrons, then we will need to provide an energy Eb to the particles.

Binding Energy per Nucleon

A more useful measure of the binding between protons and neutrons is

the binding energy per nucleon or Ebn. It is the ratio of the binding

energy of a nucleus to the number of nucleons in the nucleus:

Ebn = Eb/A … (4)

where Eb is the binding energy of the nucleus and A is the number of

nucleons in it. So, the binding energy per nucleon is the average

energy per nucleon needed to separate a nucleus into its individual

nucleons. Let’s look at a plot of the binding energy per nucleon versus

the mass number for a large number of nuclei:

https://www.toppr.com/guides/business-mathematics-and-statistics/business-mathematics/ratios/

Here is what we can observe:

● The maximum binding energy per nucleon is around 8.75 MeV

for mass number (A) = 56.

● The minimum binding energy per nucleon is around 7.6 MeV

for mass number (A) = 238.

● For 30 < A < 170, Ebn is nearly constant.

● Ebn is low for both light nuclei (A < 30) and heavy nuclei (A >

170)

Conclusion 01

The force is attractive in nature and very strong producing a binding

energy of around a few MeV per nucleon.

Conclusion 02

● Why is Ebn nearly constant in the range 30 < A < 170? The

answer is simple – the nuclear force is short-ranged. Now,

imagine a really large nucleus. A nucleon (NA) in this nucleus

will be influenced only by some of its neighbours. These

neighbours are those which lie in the short-range of the nuclear

force.

● This means that all nucleons beyond the range of the nuclear

force form NA will have no influence on the binding energy of

NA. So, we can conclude that if a nucleon has ‘p’ neighbours

within the range of the nuclear force, then its binding energy is

proportional to ‘p’.

● In the same large nucleus, if we increase the mass number (A)

by adding nucleons, it will not change the binding energy of

NA. Why? Because, in a large nucleus, most of the nucleons lie

inside it and not on the surface. Hence, the change in binding

energy, if any, would be negligibly small.

● Remember, the binding energy per nucleon is a constant and is

equal to pk, where k is a constant having the dimensions of

energy. Also, the property that a given nucleon influences only

nucleons close to it is also referred to as saturation property of

the nuclear force.

Conclusion 03

Next, imagine a very heavy nucleus having A = 240. This has a low

binding energy. Therefore, if a nucleus A = 240 breaks down into two

A = 120 nuclei, then the nucleons get bound more tightly. Right?

Also, in the process energy is released. This concept is used in

Nuclear Fission.

Conclusion 04

On the other hand, imagine two very light nuclei with A < 10. If these

two nuclei were to join to form a heavier nucleus, then the binding

energy per nucleon of the fused and heavier nucleus is more than the

Ebn of the lighter nuclei. Right? So, the nucleons are more tightly

bound post fusion. And, energy is released in the process. This is how

the Sun works!

Solved Examples for You

Question: Obtain the binding energy (in MeV) of a nitrogen nucleus

(147N) , given m (147N) =14.00307u.

Solution: A nucleus of Nitrogen contains 7 protons and 7 neutrons.

Therefore, the mass defect of this nucleus is,

ΔM = 7mp + 7mn – m (147N)

Now,

Mass of a proton = mp = 1.007825u

Mass of a neutron = mn = 1.008665u

Therefore,

ΔM = (7 x 1.007825u) + (7 x 1.008665u) – 14.00307u = 0.11236u

Now, we know that 1u = 931.5 MeV/c2. Therefore,

ΔM = 0.11236 x 931.5 MeV/c2

The binding energy of the nucleus is,

Eb = ΔMc2

= 0.11236 x 931.5 (MeV/c2) x c2

= 0.11236 x 931.5 MeV

= 104.66334 MeV.

Nuclear Force

Coulomb force determines the motion of electrons in an atom. By

now, we already know that the binding energy per nucleon is around 8

MeV, for average mass nuclei. This is much larger than the binding

energy in atoms. Hence, the nuclear force required to bind a nucleus

together must be very strong and of a different type. It should

overcome the repulsion between like-charged protons and should bind

the protons and neutrons into a really small nuclear volume. In this

article, we will look at the features of this force, also called the nuclear

binding force.

Nuclear Force

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The nucleus of all atoms (except hydrogen) contain more than one

proton. Also, protons carry a positive charge. And, like charges repel

each other. Then why or how do these nucleons stay together in a

nucleus? They should repel each other, right? This is where the strong

nuclear force comes into play.

Source: Wikipedia

Features of Nuclear Force

Between 1930 and 1950, many experiments were conducted to

understand nuclear force. Some key observations and/ or features are

listed below:

● The nuclear force acts between the charges and functions as the

gravitational force between masses. This is much stronger than

the Coulomb force. This is because the nuclear force needs to

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overpower the Coulomb repulsive force between the

like-charged protons inside the nucleus. Hence, the nuclear

force > the Coulomb force. Also, the gravitational force much

weaker than the Coulomb force.

● The distance between two nucleons is measured in femtometers

(1fm = 10–15m). The nuclear force is really attractive when the

distance between two nucleons is around 1fm. As the distance

increases beyond 2.5fm, this attractive force starts decreasing

rapidly. Hence, for a medium to a large-sized nucleus, the

forces get saturated leading to the constancy of the binding

energy per nucleon. Also, if the distance falls below 0.7fm,

then this force becomes repulsive. A rough plot of the potential

energy between two nucleons as a function of distance is

shown below.

https://www.toppr.com/guides/quantitative-aptitude/number-series/heights-and-distances/

● Finally, the nuclear force between two neutrons, two protons

and a neutron and a proton is nearly the same. It is important to

note that the nuclear force is independent of the electric charge

of the neurons. Further, unlike Coulomb’s Law or Newton’s

Law of Gravitation, Nuclear force does not have a simple

mathematical form.

There are four basic forces in nature:

1. Gravitational Force

2. Electromagnetic Force

3. Strong Nuclear Force

4. Weak Nuclear Force

The strong nuclear force is what keeps the nucleons together despite

having a similar charge.

Solved Examples for You

Question:

Assertion: More energy is released in fusion than fission.

Reason: More number of nucleons take part in fission.

A. Both Assertion and Reason are correct and Reason is the

correct explanation for Assertion.

B. Both Assertion and Reason are correct but Reason is not the

correct explanation for Assertion.

C. Assertion is correct but Reason is incorrect.

D. Both Assertion and Reason are incorrect.

Solution: Option B. Since energy released per mass is more in fusion

as compared to fission, more energy is released in it. More number of

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neutrons are released in fission as compared to fusion and hence, more

nucleons take part in fission.

Question: In nuclear reaction: 4Be9+2He4→6C12+X, X will be:

A. Proton

B. Neutron

C. β-particle

D. α-particle

Solution: 4Be9+2He4→6C12+X

No. of protons is already balanced with C.

∴ X carries no charge.

and 9+4 > 12 by 1 unit mass.

∴ X carries one unit mass.

It’s a neutron.

Nuclear Energy – Nuclear Fusion

When we talk about nuclear energy, understanding nuclear fusion is

essential. It helps us gain insight into how a massive amount of energy

is released when two light nuclei combine to form a single larger

nucleus.

Some examples of nuclear fusion:

11H + 11H → 21H + e– + v + 0.42 MeV … (1)

Here two protons combine to form a deuteron and a positron

releasing 0.42 MeV of energy.

21H + 21H → 32He + n + 3.27 MeV … (2)

Here two deuterons combine to form the light isotope of Helium

releasing 3.27 MeV of energy.

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21H + 21H → 31H + 11H + 4.03 MeV … (3)

In this case, two deuterons combine to form a triton and a proton

releasing 4.03 MeV of energy.

Note that in all these examples, two positively charged particles

combine to form a larger nucleus. Further, the Coulomb repulsion

hinders the process of fusion since it acts to prevent two positively

charged particles from getting within the range of their attractive

nuclear forces.

Now, the height of the Coulomb barrier depends on the charges and

the radii of the two interacting nuclei. For example, the barrier height

for two protons is around 400 keV. Also, higher charged nuclei have a

higher barrier height. The temperature at which the protons in a proton

gas have enough energy to cross the coulomb’s barrier is around 3 x

109 K.

Now, to generate any useful amount of energy, nuclear fusion must

occur in bulk matter. To achieve this, the temperature of the material

can be raised until the particles have enough energy to cross the

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coulomb barrier due to their thermal motions. This is thermonuclear

fusion.

Nuclear Fusion in the Sun

The temperature of the core of the Sun is around 1.5 x 107 K. Hence,

even in the sun fusion occurs only when protons having energies

above the average energy are involved. In simple words, for

thermonuclear fusion to occur extreme temperature and pressure

conditions are needed. This is only possible in the interiors of the Sun

and other stars.

Source: Wikipedia

In the Sun, nuclear fusion occurs in a multi-step process where

hydrogen is burned into helium. Here, hydrogen is the fuel and

helium, the ash. The process is represented as follows:

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11H + 11H → 21H + e– + v + 0.42 MeV … (4)

e– + e+ → γ + γ + 1.02 MeV … (5)

21H + 11H → 32He + γ + 5.49 MeV … (6)

32He + 32He → 42He + 11H + 11H + 12.86 MeV … (7)

For the fourth reaction to occur, the first three need to occur twice.

Thereby, two light helium nuclei unite to form a normal helium

nucleus. This four-step process can be summarised as:

411H + 2e– → 42He + 2v + 6γ + 26.7 MeV … (8)

Hence, we can see that four hydrogen nuclei combine to form a

helium nucleus and release 26.7 MeV of energy.

Some Trivia

Hydrogen has been burning in the Sun’s core for around 5 x 109

years! According to calculations, this will continue for another 5 x 109

years in the future too. In about 5 billion years, the Sun’s core will be

primarily helium. This is when it will start to cool down and collapse

under its own gravity. Eventually, this will raise the core temperature

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and cause the outer envelope to expand. The sun will then look like a

red giant! Once the core temperature increases to 108 K, energy

production will resume through fusion again. However, this time

helium will be burned to make carbon.

Controlled Thermonuclear Fusion

On November 01, 1952, the USA exploded a nuclear fusion device

generating energy equivalent to 10 million tons of TNT. Just to put it

into perspective, one ton of TNT produces 2.6 x 1022 MeV of energy

on explosion. This was the first thermonuclear reaction on earth. It is

very difficult to achieve a sustained and controlled source of fusion

power. However, it is regarded as the power source of the future and

most countries are pursuing it vigorously.

Solved Question for You

Question: How long can an electric lamp of 100W be kept glowing by

fusion of 2.0 kg of deuterium? Take the fusion reaction as:

21H + 21H → 32He + n + 3.27 MeV

https://www.toppr.com/guides/physics/work-energy-and-power/power/https://www.toppr.com/guides/chemistry/chemical-reactions-and-equations/types-of-reactions/

Solution: We know that 1 mole or 2 grams of deuterium contains

6.023 x 1023 atoms. Therefore, 2 kg of deuterium will contain:

{(6.023 x 1023)/2} x 2000 = 6.023 x 1026 atoms.

From the given reaction, we know that 3.27 MeV of energy is released

when 2 deuterium nuclei fuse. Therefore, the total energy released per

nucleus is:

(3.27/2) x 6.023 x 1026 MeV

= (3.27/2) x 6.023 x 1026 x 1.6 x 10–19 x 106 Joules

= 1.576 x 1014 J

Power of the lamp is 100 W = 100 J/s. So, in one second the lamp

utilizes 100 J. Therefore, the total time for which the lamp will burn is

the time taken to utilize 1.576 x 1014 J. Here is the calculation:

(1.576 x 1014/100) seconds

= (1.576 x 1014/100 x 60 x 60 x 24 x 365) years

= 4.9 x 104 years

Nuclear Energy – Nuclear Fission

Binding Energy, as we know, is the energy released in the processes in

which the light nuclei fuse or the heavy nuclei split to form a

transmuted nucleus. This energy is made available as nuclear energy.

Let’s learn about the nuclear fission in section below.

While studying about Binding energy we observed that the binding

energy per nucleon (Ebn) is nearly constant (around 8 MeV) for mass

numbers in the range 30 < A < 170. Also, for lighter nuclei (A < 30)

and heavier nuclei (A > 170), Ebn is less than 8 MeV. And, the nuclei

in the range 30 < A < 170 are more tightly bound than those below or

above the range specified.

Further, in conventional energy sources like coal or petroleum, energy

is released through chemical reactions. However, for the same

quantity of matter, nuclear sources provide million times larger energy

than these conventional sources. For example, one kilogram of coal

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gives 107J of energy on burning. On the other hand, one kilogram of

uranium creates 1014J of energy on fission.

Nuclear Fission

Post the discovery of the neutron, Enrico Fermi found that on

bombarding a neutron different elements produced new radioactive

elements. However, on bombarding a uranium target, the nucleus

broke into two nearly equal fragments and released a great amount of

energy. An example of the same is:

10n + 23592U → 23692U → 14456Ba + 8936Kr + 310n

Fission does not always produce Barium and Krypton. Here is another

example:

10n + 23592U → 23692U → 13351Sb + 9941Nb + 410n

One more example:

10n + 23592U → 23692U → 14054Xe + 9438Sr + 210n

All the fragmented nuclei produced in fission are neutron-rich and

unstable. Also, they are radioactive and emit beta articles until they

reach a stable end-product.

Nuclear Energy Released in a Fission Reaction

In the example above, the nuclear energy released is of the order of

200 MeV per nucleus undergoing a fission. Here is how it is

calculated: Imagine a heavy nucleus having A = 240. Now, this

nucleus breaks into two nuclei with A = 120 each. Therefore,

Ebn (A = 240 nucleus) = 7.6 MeV

Ebn (each of A = 120 nuclei) = 8.5 MeV

Hence, the gain in binding energy per nucleon is about 0.9 MeV (8.5 –

7.6). Therefore, the total gain in binding energy is 240×0.9 = 216

MeV.

The disintegration energy in fission events initially appears as the

kinetic energy of the fragments and neutrons. Subsequently, the

surrounding matter starts heating up. The source of energy in nuclear

reactors, which produce electricity, is nuclear fission. The enormous

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energy released in an atom bomb comes from uncontrolled nuclear

fission.

Nuclear Reactor

When 23592U undergoes a fission after being bombarded by a neutron,

it splits into two nuclei and releases a neutron. This extra neutron now

initiates the fission of another 23592U nucleus. For that matter, 2.5

neutrons are released per fission of a uranium nucleus. Also, fission

produces more neutrons than what can be consumed.

This increases the chances of a chain reaction with each neutron that is

produced, triggering another fission. If this chain reaction is

uncontrolled, then it can lead to destruction (like a nuclear bomb). On

the other hand, in a controlled manner, it can be harnessed to generate

electric power.

However, there was a small problem. The neutrons generated in

fission were highly energetic. They would escape rather than trigger

another fission reaction. Also, it was observed that slow neutrons have

a higher possibility of inducing fission in 23592U than their faster

counterparts.

Now, the energy of a neutron produced in fission of 23592U is around 2

MeV. Unless these neutrons are slowed down, they tend to escape

without inducing fission. This simply means that we need a lot of

fissionable material to sustain the chain reaction.

Slowing Down Fast Neutrons in a Nuclear Reactor to Generate Nuclear Energy

These fast neutrons are slowed down by elastic scattering with a light

nuclei. Chadwick, in his experiments, showed that in an elastic

collision with hydrogen, these neutrons almost come to rest and the

protons carry away the energy.

To understand this, imagine two marbles. One at rest and one

travelling fast. When the fast-moving marble hits the stationary

marble head-on, it comes to rest.

Therefore, in nuclear reactors, light nuclei are provided along with the

fissionable nuclei to slow down the fast neutrons. These light nuclei

are the ‘moderators’. The most commonly used moderators are water,

heavy water (D2O) and graphite.

Moderators

Moderators can ensure that the number of neutrons generated by a

given generation is greater than those produced by the preceding

generation. Hence, the multiplication factor ‘K’ or the measure of the

growth rate of neutrons in the reactor can be in the range 1 < K < 1.

Here is what it means:

● K < 1 means that the neutrons are not increasing in number

over the previous generations. Or, they are escaping without

inducing fission.

● K> 1 means that the reaction rate and nuclear power in the

reactor is increasing exponentially and can even explode. It

needs to be brought down to as close to unity as possible.

● K = 1 means that the reactor will be able to generate steady

power.

This control is managed by using cadmium rods which can absorb

neutrons. So, there is a dual control on a nuclear fission reaction.

● Slowing down the fast neutrons so that the induce fission and

start a chain reaction.

● Introducing neutron absorbing rods in the reactor to ensure that

the value of ‘K’ stays as close to 1 as possible.

Another interesting isotope is 23892U. This isotope is available in

abundance and does not fission. Also, it forms plutonium by capturing

neutrons. Here’s how:

23892U + 10n → 23992U → 23993Np + e– + v–

23993Np → 23994Pu + e– + v–

Plutonium is highly radioactive and can also undergo nuclear fission

under bombardment by slow neutrons.

A Nuclear Power Plant

The image above outlines a typical nuclear power plant based on a

pressurized water reactor. In such reactors, water is used as a

moderator and heat transfer medium. In the primary-loop, water is

circulated through the reactor vessel and transfers energy at high

temperature and pressure (at about 600 K and 150 atm) to the steam

generator, which is part of the secondary loop.

In the steam generator, evaporation provides high-pressure steam to

operate the turbine that drives the electric generator. The low-pressure

steam from the turbine is cooled and condensed to water and forced

back into the steam generator.

Since the energy released in nuclear reactors is very high, they need

very less fuel. However, in these reactors, highly radioactive elements

are produced continually. Hence, this radioactive waste needs to be

accumulated regularly. This includes both fission products and heavy

transuranic elements like plutonium and americium.

Solved Question for You

Question: The fission properties of 23994Pu are very similar to those of

23592U. The average energy released per fission is 180 MeV. How

much energy, in MeV, is released if all the atoms in 1 kg of pure

23994Pu undergo fission?

Solution:

Average energy released per fission of 23994Pu, Eav = 180 MeV

Amount of pure 23994Pu, m = 1 kg = 1000 grams

Avogadro’s number = NA = 6.023 x 1023

Mass number of 23994Pu = 239

Now, 1 mole of 23994Pu contains NA atoms. Therefore, ‘m’ grams of

23994Pu will contain {(NA/Mass number) x m} number of atoms,

= [(6.023 x 1023)/239] x 1000 = 2.52 x 1024 atoms.

Hence, total energy released during the fission of 1 kg of 23994Pu is:

E = Eav x 2.52 x 1024

= 180 x 2.52 x 1024 = 4.536 x 1026 MeV.

Radioactivity – Law of Radioactive Decay

In 1896, A.H. Becquerel accidentally discovered radioactivity. He was

studying the fluorescence and phosphorescence of compounds

irradiated with visible light. This is when he observed something

interesting. Let’s learn more about the law of radioactive decay in the

section below.

He illuminated some pieces of Uranium-Potassium-Sulphate with

visible light. Next, he wrapped these pieces in black paper and

separated it from a photographic plate by a piece of silver. He left it

for several hours. When he developed the photographic plate, he

found that there was blackening on the plate.

This meant that something was emitted by the compound which

penetrated the silver and black paper and hit the plate. Subsequent

experiments show that radioactivity is a nuclear phenomenon which

occurs when an unstable nucleus undergoes a decay. This is called

Radioactive Decay.

Radioactive Decay

There are three types of radioactive decays in nature:

● α-decay –a helium nucleus (42He) is emitted

● β-decay – where electrons or positrons (particles with the same

mass as electrons, but with a charge exactly opposite to that of

an electron) are emitted;

● γ-decay – high energy (hundreds of keV or more) photons are

emitted.

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Law of Radioactive Decay

When a radioactive material undergoes α, β or γ-decay, the number of

nuclei undergoing the decay, per unit time, is proportional to the total

number of nuclei in the sample material. So,

If N = total number of nuclei in the sample and ΔN = number of nuclei

that undergo decay in time Δt then,

ΔN/ Δt ∝ N

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Or, ΔN/ Δt = λN … (1)

where λ = radioactive decay constant or disintegration constant. Now,

the change in the number of nuclei in the sample is, dN = – ΔN in time

Δt. Hence, the rate of change of N (in the limit Δt → 0) is,

dN/dt = – λN

Or, dN/N = – λ dt

Now, integrating both the sides of the above equation, we get,

NN0∫ dN/N = λ tt0∫ dt … (2)

Or, ln N – ln N0 = – λ (t – t0) … (3)

Where, N0 is the number of radioactive nuclei in the sample at some

arbitrary time t0 and N is the number of radioactive nuclei at any

subsequent time t. Next, we set t0 = 0 and rearrange the above

equation (3) to get,

ln (N/N0) = – λt

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Or, N(t) = N0e– λt … (4)

Equation (4) is the Law of Radioactive Decay.

The Decay Rate

In radioactivity calculations, we are more interested in the decay rate

R ( = – dN/dt) than in N itself. This rate gives us the number of nuclei

decaying per unit time. Even if we don’t know the number of nuclei in

the sample, by simply measuring the number of emissions of α, β or γ

particles in 10 or 20 seconds, we can calculate the decay rate. Let’s

say that we consider a time interval dt and get a decay count ΔN (=

–dN). The decay rate is now defined as,

R = – dN/dt

Differentiating equation (4) on both sides, we get,

R = λ N0 e−λt

Or, R = R0e−λt … (5)

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Where, R0 is the radioactive decay rate at the time t = 0, and R is the

rate at any subsequent time t. Equation (5) is the alternative form of

the Law of Radioactive Decay. Now we can rewrite equation (1) as

follows,

R = λN … (6)

where R and the number of radioactive nuclei that have not yet

undergone decay must be evaluated at the same instant.

Half-Life and Mean Life

The total decay rate of a sample is also known as the activity of the

sample. The SI unit for measurement of activity is ‘becquerel’ and is

defined as,

1 becquerel = 1 Bq = 1 decay per second

An older unit, the curie, is still in common use:

1 curie = 1 Ci = 3.7 × 1010 Bq (decays per second)

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There are two ways to measure the time for which a radionuclide can

last.

● Half-life T1/2 – the time at which both R and N are reduced to

half of their initial values

● Mean life τ – the time at which both R and N have been

reduced to, e-1 of their initial values.

Calculating Half-Life

Let’s find the relation between T1/2 and the disintegration constant λ.

For this, let’s input the following values in equation (5),

R = (1/2)R0 and t = T1/2

So, we get T1/2 = (ln2)/ λ

Or, T1/2 = 0.693/ λ … (7)

Calculating Mean life

Next, let’s find the relation between the mean life τ and the

disintegration constant λ. For this, let’s consider equation (5),

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● The number of nuclei which decay in the time interval: ‘t’ to ‘t

+ Δt’ is: R(t)Δt = (λN0e–λt Δt).

● Each of them has lived for time ‘t’.

● Hence, the total life of all these nuclei is tλN0e–λt Δt

Hence, to obtain the mean life, we integrate this expression over all

the times from 0 to ∞ and divide by the total number of nuclei at t = 0

(which is N0).

τ = (λN0 0∞∫ te–λtdt)/N0

= λ0∞∫ te–λtdt

On solving this integral, we get

τ = 1/λ

Therefore, we can summarise the observations as follows:

T1/2 = (ln2)/λ = τ ln 2 … (8)

Solved Examples for You

Question: The half-life of 23892U undergoing α-decay is 4.5 × 109

years. What is the activity of 1g sample of 23892U ?

Answer: T1/2 = 4.5 x 109 years = 4.5 x 109years x 3.6 x 107

seconds/year = 1.42 x 1017 seconds

We know that 1 k mol of any isotope contains Avogadro’s number of

atoms. Hence, 1g of 23892U contains, {1/(238 x 10-3)} x 6.025 x 1026 =

25.3 x 1020 atoms. Therefore, the decay rate R is,

R = λN

= (0.693/T1/2)N

= (0.693 x 25.3 x 1020) / (1.42 x 1017)

Therefore, R = 1.23 x 104 Bq

Radioactivity – Types of Radioactive Decay

In this article, we will look at the three types of radioactive decay

namely, alpha, beta, and gamma decay. We will try to understand how

these particles are emitted and its effects on the emitting nucleus.

Alpha Radioactive Decay

An alpha particle is a helium nucleus 42He. Whenever a nucleus goes

through alpha decay, it transforms into a different nucleus by emitting

an alpha particle. For example, when 23892U undergoes alpha-decay, it

transforms into 23490Th.

23892U → 23490Th + 42He … (1)

Now, 42He contains two protons and two neutrons. Hence, after

emission, the mass number of the emitting nucleus reduces by four

and the atomic number reduces by two. Therefore, the transformation

of AZX nucleus to A-4Z-2X nucleus is expressed as follows,

AZX → A-4Z-2X + 42He … (2)

where AZX is the parent nucleus and A-4Z-2X is the daughter nucleus. It

is important to note that the alpha decay of 23892U can occur without

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an external source of energy. This is because of the total mass of the

decay products (23490Th and 42He) < the mass of the original 23892U

Or, the total mass-energy of the decay products is less than that of the

original nuclide. This brings us to the concept of ‘Q value of the

process’ or ‘Disintegration energy’ which is the difference between

the initial and final mass-energy of the decay products. For an alpha

decay, the Q value is expressed as,

Q = (mX – mY – mHe) c2 … (3)

This energy is shared between the daughter nucleus, A-4Z-2X and the

alpha particle, 42He in the form of kinetic energy. Also, alpha decay

obeys the radioactive laws.

Beta Radioactive Decay

Beta decay is when a nucleus decays spontaneously by emitting an

electron or a positron. This is also a spontaneous process, like the

alpha decay, with a definite disintegration energy and half-life. And, it

follows the radioactive laws. A Beta decay can be a beta minus or a

beta plus decay.

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Source: Wikipedia

In a Beta minus (β−) decay, an electron is emitted by the nucleus. For

example,

3215P → 3216S + e– + v– … (4)

where v- is an antineutrino, a neutral particle with little or no mass.

Also, T1/2 = 14.3 days.

Source: Wikipedia

In a Beta plus decay, a positron is emitted by the nucleus. For

example,

2211Na → 2210Ne + e+ + v … (5)

where v is a neutrino, a neutral particle with little or no mass. Also,

T1/2 = 2.6 years. The neutrinos and antineutrinos are emitted from the

nucleus along with the positron or electron during the beta decay

process. Neutrinos interact very weakly with matter. Hence, they were

undetected for a very long time.

Further, in a Beta minus decay, a neutron transforms into a proton

within the nucleus:

n → p + e– + ν– … (6)

Also, in a Beta plus decay, a proton transforms into a neutron (inside

the nucleus):

p → n + e+ + ν … (7)

Hence, we can see that the mass number (A) of the emitting nuclide

does not change. As shown in equations (6) and (7), either a proton

transforms into a neutron or vice versa.

Gamma Radioactive Decay

We know that atoms have energy levels. Similarly, a nucleus has

energy levels too. When a nucleus is in an excited state, it can

transition to a lower energy state by emitting an electromagnetic

radiation. Further, the difference between the energy states in a

nucleus is in MeV. Hence, the photons emitted by the nuclei have

MeV energies and called Gamma rays.

Source: Wikipedia

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After an alpha or beta emission, most radionuclides leave the daughter

nucleus in an excited state. This daughter nucleus reaches the ground

state by emitting one or multiple gamma rays. For example,

6027Co undergoes a beta decay and transforms into 6028Ni. The

daughter nucleus (6028Ni) is in its excited state. This excited nucleus

reaches the ground state by the emission of two gamma rays having

energies of 1.17 MeV and 1.33 MeV. The energy level diagram shown

below depicts this process.

Solved Examples for You

Question: Find the Q value and kinetic energy of the emitted alpha

particle in:

● 22688Ra

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● 22086Rn

Where,

226. m (22688Ra) = 226.02540u

227. m (22086Rn) = 220.01137u

● m (22286Rn) = 222.01750u

216. m (21684Po) = 216.00189u

Answer:

(a) After emitting an alpha particle (helium nucleus), the mass number

of 22688Ra reduces to 222 (226 – 4) and the atomic number reduces to

86 (88 – 2).Therefore, we have

22688Ra → 22286Rn + 42He

Source: Wikipedia

Now, Q value of the emitted alpha particle is,

Q = (mx – my – mHe) c2 … from equation (3) above)

= {m (22688Ra) – m (22286Rn) – m (42He)} c2

We know that,

226. m (22688Ra) = 226.02540u

227. m (22286Rn) = 222.01750u

● m (42He) = 4.002603u

Therefore,

Q-value = [226.02540u – 222.01750u – 4.002603u] c2 = 0.005297uc2

We know that, 1u = 931.5 MeV/c2. Hence,

Q = 0.005297 x 931.5 ~ 4.94 MeV

And, the Kinetic Energy of the alpha particle,

= {(Mass number after decay) / (Mass number before decay)} x Q

= (222/226) x 4.94 = 4.85 MeV.

(b) After emitting an alpha particle (helium nucleus), the mass number

of 22086Rn reduces to 216 (220 – 4) and the atomic number reduces to

84 (86 – 2).Therefore, we have

22086Rn → 21684Po + 42He

Now, Q value of the emitted alpha particle is,

Q = (mx – my – mHe) c2 … from equation (3) above)

= {m (22086Rn) – m (21684Po) – m (42He)} c2

We know that,

220. m (22086Rn) = 220.01137u

221. m (21684Po) = 216.00189u

● m (42He) = 4.002603u

Therefore,

Q-value = [220.01137u – 216.00189u – 4.002603u] c2 = 0.006877uc2

We know that, 1u = 931.5 MeV/c2. Hence,

Q = 0.006877 x 931.5 ~ 6.41 MeV

And, the Kinetic Energy of the alpha particle,

= {(Mass number after decay) / (Mass number before decay)} x Q

= (216/220) x 6.41 = 6.29 MeV.