Radioactivity

The nucleus is at the centre of an atom whose size is of the order of 10^-15 m to 10^-14 m (i.e., 10^-5 to 10^-4 times the size of the atom). It consists of protons and neutrons.

If Z = atomic number and A = mass number of an atom, then the atom contains,

Number of protons = Z

Number of neutrons = A – Z

(a) The three types of radiations are α, β and γ .

(b) The radiations which are deflected by the electric field are α and β radiations.

(c) The most penetrating radiations is γ radiations.

(d) The radiation which travels with the speed of light is γ radiation.

(e) The radiation which has the highest ionizing power is α radiations.

(f) The radiations consisting of the same kind of particles as the cathode rays are β radiations.

(a) The radiation of zero mass are γ radiation.

(b) The radiation which has the lowest ionising power is the γ radiation.

(c) The radiation which has the lowest penetrating power is the α radiation.

(d) α particles have a charge of +3.2 × 10^-19 C (twice the charge of a proton) and rest mass of 6.68 × 10^-27 kg (four times the mass of proton)

(e) When α particles become neutral, it is found to be the atom of Helium gas. Below is the model of its neutral atom:

(f) The radiations come from the nucleus of the atom

(a) The radiations are —

• A → γ radiation
• B → α radiation
• C → β radiation

(b) The radiation labelled as A pass undeviated which means that they are uncharged (or neutral) so they must be γ radiations.
The radiation labelled as B turn to the left which means that they are positively charged so they must be α radiations.
The radiation labelled as C turn to the right which means that they are negatively charged so they must be β radiations.

(a) Below diagram shows the deflection of the radioactive particles in the magnetic field:

(b) The direction of deflection in a magnetic field is given by Fleming’s left hand rule.

(a) Below diagram shows the paths of α, β and γ radiations:

(b) The source S is kept in a thick lead walled container with a narrow opening so that the walls of the container absorb the radiations which strike on them from the inside and thus radiations only come out through the narrow opening.

(c) γ radiation is unaffected by the electrostatic field.

(d) The β radiation is deflected the most.
α particle is a helium nucleus consisting of two protons and two neutrons. Beta particles are fast-moving electrons with a very low mass. α particles are heavier than β particles so β particles have a high charge to mass density as compared to α particles. Due to this reason, β particles are deflected much more than the α particles.

(e) α radiation causes the least biological damage.

α Radiation

• Nature — Stream of positively charged particles i.e., helium nucleus
• Charge — Positive charge (two times that of a proton) = +3.2 × 10^-19 C (or +2e)
• Mass — Four times the mass of proton i.e., 6.68 × 10^-27 kg
• Effect of Electric Field — Less deflected

β Radiation

• Nature — Stream of negatively charged particles, i.e., energetic electrons
• Charge — Negative charge = -1.6 × 10^-19 C (or –e)
• Mass — Equal to the mass of electron i.e., 9.1 × 10^-31 kg
• Effect of Electric Field — More deflected than α radiations, but in a direction opposite to those of α radiations

γ Radiation

• Nature — Highly energetic electromagnetic radiation
• Charge — No charge (Neutral)
• Mass — No mass (rest mass is zero)
• Effect of Electric Field — Undeviated

γ radiations are produced when the daughter or the parent nucleus is in a state of excitation ( i.e., when it has an excess of energy). This extra energy is released in the form of γ radiation.

Two common properties of γ radiations and visible light are as follows —

(i) They both are not deflected by the electric and magnetic fields since they are uncharged particles.

(ii) Speed of γ radiations is same as the speed of light (i.e., 3 x 10⁸ m s^-1)

(a) In emitting a β particle, the number of nucleons in the nucleus remains same, but the number of neutrons is decreased by one and the number of protons is increased by one.

If a radioactive nucleus P with mass number A and atomic number Z emits a β particle to form a daughter nucleus Q with mass number A and atomic number Z + 1, the change can be represented as follows —

$\begin{matrix} _{Z}^{A}\text{P} & \longrightarrow & _{Z+1}^{\space \space \space \space \space A}\text{Q} & + & \space _{-1}^{\space \space \space 0}\text{e} \\ \footnotesize{\text{Parent nucleus}} & & \footnotesize{\text{Daughter nucleus}} & & \footnotesize{\text{(Beta-particle)}} \end{matrix}$

(a) Atomic number Z is increased by 1

(b) Mass number remains unchanged

(a) The composition of B is 82 protons and 126 neutrons.
Explanation: When the nucleus A emits an α-particle, the resulting nucleus B has 2 protons and 2 neutrons less than the original nucleus.

(b) The composition of C is 83 protons and 125 neutrons.
Explanation: When the nucleus B emits a β-particle, the resulting nucleus C has 1 proton more and 1 neutron less than the nucleus B.

(c) The mass number of nucleus A = number of protons + number of neutrons = 84 + 128 = 212.

(d) No, the composition of nucleus C does not change on emitting γ radiations as there is no change in the mass number A and atomic number Z of the nucleus in gamma emission.

(a) The particle emitted is α-particle.

(b) As the atomic number is decreased by 2 and the mass number is decreased by 4 hence, we can say that an α-particle is emitted.

(c) This change can be expressed in the form of a reaction as follows:

$\begin{matrix} _{92}^{238}\text{A} & \longrightarrow & _{90}^{234}\text{B} & + & \space _{2}^{4}\text{He} \\ \footnotesize{\text{Parent nucleus}} & & \footnotesize{\text{Daughter nucleus}} & & \footnotesize{\text{(α-particle)}} \end{matrix}$

(a) The number 24 is called the mass number and the number 11 is called the atomic number.

(b) The equation representing β-decay is as follows —

$_{11}^{24}\text{Na} \longrightarrow \space _{12}^{24}\text{Mg} + _{-1}^{\space \space\space 0}\text{e}$

(c) The product nucleus is called the isobar because they have same mass number and different atomic number.

(a) Atomic number and mass number are as follows:

Atomic number = no. of protons = 15
Mass number = no. of protons + no. of neutrons = 15 + 16 = 31

(b) Atomic number = 15 and Mass number = 15 + 17 = 32

(c) Atomic number = 16 and Mass number = 32.
Explanation: After emission of β-particle, number of neutrons is decreased by one and number of protons is increased by one. So, the mass number (A) does not change whereas the atomic number (Z) increases by one.

As we know, the atoms of same element, having same atomic number Z, but different mass number A, are called isotopes.

When the element P disintegrates by α emission, atomic number of P decreases by 2 and mass no. decreases by 4 then the atomic number increases by 1 due to first β emission and again increases by 1 due to second β emission and we get Q. We observe that atomic number of Q becomes the same as it was before the emission (i.e. P). Hence, P and Q are the isotopes.

both (1) and (3)

Reason — In a neutral atom, the number of protons is equal to the number of electrons to maintain electrical neutrality.

The total number of protons in a neutral atom is known as the atomic number.

the number of neutron + number of protons

Reason — Nucleons include both protons and neutrons, which are found in the nucleus of an atom. Therefore, the total number of nucleons is the sum of the number of protons and the number of neutrons in the nucleus.

Number of protons = A - Z

Reason — The number of protons in an atom is equal to its atomic number (Z), not the difference between the mass number (A) and the atomic number (Z).

8

Reason — The number of neutrons in a nucleus can be calculated using the formula:

Number of neutrons = Mass number (A) - Atomic number (Z)

Given,

Mass number (A) = 15
Atomic number (Z) = 7
So, number of neutrons = 15 - 7 = 8

Isotopes

Reason — Isotopes are atoms of the same element that have the same atomic number (Z) but different mass numbers (A). They have the same number of protons but different numbers of neutrons.

Tin

Reason — Tin has largest number of isotopes i.e., 10

Isotones

Reason — ³⁹₁₉K and ⁴⁰₂₀Ca are isotones as they both have the same number of neutrons in the nucleus (each having 20 neutrons in the nucleus).

All of the above

Reason — All three elements listed—uranium, polonium, and actinium—are radioactive substances. They undergo radioactive decay, emitting radiation in the form of alpha particles, beta particles, or gamma rays.

α radiation

Reason — The mass of an alpha particle is roughly four times the mass of a proton i.e., 6.68 x 10^-27 kg and is the heaviest among the given options.

carbon

Reason — To study the age of excavated materials of archeological significance, we study the rate of decay of an isotope of carbon. The process is known as carbon dating.

one place higher

Reason — By emission of a beta particle, the mass number A does not change but the atomic number is increased by one. It changes into a new nucleus of element with properties similar to those of an element one place higher in the periodic table.

equal to the speed of light, i.e., 3 x 10⁸ ms^-1

Reason — Gamma (γ) radiation consists of high-energy photons emitted from the nucleus during radioactive decay. These photons travel at the speed of light i.e., 3 x 10⁸ ms^-1 in vacuum because they are electromagnetic waves.

$_{Z}^{A}\text{P} \longrightarrow \space _{Z}^{A}\text{P} + γ$

Reason — The correct representation of γ emission is:

$_{Z}^{A}\text{P} \longrightarrow \space _{Z}^{A}\text{P} + γ$

This represents a gamma (γ) decay, where there is no change in mass number A and atomic number Z of the nucleus.

assertion is false but reason is true.

Explanation

Assertion (A) is false. Out of α, β and γ radiation, γ-particles have maximum penetrating power as the penetrating power of a particle depends upon the energy carried by it, its mass and the charge carried by it. A particle with a greater energy can use it to force itself into another medium and thereby penetrate it. α-particles rapidly dissipates its energy as it moves through the medium and therefore its penetrating power is quite small.

Reason (R) is true. An α particle contains two protons and two neutrons, β particle has mass of an electron and γ particles are massless so an α particle is heavier than β and γ particles.

Both A and R are true and R is the correct explanation of A.

Explanation

Assertion (A) is true. Isobars have the same mass number but different atomic numbers. Both ¹⁴₆C (carbon-14) and ¹⁴₇N (nitrogen-14) have a mass number of 14, making them isobars.

Reason (R) is true. ¹⁴₆C has 6 protons and 8 neutrons, while ¹⁴₇N has 7 protons and 7 neutrons both have mass number equal to 14, hence are isobars.

The three constituents of an atom are electrons, protons and neutrons.

The mass of an electron is 9.1 × 10^-31 kg and charge is -1.6 × 10^-19 C.

The mass of a proton is 1.67 × 10^-27 kg and charge is +1.6 × 10^-19 C.

The mass of a neutron is 1.67 × 10^-27 kg and charge is zero.

The protons and neutrons reside inside the nucleus of the atom which is at it's centre, while the electrons revolve around the nucleus in some specific orbits.

(a) Atomic number (Z) — The atomic number of an atom is equal to the number of protons in it's nucleus (which is same as the number of electrons in a neutral atom). i.e.,
Z = number of protons in the nucleus of an atom.

(b) Mass number (A) — The mass number of an atom is equal to the total number of nucleons (i.e., the sum of the number of protons and the number of neutrons ) in it's nucleus. i.e.,
A = number of protons + number of neutrons in the nucleus of an atom.

The Atomic Number (Z) of Na = 11

The Mass Number (A) of Na = 23

Number of Neutrons = A - Z = 23 - 11 = 12

Its atomic model is shown below:

The atoms of the same element, having same atomic number Z, but different mass number A, are called isotopes.

Example — Hydrogen has three isotopes namely protium $_{1}^{1}\text{H}$, deuterium $_{2}^{1}\text{H}$ and tritium $_{3}^{1}\text{H}$.

The atoms of different elements which have the same mass number A, but different atomic number Z are called isobars.

Example — $_{11}^{23}\text{Na}$ and $_{12}^{23}\text{Mg}$ are isobars.

The atoms of a substance having same atomic number, but different mass number are known as isotopes.

Example — Hydrogen has three isotopes namely protium $_{1}^{1}\text{H}$, deuterium $_{1}^{2}\text{H}$ and tritium $_{1}^{3}\text{H}$. Each isotope in it's nucleus has one proton (Z = 1) but protium ($_{1}^{1}\text{H}$) has no neutron, deuterium ($_{1}^{2}\text{H}$) has one neutron and tritium ($_{1}^{3}\text{H}$) has two neutrons. There is one electron outside the nucleus in the atom of each isotope. Thus, the structures of these isotopes differ by the number of neutrons in their nuclei.

Radioactivity is a nuclear phenomenon. It is the process of spontaneous emission of α or β and γ radiations from the nucleus of atoms during their decay.

Uranium and Polonium are radioactive substances.

There will be no change in the nature of radioactivity when a radioactive substance is oxidized because radioactivity is a nuclear phenomena.

α and β radiations are charged particles (i.e., α radiation is positively charged and β radiation is negatively charged) therefore, they are deflected in an electric or magnetic field. γ radiations are uncharged particles hence, they do not show deflection in an electric or magnetic field.

No, it is not possible to deflect γ radiations as α and β particles, using the electric or magnetic field because α and β particles are charged particles and so they show deflection whereas γ radiations are uncharged particles hence, they do not deflect using electric or magnetic field.

(a) α radiation consists of two protons and two neutrons. β radiation is composed of fast moving electrons emitted from the nucleus of an atom while γ radiation is photon or electromagnetic waves like X-rays and light.

(b) No, hydrogen ($_1^1\text{H}$) nucleus cannot emit an alpha particle because an alpha particle consists of two protons and two neutrons whereas hydrogen ($_1^1\text{H}$) nucleus does not have two protons and two neutrons.

(c) α radiations have least penetrating power.

(i) When an α particle captures one electron it changes to a singly ionized helium He⁺.

(ii) When an α particle captures two electrons it changes to a neutral helium atom.

Any physical change (such as change in pressure and temperature) or chemical change (such as excessive heating, freezing, action of strong electric and magnetic fields, chemical treatment, oxidation etc.) does not change the rate of decay and the nature of radiation emitted by the substance.

This clearly shows that the phenomenon of radioactivity cannot be due to the orbital electrons which could easily be affected by such changes. Therefore the radioactivity should be the property of the nucleus. Thus, radioactivity is a nuclear phenomenon.

(a) When the nucleus of an element emits an α particle, the position of the resultant element changes to two places earlier (i.e., to the left of the parent element in the periodic table).

This happens because the atomic number decreases by 2, hence the new element shifts two places to the left on emission of an α particle.

(b) When the nucleus of an element emits a β particle, the position of the resultant element changes to one place after (i.e., one place to the right ).

This happens because the atomic number increases by 1, hence the new element shifts one place to the right on emission of a β particle.

(c) When the nucleus of an element emits a γ radiation, there is no change in the position of the element in the periodic table.

This happens because there is no change in the atomic number, hence there is no change in the position of the element on emission of γ radiations.

(a) If the nucleus X of a radioactive element of mass number A and atomic number Z emits an α particle, the daughter nucleus Y of a new element has mass number equal to (A - 4) and atomic number equal to (Z - 2).
Thus, atomic number decreases by 2 units and mass number decreases by 4 units.

$\begin{matrix} _{Z}^{A}\text{X} & \longrightarrow & _{Z-2}^{A-4}\text{Y} & + & \space _{2}^{4}\text{He} \\ \footnotesize{\text{Parent nucleus}} & & \footnotesize{\text{Daughter nucleus}} & & \footnotesize{\text{(α-particle)}} \end{matrix}$

Example — When a radioactive uranium nucleus $_{92}^{238}\text{U}$ emits an α particle, a new nucleus thorium $_{90}^{234}\text{Th}$ is formed and the change is represented as follows —

$_{92}^{238}\text{U} \longrightarrow \space _{90}^{234}\text{Th} + \space _{2}^{4}\text{He}$

(b) In emitting a β particle, the number of nucleons in the nucleus remains same, but the number of neutrons is decreased by one and the number of protons is increased by one.
If a radioactive nucleus P with mass number A and atomic number Z emits a β particle to form a daughter nucleus Q with mass number A and atomic number Z + 1, the change can be represented as follows:

$\begin{matrix} _{Z}^{A}\text{P} & \longrightarrow & _{Z+1}^{\space \space \space \space \space A}\text{Q} & + & \space _{-1}^{\space \space \space 0}\text{e} \\ \footnotesize{\text{Parent nucleus}} & & \footnotesize{\text{Daughter nucleus}} & & \footnotesize{\text{(β-particle)}} \end{matrix}$

Example — A radioactive carbon nucleus $_{6}^{14}\text{C}$, having 14 nucleons (6 protons and 8 neutrons) emits a β particle and changes to a new nucleus nitrogen $_{7}^{14}\text{N}$ having 14 nucleons (7 protons and 7 neutrons).

The change is represented as follows —

$_{6}^{14}\text{C} \longrightarrow \space _{\space \space 7}^{14}\text{N} + _{-1}^{\space \space \space 0}\text{e}$

(c) There is no change in the mass number A and atomic number Z of the nucleus in gamma emission. It is represented as follows:

$\begin{matrix} _{Z}^{A}\text{X*} & \longrightarrow & _{Z}^{A}\text{X} & + & \space \text{γ} \\ \footnotesize{\text{Parent nucleus}} & & \footnotesize{\text{Daughter nucleus}} & & \footnotesize{\text{(gamma radiation)}} \end{matrix}$

Here the star indicates the excited state of the nucleus. Thus in gamma emission, the excited nucleus comes to its ground state.

$_{85}^{222}\text{R}$ before the emission of β particle, would have atomic number 84, and before 2 α emissions, atomic number would have been 88. Hence, atomic number (Z) = 88.

$_{85}^{222}\text{R}$ before the emission of β particle, would have mass number 222, and before 2 α emissions, mass number would have been 230. Hence, mass number (A) = 230.

One example of radio isotopes is Cobalt $_{27}^{60}\text{Co}$ (Z = 27 and A = 60).

Radio isotopes are used in the Medical field. Many diseases such as leukemia, cancer, etc. are cured by radio therapy with the help of radio isotopes.

The alpha particles are not used in radio therapy because they cannot penetrate the human skin.

The nucleus of an atom becomes radioactive when number of neutrons in the nucleus exceeds the number of protons in it.

1. Many diseases such as leukemia, cancer, etc., are cured by radiation therapy. γ radiations from cobalt-60 ($_{27}^{60}\text{Co}$) are used to treat cancer by killing the cells in the malignant tumor of the patient.
2. The salt of weak radioactive isotopes such as radio-sodium chloride, radio-iron and radio-iodine are used for diagnosis. Such radio isotopes are called the tracers. These tracers are used to detect the suspected brain tumors and blood clots before they become dangerous.
3. γ rays emitted by the radio isotopes are used to sterilize bandages, dressings, syringes and other equipments to make them free from germs.

α < β < γ

An α-particle rapidly dissipates it's energy as it moves through a medium and therefore it's penetrating power is quite small. It can penetrate only through 3-8 cm in air. It can easily be stopped by a thin card sheet or a thick paper. As it is least penetrating so it causes least biological damage.

The penetrating power of β-particles is more than that of the α-particles but less than that of γ-particles. They can travel through nearly 5 m in air and pass through thin card sheet, and even through thin aluminium foil, but a 5 mm thick aluminium sheet can stop the beta particles. As the penetrating power is in between α and γ, hence the biological damage cause by β radiation is also in between α and γ.

The penetrating power of γ-rays is high. It is about (100,00) times that of α-particles and (100) times that of β-particles. They can pass through 500 m in air or through 30 cm thick sheet of iron. A thick sheet of lead is required to stop them. As penetrating power is most hence biological damage caused by γ-rays is most.

Two main sources of nuclear radiations are:

1. Radioactive fall out from nuclear plants — The nuclear power plants are now a major source of electricity in the world. If somehow, there is an accident in the reactor of power plant, a large amount of radioactive material and radiations will escape out into the atmosphere. This will not only affect the population around the plant, but will also affect the life at far off places where they will reach due to air currents.
2. Nuclear waste — The nuclear waste are still very radioactive and are a source of harmful radiations therefore, they should not be dumped in the open garbage. They can contaminate water and soil and affect the human and living organisms.

The harmful biological effects of nuclear radiations are of three types:

1. Short term recoverable effects like diarrhea, sore throat, loss of hair, nausea, etc.
2. Long term irrecoverable effects like leukemia and cancer.
3. Genetic effects that appear in the later generations of the person exposed to the radiation as the genes in the living cell get modified.

Two safety measures to be taken while establishing a nuclear power plant are:

1. The nuclear reactor of the power plant must be shielded with lead and steel walls so as to stop radiations from escaping out to the environment during it's normal operation.
2. The nuclear reactor must be housed in an air tight building of strong concrete structure which can withstand earthquakes, fires and explosion.

The fuel rods used in the nuclear power reactors are rejected when their activity decreases below a certain level. They then become a nuclear waste.

These rods are still quite radioactive and are the source of harmful radiations therefore they should not be dumped in open garbage. They can contaminate water and soil and affect the human and living organism.

Safety precautions that people should take while handling the radioactive substances are —

1. They should put on special lead lined aprons and lead gloves.
2. They should handle the radioactive materials with long lead tongs.
3. The radioactive substances must be kept in a thick lead container with a very narrow opening, so that the walls of the container absorb the radiations which strike on them from the inside and thus radiations only come out through the narrow opening.

Radioactive substance should not be touched by hands because these radiations are harmful and can damage the living tissues and cause radiation burns.

Background radiations are the radioactive radiations (such as α, β and γ) to which we all are exposed, even in the absence of an actual visible radioactive source.

The sources of background radiation are:

1. Internal source — the radioactive substances such as potassium (K-40), carbon (C-14) and radium are present inside our body.
2. External source — cosmic rays, naturally occurring radioactive elements such as radon-222 and solar radiation.

It is not possible for us to keep ourselves away from the background radiations.

The name given to elements with same mass number A, and different atomic number Z is isobars.

(i) Ionizing power of α, β and γ radiations arranged in ascending order is as follows:
γ < β < α

(ii) Penetrating power of α, β and γ radiations arranged in ascending order is as follows:
α < β < γ

The speeds are:

• α Radiation — 10⁷ m s^-1
• β Radiation — 2 x 10⁸ m s^-1
• γ Radiation — 3 x 10⁸ m s^-1

A β particle leaves it's mass number unchanged, but increases it's atomic number by one.
The symbol of β particle is given by $_{-1}^{ \space \space \space 0}\text{e}$ or $_{-1}^{ \space \space \space 0}\text{β}$.

(i) Atomic number

• When an α particle is emitted atomic number decreases by 2.
• When a β particle is emitted atomic number increases by 1.
• When γ radiation is emitted there is no change in atomic number.

(ii) Mass number

• When an α particle is emitted mass number decreases by 4.
• When a β particle is emitted there is no change in the mass number.
• When a γ radiation is emitted there is no change in the mass number.

(a) The above reaction is allowed as the mass number is conserved.

(b) The above reaction is not allowed as the mass number is not conserved.

(a) The mass number and atomic number of an element are not changed when it emits γ radiations.

(b) The atomic number of a radioactive element is not changed when it emits γ radiations .

(c) During the emission of a beta particle, the mass number remains same.

(a) Due to emission of a β particle, the atomic number increases by 1 and the mass number is unchanged. Hence, the nuclear change is —

$_{x}^{a}\text{P} \longrightarrow _{x + 1}^{a}\text{Q} + _{-1}^{0}\text{β}$

(b) Due to α emission, atomic number decreases by 2 and mass number decreases by 4. Also, an alpha particle is same as a Helium nucleus with 2 protons and 2 neutrons. Hence, we get the nuclear change as —

$_{92}^{238}\text{U} \longrightarrow _{90}^{234}\text{Th} + _{2}^{4}\text{He} + \text{energy}$

(c) When there is an α emission from $_{92}^{238}\text{P}$ , then the atomic number decreases by 2 and mass number decreases by 4. Hence, we get $_{90}^{234}\text{Q}$ .

Now, when $_{90}^{234}\text{Q}$ undergoes β emission, the atomic number increases by 1 and the mass number is unchanged. Hence, we get $_{91}^{234}\text{R}$.

In the final step when $_{91}^{234}\text{R}$ undergoes β emission again the atomic number increases by 1 and the mass number is unchanged. Hence we get $_{92}^{234}\text{S}$.

So the complete nuclear change is as follows —

$_{92}^{238}\text{P} \overset{\alpha} \longrightarrow \space _{90}^{234}\text{Q} \overset{\beta} \longrightarrow \space _{91}^{234}\text{R} \overset{\beta} \longrightarrow \space _{92}^{234}\text{S}$

(d) When there is an α emission from $_{Z}^{A}\text{X}$ , then the atomic number decreases by 2 and mass number decreases by 4. Hence, we get $_{Z - 2}^{A - 4}\text{X}_{1}$

After the γ emission, there is no change in atomic number and mass number. Hence, we get $_{Z - 2}^{A - 4}\text{X}_{2}$

In the final step when there are two β emissions, the atomic number increases by 2 and the mass number is unchanged and we get $_{Z}^{A - 4}\text{X}_{3}$

Hence, the nuclear change is as follows —

$_{Z}^{A}\text{X} \overset{\alpha} \longrightarrow _{Z - 2}^{A - 4}\text{X}_{1} \overset{\gamma} \longrightarrow _{Z - 2}^{A - 4}\text{X}_{2} \overset{2\beta} \longrightarrow _{\space \space \space \space \space Z}^{A - 4}\text{X}_{3}$

(e) The daughter nucleus $_{\space \space 69}^{172}\text{X}_{3}$ is formed after α decay of $\text{X}_{2}$, so atomic number of $\text{X}_{2}$ will be 2 units more and it's mass number will be 4 units more i.e., $_{\space \space 71}^{176}\text{X}_{2}$

The nucleus $_{\space \space 71}^{176}\text{X}_{2}$ is formed after α decay of $\text{X}_{1}$ so atomic number of $\text{X}_{1}$ will be 2 units more and it's mass number will be 4 units more i.e., $_{\space \space 73}^{180}\text{X}_{2}$

The nucleus $_{\space \space 73}^{180}\text{X}_{2}$ is formed after β decay of X, the atomic number will be one unit less and the mass number is unchanged i.e., $_{\space \space 72}^{180}\text{X}$

Hence, the nuclear change is as follows —

$_{\space \space 72}^{180}\text{X} \overset{\beta} \longrightarrow \space _{\space \space 73}^{180}\text{X}_{1} \overset{\alpha} \longrightarrow \space _{\space \space 71}^{176}\text{X}_{2} \overset{\alpha} \longrightarrow \space _{\space \space 69}^{172}\text{X}_{3}$

We usually use isotopes emitting γ radiations as radioactive tracers in medical science, because they are most penetrating and can easily pass through the human body.

The following are radioisotope —

(a) $_{6}^{14}\text{C}$

(b) $_{15}^{32}\text{P}$

(c) $_{19}^{40}\text{K}$

The above mentioned elements are isotopes because the number of neutrons exceeds the number of protons.

(a) The ore of uranium found in nature contains two isotopes ($_{92}^{238}\text{U}$ and $_{92}^{235}\text{U}$). Both are fissionable.

(b) Although both the isotopes are fissionable, but experimentally it is found that the isotope $_{92}^{235}\text{U}$ is more fissionable than the isotope $_{92}^{238}\text{U}$.
The reason is that the fission of $_{92}^{238}\text{U}$ nucleus is possible only by fast neutron, while the fission of $_{92}^{235}\text{U}$ nucleus can occur even by the slow neutrons.

(c) Either of slow or fast neutron can be used for the fission reaction of $_{92}^{235}\text{U}$.

931 MeV

Reason — 1 atomic mass unit (amu) is equivalent to 931 MeV (mega-electronvolts).

9.1 x 10^-31

Reason — The mass of an electron is 9.1 x 10^-31.

neutron

Reason — The particle used in nuclear fission for bombardment is neutron.

E = Δmc²

Reason — According to the mass energy equivalence :

E = Δmc²

Here, Δm is the loss in mass in kg, c is the speed of light (= 3 x 10⁸ m s^-1) and E is the energy in joule (J).

both (1) and (2)

Reason — In each fission reaction, the atomic number and mass number remains conserved.

Both (2) and (3)

Reason — The chain reaction is a nuclear reactor are controlled by absorbing some of the neutrons emitted in the fission process by means of the cadmium rods and then making them slow by the moderators (such as graphite, heavy water, etc). This way, the energy obtained in fission reaction can be controlled and utilised for constructive purposes.

hydrogen

Reason — The source of energy in the Sun and stars is obtained from the process of nuclear fusion, specifically the fusion of hydrogen nuclei (protons) into helium nuclei. This process releases a tremendous amount of energy.

10⁷ K

Reason — Nuclear fusion is not possible at ordinary temperature and ordinary pressure because when two nuclei approach each other, due to their positive charge, the electrostatic force of repulsion becomes too strong between them that they do not fuse. Hence, to make the fusion possible, a high temperature of (approximately 10⁷ K) and high pressure is required. Due to thermal agitations both nuclei acquire sufficient kinetic energy at such a high temperature so as to overcome the force of repulsion between them when they approach each other and so they get fused.

Given,

loss in mass (∆m) = 0.2 a.m.u.

We know that,

1 a.m.u = 931 MeV

Therefore, E = 0.2 × 931 = 186.2 MeV

Hence, the energy released = 186.2 MeV

Given,

loss in mass (∆m) = 0.0265 a.m.u.

We know that,

1 a.m.u = 931.5 MeV

Therefore, E = 0.0265 a.m.u. × 931.5 MeV = 24.7 MeV

Hence, the energy released = 24.7 MeV

In a nuclear change due to a radioactive phenomenon (such as decay, fission or fusion), the total sum of masses of product nuclei is always less than the total sum of the masses of reactant nuclei. Thus, there ia a loss of mass, due to which the energy is released. The energy so obtained is called the nuclear energy.

The radioactive phenomenon responsible for the release of nuclear energy are fission and fusion.

Einstein’s mass-energy equivalence relation is E = (∆m)c².

Here,

∆m is the loss in mass in kg,
c is the speed of light (= 3 × 10⁸ m s^-1) and
E is the energy in joule (J).

The mass of atomic particles is expressed in atomic mass unit (a.m.u).

1 a.m.u. = 931 MeV

The mass of a proton = 1.00727 a.m.u.
The mass of a neutron = 1.00865 a.m.u.
The mass of an electron = 0.00055 a.m.u.

Nuclear fission is the process in which a heavy nucleus splits into two lighter nuclei of nearly the same size, when bombarded with slow neutrons. In each fission reaction, a tremendous amount of energy is released.

Uranium is one substance that can be used for nuclear fission.

Example of fission reaction is —

$\begin{matrix} _{92}^{235}\text{U} & + & _{0}^{1}\text{n} & \longrightarrow & (_{92}^{236}\text{U}) & \longrightarrow & _{56}^{144}\text{Ba} \\ \footnotesize{\text{(uranium)}} & & \footnotesize{\text{(neutron)}} & & & & \footnotesize{\text{(barium)}} \\ & + & _{36}^{89}\text{Kr} & + & 3 \space _{0}^{1}\text{n} & + & \text{energy} \\ & & \footnotesize{\text{(krypton)}} & & \footnotesize{\text{(neutrons)}} \end{matrix}$

Approximately 190 MeV energy is released in the fission of one nucleus.

The reason for the energy release is the loss in mass. In a nuclear change due to a radioactive phenomenon (such as decay, fission or fusion), the total sum of masses of product nuclei is always less than the total sum of the masses of reactant nuclei. Thus, there ia a loss of mass, due to which the energy is released.

When slow neutrons are bombarded on uranium-235 ($_{92}^{235}\text{U}$), each uranium nucleus splits into two nearly equal fragments ($_{56}^{144}\text{Ba}$ and $_{36}^{89}\text{Kr}$) with the release of three neutrons and tremendous amount of energy (nearly 190 MeV). These new neutrons can cause fission in the other uranium nuclei under the suitable conditions. Thus, a chain of fission of nuclei is formed which once started, continues till the entire uranium is consumed.

The chain reaction is controlled by absorbing some of the neutrons emitted in the fission process by means of the cadmium rods and then making them slow by the moderators (such as graphite, heavy water, etc). This way, the energy obtained in fission reaction can be controlled and utilised for the constructive purposes.

1. For destructive use — Nuclear fission is used in nuclear bomb. The energy released is fast and uncontrolled.
2. For constructive use — The fission process is used in a nuclear reactor, where the rate of release of energy is slow and controlled. This energy is used to generate the electric power.

(a) Nuclear fusion is the process in which two light nuclei combine to form a heavy nucleus. In this process, huge amount of energy is released.

Example —

When two deuterium nuclei ($_{1}^{2}\text{H}$) fuse, 3.3 MeV energy is released and the nucleus of helium isotope ($_{2}^{3}\text{He}$) is formed. This helium isotope again gets fused with one deuterium nucleus to form a helium nucleus ($_{2}^{4}\text{He}$) and 18.3 MeV energy is released in this process. The nuclear reaction are —

$_{1}^{2}\text{H} + \space _{1}^{2}\text{H} \longrightarrow \space _{2}^{3}\text{He} + \space _{0}^{1}\text{n} + 3.3 \text{ MeV}$

$_{2}^{3}\text{He} + \space _{1}^{2}\text{H} \longrightarrow \space _{2}^{4}\text{He} + \space _{1}^{1}\text{H} + 18.3 \text{ MeV}$

Thus in all, three deuterium nuclei fuse to form a helium nucleus with a release of 21.6 MeV energy.

(b) Nuclear fusion is also called as thermo-nuclear reaction because nuclear fusion takes place at very high temperature.

Nuclear fusion is not possible at ordinary temperature and ordinary pressure because when two nuclei approach each other, due to their positive charge, the electrostatic force of repulsion becomes too strong between them that they do not fuse. Hence, to make the fusion possible, a high temperature of (≈ 10⁷ K) and high pressure is required. Due to thermal agitations both nuclei acquire sufficient kinetic energy at such a high temperature so as to overcome the force of repulsion between them when they approach each other and so they get fused.
That is why the fusion reaction is also called the thermo nuclear reaction.
The temperature required is ≈ 10⁷ K.

(a) A nuclear fusion reaction is given below:

$\begin{matrix} _{1}^{2}\text{H} & + & _{1}^{2}\text{H} & \longrightarrow & _{2}^{3}\text{He} & + & _{0}^{1}\text{n} & + & 3.3\text{ MeV} \\ \footnotesize{\text{(deuterium)}} & & \footnotesize{\text{(deuterium)}} & & \footnotesize{\text{(helium isotope)}} & & \footnotesize{\text{(neutron)}} \end{matrix}$

$\begin{matrix} _{2}^{3}\text{He} & + & _{1}^{2}\text{H} & \longrightarrow & _{2}^{4}\text{He} & + & _{1}^{1}\text{H} & + & 18.3\text{ MeV} \\ \footnotesize{\text{(helium isotope)}} & & \footnotesize{\text{(deuterium)}} & & \footnotesize{\text{(helium)}} & & \footnotesize{\text{(proton)}} \end{matrix}$

In this reaction, three deuterium nuclei fuse to form a helium nucleus.

(b) Approximately, 21.6 MeV energy is released in the reaction.

(c) When two deuterium nuclei fuse, 3.3 MeV energy is released and the nucleus of helium isotope ($_{2}^{3}\text{He}$ ) is formed. This helium again gets fused with one deuterium nucleus to form a helium nucleus ($_{2}^{4}\text{He}$) and 18.3 MeV is released.

(a) In both the process nuclear fission and fusion, the similarity is the release of energy due to loss in mass.

(b) Differences between nuclear fission and fusion are as follows —

Two examples of nuclear fusion are given below:

$\begin{matrix} _{1}^{1}\text{H} & + & _{1}^{1}\text{H} & \longrightarrow & _{1}^{2}\text{H} & + & 0.42\text{ MeV} \\ \footnotesize{\text{(proton)}} & & \footnotesize{\text{(proton)}} & & \footnotesize{\text{(deuterium)}} \end{matrix}$

$\begin{matrix} _{1}^{2}\text{H} & + & _{1}^{2}\text{H} & \longrightarrow & _{1}^{3}\text{H} & + & _{1}^{1}\text{H} & + & 4.0\text{ MeV} \\ \footnotesize{\text{(deuterium)}} & & \footnotesize{\text{(deuterium)}} & & \footnotesize{\text{(tritium)}} & & \footnotesize{\text{(proton)}} \end{matrix}$

(a) $_{92}^{235}\text{U}$ + $_{0}^{1}\text{n}$ ⟶ $_{56}^{141}\text{Ba}$ + $_{36}^{92}\text{Kr}$ + 3 $_{0}^{1}\text{n}$ + energy

(b) $_{92}^{235}\text{U}$ + $_{0}^{1}\text{n}$ ⟶ $_{57}^{148}\text{La}$ + $_{35}^{85}\text{Br}$ + 3 $_{0}^{1}\text{n}$ + energy

(a) $_{2}^{3}\text{He}$ + $_{1}^{2}\text{H}$ ⟶ $_{2}^{4}\text{He}$ + $_{1} ^{1}\text{H}$ + energy

(b) $_{1}^{2}\text{H}$ + $_{1}^{2}\text{H}$ ⟶ $_{2}^{3}\text{He}$ + $_{0}^{1} \text{n}$ + energy

(a) In nuclear fusion, the energy released per unit mass is more as compared to nuclear fission.

(b) Nuclear fission is possible at ordinary temperature.

Nuclear Fusion is the source of energy of sun or stars.

(a) The reaction taking place is Nuclear Fission.

(b) The reaction taking place is Nuclear Fusion.