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# Radioactive Decay - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

## Quick Facts

• Radioactivity - (I), Radioactivity - (II) are considered the most difficult concepts.

• 42 Questions around this concept.

## Solve by difficulty

A radioactive nucleus (initial mass number A and atomic number Z ) emits $\dpi{100} 3\alpha -$ particles and 2 positrons. The ratio of the number of neutrons to that of protons in the final nucleus will be

When  $\dpi{100} _{3}Li^{7}$ nuclei are bombarded by protons, and the resultant nuclei are $\dpi{100} _{4}Be^{8}$ the emitted particles will be :

The energy spectrum of    emitted  from a  radioactive source is

Which of the following cannot be emitted by radioactive substances during their decay?

Which of the following radiations has the least wavelength?

In gamma-ray emission from a nucleus :

## Concepts Covered - 0

A. H. Becquerel, discovered radioactivity purely by accident. While doing an experiment related to the fluorescence and phosphorescence of compounds irradiated with visible light and during this experiment, he observed an interesting phenomenon. After illuminating some pieces of uranium-potassium sulphate with visible light, Becquerel wrapped them in black paper and then separated the package from a photographic plate with a piece of silver. After several hours of exposure, the photographic plate was developed. Blackening occurs due to something that must have been emitted by the compound and was able to penetrate both black paper and silver. Day by day many experiments have been performed and subsequently showed that radioactivity was a nuclear phenomenon in which an unstable nucleus undergoes a decay. This is referred to as radioactive decay.

Three types of radioactive decay occur in nature :

(i) α-decay in which a helium nucleus $_{4}^{2}\textrm{He}$ is emitted;

(ii) β-decay in which electrons or positrons (positrons is the particles with the same mass as electrons but with a charge exactly opposite to that of the electron) are emitted;

(iii) γ-decay in which high frequency and energy (hundreds of keV or more) photons are emitted. Each of these decay will be considered in subsequent sub-sections.

An alpha particle is a helium nucleus, it means that it can be represented as 42He. So, whenever a nucleus goes through alpha decay, then it gets transformed into a different nucleus by emitting an alpha particle. Let us take an example when 23892U undergoes alpha-decay, it gets transformed into 23490Th, which is shown as -

$^{238}_{92} \mathrm{U} \rightarrow^{234}_{90} \mathrm{Th}+_{2}^{4} \mathrm{He}$

Now we have seen that 42He contains two protons and two neutrons. Hence when the alpha particle gets emitted, the mass number of the emitting nucleus reduces by four and similarly the atomic number reduces by two. Therefore, in general, we can write that AZX nucleus to A-4Z-2X nucleus is expressed as follows,

$^{\mathrm{A}}_{\mathrm{Z}} \mathrm{X} \rightarrow \ \ \ ^{\mathrm{A}-4}_\mathrm{{Z}-2} \mathrm{X}+_{2}^{4} \mathrm{He}$

where AZX is the parent nucleus and A-4Z-2X is the daughter nucleus. One very important point to be noted is that the alpha decay of 23892U can occur without an external source of energy. The reason behind this is that the total mass of the decay products i.e., (23490Th and 42He) < the mass of the original 23892U.

Or, we can say that the total mass energy of the decay products is less than that of the original nuclide. This gives rise to a new term called ‘Q value of the process’ or ‘Disintegration energy’ which is the difference between the initial and final mass-energy of the decay products. So, for alpha decay, the Q value is expressed as -

$\mathrm{Q}=\left(\mathrm{m}_{\mathrm{X}}-\mathrm{m}_{\mathrm{Y}}-\mathrm{m}_{\mathrm{He}}\right) \mathrm{c}^{2}$

This disintegration energy is shared between the daughter nucleus, A-4Z-2X and the alpha particle both and for 42He it is in the form of kinetic energy.

Note-

Alpha decay obeys the radioactive laws.

Nuclear Reactions-

The process by which the identity of a nucleus is changed when it is bombarded by an energetic particle is called a nuclear reaction.

The general expression for the nuclear reaction is as follows.

$X+Y\rightarrow Z+Q$

where X and Y are known as reactants and Z is known as products.

and Q is the energy of the nuclear reaction (i.e Q value)

Q value-

The energy absorbed or released during the nuclear reaction is known as the Q-value of nuclear reaction.

$Q \text { -value }= (Mass \ of \ reactants - mass \ of \ products)c^{2} \ Joules$

for the below reaction

$X+Y\rightarrow Z+Q$

$Q=(M_{x}+M_{y}-M_{z})C^{2}$

where

$M_{x}$ and $M_{y}$ are mass of reactant

$M_{z}$  is mass of product

• If Q < 0, The nuclear reaction is known as endothermic. (The energy is absorbed in the reaction)

• If Q> 0, The nuclear reaction is known as exothermic (The energy is released in the reaction)

Law of conservation in nuclear reactions-

Following quantities are conserved in nuclear reactions

• The mass number  (A)
• The charge number (Z)
• Linear momentum/angular momentum
• Total energy

Beta decay happens when a nucleus decays spontaneously by emitting an electron or a positron. Like alpha decay it is also a spontaneous process, and it is also having definite disintegration energy and half-life as well. It is also following the radioactive laws. A Beta decay can either be a beta minus or a beta plus decay. Let us discuss one by one -

In a Beta minus (β) decay, as the name indicates that an electron is emitted by the nucleus. Let us take an example,

$^{32}_{15} \mathrm{P} \rightarrow ^{32}_{16} \mathrm{S}+\mathrm{e}^{-}+\mathrm{v}^{-}$

where, $\fn_cm v^-$ is an antineutrino,

Note- Property of neutrino

1. Zero electric charge
2. It's mass much less than the mass of the electron
3.  Very weak matter making it quite difficult to detect

i.e For

• β Minus - decay

$^{A}_{Z}X\rightarrow _{Z+1}^{A}Y+e ^{-}+ {\nu }^{-}+Q\ value$

$Q\: value =\left [ M_{X}-M_{Y} \right ]c^{2}$

In a Beta plus decay, as the name indicates that it is a positron is emitted by the nucleus. Let us take an example,

$^{22}_{11} \mathrm{Na} \rightarrow ^{22}_{10} \mathrm{Ne}+\mathrm{e}^{+}+\mathrm{v}$

where v is a neutrino, which is a neutral particle with negligible or no mass.  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.

i.e For

• β plus decay

$^{A}_{Z}X\rightarrow _{Z-1}^{A}Y+e ^{+}+ {\nu }+Q \ value$

$Q\: value =\left [ M_{X}-M_{Y}-2M_{e} \right ]c^{2}$

Further, in a Beta minus decay, a neutron transforms into a proton (inside the nucleus):

$\mathrm{n} \rightarrow \mathrm{p}+\mathrm{e}^{-}+\mathrm{v}^{-}$

Also, in a Beta plus decay, a proton transforms into a neutron within the nucleus:

$p \rightarrow n+e^{+}+v$

Here, the point to be noted that the mass number (A) of the emitting nuclide does not change. From the above equation, either a neutron transforms into a proton or vice versa.

From the previous chapters as well as from chemistry, we know that atoms have energy levels. Like the atom, a nucleus has energy levels too. When a nucleus is in an excited state then to make itself stable, transition occurs to a lower energy state by emitting an electromagnetic radiation. The energy difference between the states in a nucleus is in MeV. Hence, the photons emitted by the nuclei have MeV energies and called Gamma rays.

As we have seen that after an alpha or beta emission, most radionuclides leave the daughter nucleus in an excited state. To reach the ground state, this daughter nucleus emits one or multiple gamma rays. Let us take an example,

6027Co undergoes a beta decay and then transforms into 6028Ni. Then this 6028Ni become daughter nucleus. 6028Ni is in its excited state and 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 shows this process clearly -

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