Radioactive Decay

Radioactive decay is a natural process in which unstable atomic nuclei release energy to become more stable over time. Commonly observed in elements such as uranium, radium, and carbon, radioactive decay plays a crucial role in fields like nuclear energy, medicine, archaeology, and environmental science. Understanding how radioactive decay works not only helps explain the behavior of radioactive materials, but also supports important applications such as carbon dating, radiation therapy, and nuclear power generation. In this guide, we’ll break down the types of radioactive decay, why it occurs, and how it impacts both science and everyday life.

1.0Radioactive Decay Definition

• Radioactive decay is the process in which an unstable atomic nucleus releases energy and changes into a more stable form.
• In radioactive decay, an unstable nucleus emits particle or particle. After emission of or   the remaining nucleus may emit -particle, and converts into more stable nucleus.

$\alpha -\text{Particle}$

• $$\begin{gathered} \text{It is a doubly charged helium nucleus. It contains two protons and two neutrons.} \\ \text{Mass}of\alpha -\text{particle}=\text{Mass of }{}_2^4\mathrm{He}\text{atom}-2m_e\approx 4m_p \\ \text{Charge of}\alpha -\text{particle = +2e} \end{gathered}$$

$\beta -\text{Particle}$

$$\begin{gathered} \text{(a)}{\beta }^{-}(electron) \\ \text{Mass}=m_e;\text{charge}=-e \\ \text{(b)}{\beta }^{+}(positron) \\ \text{Mass}=m_e;\text{charge}=+e \\ \text{Positron is an antiparticle of electrons.} \end{gathered}$$

Antiparticle:A particle is called an antiparticle of another if on collision both can annihilate (destroy completely) and convert into energy.

For example:

$$\begin{gathered} (1)\text{ Electrons}(-e,m_e)\text{and positron}(+e,m_e)\text{ are antiparticles.} \\ (2)\text{ Neutrino}(\nu )\text{ and antineutrino}(\bar{\nu })\text{ are antiparticles.} \\ \gamma -\text{Particle - They are energetic photons of energy of the order of MeV and having rest mass zero.} \end{gathered}$$

2.0Radioactive Decay (Displacement Law)

$$\begin{gathered} \alpha -\text{Decay}\text{ - Nuclei with mass number greater than 210 undergo }\alpha -\text{ decay.} \\ {}_Z^{A}X\to {}_{Z-2}^{A-4}Y+{}_2^4\mathrm{He}+Q \end{gathered}$$

$\text{Q Value}$

$$\begin{gathered} \text{It is defined as the energy released during the decay process.} \\ Q\text{ value}=\text{rest mass energy of reactants}-\text{rest mass energy of products} \\ \text{This energy is available in the form of increase in K.E. of the products.} \\ {M}_x=\text{mass of atom }{}_Z^{A}X \\ {M}_y=\text{mass of atom }{}_{Z-2}^{A-4}Y \\ {M}_{\mathrm{He}}=\text{mass of atom }{}_2^4\mathrm{He} \\ Q\text{ value}=\left[\left(M_x-Z{m}_e\right)-\left\{\left(M_y-(Z-2)m_e\right)+\left(M_{\mathrm{He}}-2m_e\right)\right\}\right]c^2 \\ Q\text{ value}=\left[M_x-M_y-M_{\mathrm{He}}\right]c^2 \end{gathered}$$

$$\begin{gathered} \text{Considering actual number of electrons in }\alpha -\text{ decay} \\ Q\text{ value}=\left[M_x-\left(M_y+2m_e\right)-\left(M_{\mathrm{He}}-2m_e\right)\right]c^2 \\ Q\text{ value}=\left[M_x-M_y-M_{\mathrm{He}}\right]c^2 \end{gathered}$$

Calculation of Kinetic Energy of Final Products:

As atom X was initially at rest and no external forces are acting, so final momentum also has to be zero.Hence both Y and $\alpha$-particles will have the same momentum in magnitude but in opposite directions.

$$\begin{gathered} p_{\alpha }^2=p_Y^2;2m_{\alpha }{T}_{\alpha }=2m_Y{T}_Y\text{(Here we are representing T for kinetic energy)} \\ Q=T_Y+T_{\alpha };m_{\alpha }{T}_{\alpha }=m_Y{T}_Y \\ {T}_{\alpha }=\frac{m_Y}{m_{\alpha }+m_Y}Q;T_Y=\frac{m_{\alpha }}{m_{\alpha }+m_Y}Q\Rightarrow T_{\alpha }=\frac{A-4}{A}Q;T_Y=\frac{4}{A}Q \end{gathered}$$

From the above calculation, one can see that all the -particles emitted should have the same kinetic energy. Hence, if they are passed through a region of uniform magnetic field having direction perpendicular to velocity, they should move in a circle of same radius.

$\frac{mv}{qB}=\frac{mv}{2eB}=\frac{\sqrt{2Km}}{2eB}$

Experimental Observation:

Experimentally it has been observed that all the $\alpha$-particles do not move in the circle of the same radius, but they move in circles having different radii. This shows that they have different kinetic energies. But it is also observed that they follow circular paths of some fixed values of radius i.e. yet the energy of emitted -particles are not the same but they are quantized. The reason behind this is that all the daughter nuclei produced are not in their ground state but some of the daughter nuclei may be produced in their excited states and they emit photons to acquire their ground state. The only difference between Y and $Y^{\ast }$ is that $Y^{\ast }$ is in an excited state and Y is in ground state. 

$$\begin{gathered} \text{Let the energy of emitted}\gamma \text{-particles be E.} \\ \therefore Q=T_{\alpha }+T_Y+E \\ \text{Where }Q=\left[M_X-M_Y-M_{\mathrm{He}}\right]c^2 \\ {T}_{\alpha }+T_Y=Q-E \\ {T}_{\alpha }=\frac{m_Y}{m_{\alpha }+m_Y}(Q-E);T_Y \\ \frac{m_{\alpha }}{m_{\alpha }+m_Y}(Q-E) \\ {\beta }^{-}\text{-Decay} \\ {}_Z^{A}X\to {}_{Z+1}^{A}Y+{}_{-1}^{0}e+Q \end{gathered}$$

${}_{-1}{e}^0\text{ can also be written as }{}_{-1}^0\beta$

Here also one can see that by momentum and energy conservation, we will get

$T_e=\frac{m_Y}{m_e+m_Y}Q;T_Y=\frac{m_e}{m_e+m_Y}Q$

As $m_e\ll m_Y$ , we can consider that all the energy is taken away by the electron.From the above results, we will find that all the $\beta$-particles emitted will have the same energy and hence they have the same radius if passed through a region of perpendicular magnetic field. But, experimental observations were completely different.

On passing through a region of uniform magnetic field perpendicular to the velocity, it was observed that $\beta$-particles take circular paths of different radius having a continuous spectrum. To explain this, Pauling has introduced the extra particles called neutrino and antineutrino (antiparticle of neutrino).

$\bar{\nu }\to \text{antineutrino},\nu \to \text{neutrino}$

3.0Properties of Antineutrino $[(\bar{\nu })\text{ and Neutrino }(\nu )]$

$$\begin{gathered} \text{(1) They have rest mass equal to zero or, at most, the mass equivalent of a few electron-volts. Their speed is c (or nearly equal to c). Energy: }E=m{c}^2. \\ \text{(2) They are chargeless (neutral).} \\ \text{(3) They have spin quantum number}s=\pm \frac{1}{2}. \\ \text{Considering the emission of antineutrino, the equation of }{\beta }^{-}\text{decay can be written as } \\ {}_Z^{A}X\to {}_{Z+1}^{A}Y+{}_{-1}^{0}e+Q \\ \text{(4) They are not electromagnetic in nature like the photon. Hence, neutrinos can pass unimpeded through vast amounts of matter.} \\ \text{Production of antineutrino along with the electron helps to explain the continuous spectrum, because the energy is distributed randomly between the electron and the antineutrino}(\bar{\nu }).\text{ It also helps to explain the spin quantum number balance, since p, n and e each have spin quantum number}\pm \frac{1}{2}. \\ \text{During }{\beta }^{-}\text{decay, inside the nucleus a neutron is converted into a proton with emission of an electron and an antineutrino:} \\ n\to p+{}_{-1}^{0}e+\bar{\nu }\text{Let }{M}_x=\text{mass of atom }{}_Z^{A}X \\ {M}_y=\text{mass of atom }{}_{Z+1}^{A}Y \\ {M}_e=\text{mass of electron}. \\ \text{Q- value}=[(M_x-Z{m}_e)-\{M_y-(Z+1)m_e+m_e\}]c^2=(M_x-M_y)c^2 \\ \text{Considering the actual number of electrons,} \\ Q=[M_x-\{(M_y-m_e)+m_e\}]c^2=(M_x-M_y)c^2 \end{gathered}$$

Energy Spectrum of $\beta$-Particles:

The figure below shows the energy spectrum of the electrons emitted in the beta decay.

${\beta }^{+}\text{ Decay}$

$$\begin{gathered} {}_Z^{A}X\to {}_{Z-1}^{A}Y+{}_{+1}^{0}e+\nu +Q \\ \text{In }{\beta }^{+}\text{ decay, inside a nucleus a proton is converted into a neutron, a positron, and a neutrino.} \\ p\to n+{}_{+1}^{0}e+\nu \\ \text{As mass increases during the conversion of a proton into a neutron, energy is required for }{\beta }^{+}\text{ decay to take place.} \\ \therefore {\beta }^{+}\text{ decay is a rare process. It can occur only in nuclei where a proton can take energy from the nucleus itself.} \\ Q\text{value = }[(M_x-Z{m}_e)-\{(M_y-(Z-1)m_e)+m_e\}]c^2=\left[M_x-M_y-2m_e\right]c^2 \\ \text{Considering the Actual Number of Electrons} \\ Q\text{value = }[M_x-\{(M_y+m_e)+m_e\}]c^2=\left[M_x-M_y-2m_e\right]c^2 \end{gathered}$$

K Capture

It is a rare process which is found only in a few nuclei. In this process the nucleus captures one of the atomic electrons from the K shell. A proton in the nucleus combines with this electron and converts itself into a neutron. A neutrino is also emitted in the process and is emitted from the nucleus.

Electron capture is competitive with positron emission. It occurs more often than positron emission in heavy nuclides because electrons are relatively closer to the nucleus which allows more interaction.

$$\begin{gathered} p+{}_{-1}^{0}e\to n+\nu \\ \text{If X and Y are atoms, the reaction is written as:}{}_Z^{A}X\to {}_{Z-1}^{A}Y+\nu +Q+\text{characteristic X-rays of }Y \\ \text{If X and Y are taken as nuclei, then the reaction is written as:}{}_Z^{A}X+{}_{-1}^{0}e\to {}_{Z-1}^{A}Y+\nu \end{gathered}$$

$\gamma$ -Decay

Like an atom a nucleus can also exist in states whose energies are higher than that of its ground state.Excited nuclei return to their ground states by emitting photons whose energies correspond to the energy differences between the various initial and final states in the transitions involved. The photons emitted by nuclei have energy up to several MeV, and are traditionally called gamma rays.

$$\begin{gathered} {}_{12}^{27}\mathrm{Mg}\to {}_{13}^{27}\mathrm{Al}+{}_{-1}^{0}e \\ {}_{13}^{27}{\mathrm{Al}}^{\ast }\to {}_{13}^{27}\mathrm{Al}+\gamma \\ {\mathrm{Al}}^{\ast }\text{ represents an aluminium nucleus in its excited state.} \\ \text{When }\gamma \text{-rays pass through a slab, their intensity decreases exponentially with the slab thickness x.}I=I_0{e}^{-\mu x}\text{where }\mu \text{ is the absorption coefficient, which depends on the material of the slab.} \end{gathered}$$

Note:

(1) Nuclei having atomic numbers from Z = 84 to 112 shows radioactivity.

(2) Nuclei having Z = 1 to 83 are stable (only few exceptions are there)

(3) Whenever a neutron is produced, a neutrino is also produced.

(4) Whenever a neutron is converted into a proton, an antineutrino is produced.

4.0Nuclear Stability

The plot of neutron number(N) vs proton number(Z) shows a narrow band of stable nuclides.

For light nuclides, N ≈ Z, so N/Z ≈ 1.

For heavy stable nuclides, the N/Z ratio increases to about 1.6.

Nuclides with excess neutrons lie above the stability belt and are unstable, undergoing β⁻ decay.

Nuclides with excess protons lie below the belt and decay via β⁺ emission or K-capture.

5.0Nuclear Force

1. Nuclear forces are fundamentally attractive and bind nucleons together despite the repulsion between protons.
2. They are the strongest forces at nuclear distances (≈100 times stronger than electromagnetic forces).
3. They have a very short range, acting only within the nucleus.
4. They operate between all nucleon pairs—n–n, n–p, and p–p.
5. Their strength does not depend on the type of nucleons involved.
6. Nuclear forces are non-central and depend on nucleon spin orientation—stronger for parallel spins and weaker for antiparallel spins.

6.0Radioactive Decay: Statistical Law

$$\begin{gathered} \text{(Given by Rutherford and Soddy)} \\ \text{Rate of radioactive decay}\propto N \\ \Rightarrow \frac{dN}{dt}\propto N \\ \frac{dN}{dt}=-\lambda N \\ \text{where }N=\text{number of active nuclei} \\ \text{where }\lambda =\text{decay constant of the radioactive substance} \\ \text{The decay constant varies from one radioactive substance to another, but it is independent of the quantity of the substance and of time.} \\ \text{SI unit of }\lambda ={\text{s}}^{-1} \\ \text{If }{\lambda }_1>{\lambda }_2,\text{ then the first substance is more radioactive (less stable) than the second one.} \\ \text{For the case, if A decays to B with decay constant }\lambda : \\ A\stackrel{\lambda }{\to }B \\ A\to B \\ t=0N_{0}0\text{where }{N}_0=\text{number of active nuclei of }A\text{ at }t=0 \\ t=t:N{N}^{\prime }\text{where }N=\text{number of active nuclei of }t=t \\ \text{Rate of radioactive decay of A:}-\frac{dN}{dt}=\lambda N \\ -{\int }_{N_0}^N\frac{dN}{N}={\int }_0^t\lambda dt \\ N=N_0{e}^{-\lambda }t\text{(It is exponential decay)} \end{gathered}$$

$$\begin{gathered} \text{Number of nuclei decayed (i.e. the number of nuclei of B formed)} \\ {N}^{\prime }=N_0-N=N_0-N_0{e}^{-\lambda t} \\ {N}^{\prime }=N_0\left(1-e^{-\lambda t}\right) \end{gathered}$$          

$\text{Half Life}\left(T_{1/2}\right)$

$$\begin{gathered} \text{It is the time in which the number of active nuclei becomes half.} \\ N=N_0{e}^{-\lambda t} \\ \text{After one half-life,} \\ N=\frac{N_0}{2} \\ \frac{N_0}{2}=N_0{e}^{-\lambda }t\Rightarrow t=\frac{\ln 2}{\lambda }\Rightarrow t=\frac{\ln 2}{\lambda }\Rightarrow \frac{0.693}{\lambda }=t_{1/2} \\ {t}_{1/2}=\frac{\ln 2}{\lambda }\Rightarrow \frac{0.693}{\lambda }\text{(To be Remembered)} \\ \text{Number of nuclei present after n half lives i.e. after a time}t=n{t}_{1/2} \\ N=N_0{e}^{-\lambda t}=N_0{e}^{-\lambda n{t}_{1/2}}=N_0{e}^{-\lambda n\frac{\ln 2}{\lambda }} \\ =N_0{e}^{\ln 2(-n)}=N_0(2{)}^{-n}=N_0{\left(\frac{1}{2}\right)}^n=\frac{N_0}{2^n} \\ \left\{n=\frac{t}{t_{1/2}}.\text{(It may be a fraction, need not to be an integer)}\right\} \\ {N}_0\to \frac{N_0}{2}\to N_0{\left(\frac{1}{2}\right)}^2\to N_0{\left(\frac{1}{2}\right)}^3\to \cdots \cdots \cdots \cdots \cdots \cdots N_0{\left(\frac{1}{2}\right)}^n \end{gathered}$$

Activity

$$\begin{gathered} \text{Activity refers to the rate at which radioactive nuclei undergo decay. It is represented by the symbols A or R.} \\ A=\lambda N \\ \text{If a radioactive substance changes only due to decay then} \\ A=-\frac{dN}{dt} \\ \text{As in that case,} \\ N=N_0{e}^{-\lambda t} \\ A=\lambda N=\lambda N_0{e}^{-\lambda t} \\ A=A_0{e}^{-\lambda t} \\ \text{SI Unit of Activity:}\text{ The SI unit of activity is the becquerel (Bq), which corresponds to one disintegration per second (1 dps).} \\ \text{A commonly used non-SI unit is the curie (Ci), defined as:} \\ 1\text{curie}=3.7\times 10^{10}\text{disintegrations per second}, \\ \text{which represents the activity of 1 gram of radium.} \\ \text{Specific Activity: }\text{the activity measured per unit mass.} \\ \text{Average Life:} \\ {T}_{\text{avg}}=\frac{\text{Sum of ages of all the nuclei}}{N_0}=\frac{{\int }_0^{\infty }\lambda N_0{e}^{-\lambda t}dt\cdot t}{N_0}=\frac{1}{\lambda } \end{gathered}$$