Due to the radioisotope of the element having an unstable nucleus, the atom particles cannot be bonded since there is no energy. The isotopes constantly decay to stabilise themselves by releasing a significant amount of energy in the form of radiation.

Transmutation is referred to as the process of isotope transformation into an element of a stable nucleus. It can occur both in natural and artificial ways.

There are three types, namely:

1. Alpha
2. Beta
3. Gamma.

### 1. Alpha decay:

When an alpha particle emits its nucleus, the process is called alpha decay. The formula of alpha decay is given as:

 $$\begin{array}{l}E=(m_{i}-m_{f}-m_{p})c^{2}\end{array}$$

Where,

• mi is the initial mass of the nucleus
• mf is the mass of the nucleus after particles emission
• mp is the mass of the emitted particle

The nucleus of helium is taken as the very stable alpha particle. It has a group of two protons and two neutrons. For example, alpha decay of uranium-238 is shown below-

$$\begin{array}{l}U_{92}^{238}\rightarrow Th_{90}^{234}+He_{2}^{4}\end{array}$$

Transmutation is referred to as the process of isotopes transforming into an element of a stable nucleus.

### 2. Beta Decay:

A beta particle is often referred to as an electron, but it can also be a positron. If the reaction involves electrons, the nucleus sheds out neutrons one by one. Even the proton number increases accordingly. A beta decay process is shown below:

$$\begin{array}{l}Th_{90}^{234}\rightarrow Pa_{91}^{234}+e_{-1}^{0}\end{array}$$

### 3. Gamma Decay:

The nucleus has orbiting electrons, which indeed have some energy, and when an electron jumps from a level of high energy to a level of low energy, there is an emission of a photon. The same thing happens in the nucleus: whenever it rearranges into a lower energy level, a high-energy photon is shooted out, which is known as a gamma ray.

## Top 15 most important and expected questions on nuclei in Hindi

In the sample, there is a proportionality between radioactive decay per unit time and the overall number of nuclei of radioactive compounds. We can mathematically quantify the rate of this type of decay through this proportionality.

If N = number of nuclei in a sample and ΞN is the number of radioactive decays per unit time Ξt then,

ΞNΞT β N or ΞNΞT = Ξ»N

Where, Ξ» is the constant of proportionality ( or radioactive decay constant or disintegration constant).

Also, Ξ N reduces the total number of nuclei present in the sample.

According to the convention, this should be termed negative.

$$\begin{array}{l}\partial N\partial t=-\lambda N\end{array}$$

After rearranging this,

$$\begin{array}{l}\partial NN=-\lambda \partial t\end{array}$$

Integrating both sides will result in,

$$\begin{array}{l}\int NN_{0}\partial NN=-\lambda \int tt_{0}\partial t\end{array}$$
$$\begin{array}{l}InN-InN_{0}=-\lambda \left ( t-t_{0} \right )\end{array}$$

Here,

N0 represents the original number of nuclei in the sample at a time t0, i.e. t=0.

Applying that in the equation results in;

$$\begin{array}{l}InNN_{0}=-\lambda t\end{array}$$

$$\begin{array}{l}N_{t}=N_{0}^{e-\lambda t}\end{array}$$

This type of decay is exponential.

Calculating the rate of decay,

$$\begin{array}{l}R=-\partial N\partial t\end{array}$$

Substituting Nt in the equation and differentiating it,

$$\begin{array}{l}N_{t}=N_{0}^{e-\lambda t}\end{array}$$

Differentiation result is,

$$\begin{array}{l}R=-\partial N\partial t=\lambda N_{0}^{e-\lambda t}R=R_{0}^{e-\lambda t}\end{array}$$

R0 here represents the decay rate at the time, t=0.

Substituting the original equation back here,

$$\begin{array}{l}\Delta N\Delta t=\lambda N\end{array}$$

We get,

$$\begin{array}{l}R=\lambda N\end{array}$$

The total decay rate R of a radioactive sample is called the activity of that sample which is represented with the unit Becquerel, in honour of its scientist. 1 becquerel = 1 Bq = 1 decay per second Another unit is the curie. 1 curie = 1 Ci = 3.7Γ1010 Bq

### Law Of Radioactive Decay Derivation

According to the radioactive decay law, when a radioactive material undergoes either πΌ or Ξ² or β½ decay, the number of nuclei undergoing the decay per unit time is proportional to the total number of nuclei in the given sample material.Β

The radioactive decay law states that βThe probability per unit time that a nucleus will decay is a constant, independent of timeβ.

It is represented by Ξ» (lambda) and is called decay constant.

The mathematical representation of the law of radioactive decay is:

$$\begin{array}{l}\frac{\Delta N}{\Delta t}\propto N\end{array}$$

Where,
N: the total number of nuclei in the sample Ξ
N: number of nuclei that undergoes decay
Ξt: unit time

$$\begin{array}{l}\frac{\Delta N}{\Delta t}=\lambda N (eq.1)\end{array}$$

Where,
Ξ»: radioactive decay constant, also known as disintegration constant

The change in the sample with respect to the number of nuclei is given as:

$$\begin{array}{l}\frac{dN}{dt}=-\lambda N\end{array}$$
$$\begin{array}{l}\frac{dN}{N}=-\lambda dt\end{array}$$
$$\begin{array}{l}\int_{N_{0}}^{N}\frac{dN}{N}=\lambda\int_{t_{0}}^{t}dt (eq.2)\end{array}$$
$$\begin{array}{l}lnN-lnN_{0}=-\lambda (t-t_{0}) (eq.3)\end{array}$$

Where,
t0: arbitrary time

$$\begin{array}{l}ln(\frac{N}{N_{0}})=-\lambda t\end{array}$$
$$\begin{array}{l}N(t)=N_{0}e^{-\lambda t} (eq.4)\end{array}$$

Therefore, eq.4 is the law of radioactive decay.