JEE Physics 2023 Question with Solution

Given: Two radioactive elements A and B initially have the same number of atoms. The half-life of A is the same as the average life of B.

Find: The correct relation between $\lambda_A$ and $\lambda_B$.

For radioactive decay,

$$T = \frac{\ln 2}{\lambda}$$

and

$$\tau = \frac{1}{\lambda}$$

where $T$ is the half-life and $\tau$ is the average life.

Given that the half-life of A equals the average life of B,

$$T_A = \tau_B$$

So,

$$\frac{\ln 2}{\lambda_A} = \frac{1}{\lambda_B}$$

Rearranging,

$$\lambda_A = \lambda_B \ln 2$$

Therefore, the correct option is D.

Given: $T_A = \tau_B$

Find: Relation between $\lambda_A$ and $\lambda_B$.

Use the standard relations:

$$T_A = \frac{\ln 2}{\lambda_A}$$ $$\tau_B = \frac{1}{\lambda_B}$$

Now equate them:

$$\frac{\ln 2}{\lambda_A} = \frac{1}{\lambda_B}$$

Cross-multiplying,

$$\lambda_B \ln 2 = \lambda_A$$

Hence,

$$\lambda_A = \lambda_B \ln 2$$

So the correct relation is $\lambda_A = \lambda_B \ln 2$.