QJEEAdv08Phb9e9: JEE Advanced 2008 Physics - Class 12 JEEAdv08Phb9e9 Solution: A radioactive sample S 1 having activity of 5 Ci has twice the…

To solve this problem, let's begin with some fundamental concepts of radioactive decay.

The activity of a radioactive sample, which is its decay rate, is given by the equation:

$A = \lambda N$

where:

• $A$ is the activity.
• $\lambda$ is the decay constant.
• $N$ is the number of radioactive nuclei present.

Given that sample S₁ has an activity of 5 $\mu$ Ci and sample S₂ has an activity of 10 $\mu$ Ci, and that sample S₁ has twice the number of nuclei as sample S₂, we have:

$A_1 = \lambda_1 N_1$

$A_2 = \lambda_2 N_2$

We are also given that:

$N_1 = 2N_2$

Substitute $N_1 = 2N_2$ into the first equation:

$5 \, \mu\text{Ci} = \lambda_1 (2N_2)$

And for the second equation:

$10 \, \mu\text{Ci} = \lambda_2 N_2$

Dividing the first equation by the second equation to eliminate $N_2$ , we get:

$\frac{5 \, \mu\text{Ci}}{10 \, \mu\text{Ci}} = \frac{\lambda_1 (2N_2)}{\lambda_2 N_2}$

Simplifying this, we obtain:

$\frac{1}{2} = \frac{2 \lambda_1}{\lambda_2}$

Therefore:

$\lambda_2 = 4 \lambda_1$

Next, we can relate the decay constant to the half-life using the equation:

$\lambda = \frac{\ln(2)}{T_{1/2}}$

Using the relationship between the decay constants, we have:

$\frac{\lambda_2}{\lambda_1} = \frac{T_{1/2,1}}{T_{1/2,2}}$

Substituting $\lambda_2 = 4 \lambda_1$ , we get:

$4 = \frac{T_{1/2,1}}{T_{1/2,2}}$

Therefore:

$T_{1/2,1} = 4 T_{1/2,2}$

Given the options, the half-lives that satisfy this relationship are:

Option A: 20 years and 5 years, respectively.

In this case:

$\frac{20 \text{ years}}{5 \text{ years}} = 4$

Thus, option A is correct. Therefore, the half-lives of S₁ and S₂ are 20 years and 5 years, respectively.