JEE Chemistry 2023 Question with Solution

Given: Half-life of A is $15 \, \text{min}$, half-life of B is $5 \, \text{min}$, and initial concentration of B is $4$ times that of A.

Find: The time when concentrations of A and B become equal.

The decay of a substance follows:

$$N(t) = N_0 \times \left(\frac{1}{2}\right)^{t/t_{1/2}}$$

Let the initial concentration of A be $N_A$. Then the initial concentration of B is $N_B = 4N_A$.

For substance A:

$$N_A(t) = N_A \times \left(\frac{1}{2}\right)^{t/15}$$

For substance B:

$$N_B(t) = 4N_A \times \left(\frac{1}{2}\right)^{t/5}$$

At the required time, concentrations are equal:

$$N_A \times \left(\frac{1}{2}\right)^{t/15} = 4N_A \times \left(\frac{1}{2}\right)^{t/5}$$

Cancel $N_A$ from both sides:

$$\left(\frac{1}{2}\right)^{t/15} = 4 \times \left(\frac{1}{2}\right)^{t/5}$$

Rewrite $4$ as $2^2$:

$$\left(\frac{1}{2}\right)^{t/15} = \left(\frac{1}{2}\right)^{t/5 - 2}$$

Equating the exponents:

$$\frac{t}{15} = \frac{t}{5} - 2$$ $$\frac{t}{15} - \frac{t}{5} = -2$$

Multiply by $15$:

$$t - 3t = -30$$ $$-2t = -30$$ $$t = 15$$

Therefore, the time required for the concentrations to become equal is $15 \, \text{min}$.

Given: Radioactive decay of both substances starts together. Half-lives are $15 \, \text{min}$ for A and $5 \, \text{min}$ for B. Initially, B has concentration $4x$ if A has concentration $x$.

Find: Time at which both concentrations become the same.

Using the exponential decay form:

$$[A]_t = [A]_0 e^{-kt}$$

For A:

$$[A]_t = x e^{-k_A t}, \qquad k_A = \frac{\ln 2}{15}$$

So,

$$[A]_t = x e^{-\left(\frac{\ln 2}{15}\right)t}$$

For B:

$$[B]_t = 4x e^{-k_B t}, \qquad k_B = \frac{\ln 2}{5}$$

So,

$$[B]_t = 4x e^{-\left(\frac{\ln 2}{5}\right)t}$$

At equal concentrations:

$$x e^{-\left(\frac{\ln 2}{15}\right)t} = 4x e^{-\left(\frac{\ln 2}{5}\right)t}$$

Cancel $x$:

$$e^{-\left(\frac{\ln 2}{15}\right)t} = 4 e^{-\left(\frac{\ln 2}{5}\right)t}$$ $$e^{t\left(\frac{\ln 2}{5} - \frac{\ln 2}{15}\right)} = 4$$

Taking natural logarithm:

$$t\left(\frac{\ln 2}{5} - \frac{\ln 2}{15}\right) = \ln 4$$ $$t \ln 2 \left(\frac{1}{5} - \frac{1}{15}\right) = 2 \ln 2$$ $$t \left(\frac{2\ln 2}{15}\right) = 2 \ln 2$$ $$t = 15 \, \text{min}$$

Therefore, the correct answer is $15$.