CGPET 2021 Solved Paper

Given, initial number of nuclei $= N_0$ After time t, remaining number of nuclei, $N = N_0 e^{-\lambda t} \text{ (where } \lambda = \text{ decay constant)}$ Given, at time, $t = 0$ For radioactive substance A, $N_A = N_0 e^{-5\lambda t}$ Similarly, for B, $N_B = N_0 e^{-\lambda t}$ ∴ Ratio $N_A/N_B = \frac{N_0 e^{-5\lambda t}}{N_0 e^{-\lambda t}} = \frac{e^{-5\lambda t}}{e^{-\lambda t}}$ ⇒ $N_A/N_B = e^{-4\lambda t}$ .....(i) It is given that after time t, the ratio, $N_A/N_B = \left( \frac{1}{2} \right)^2$......(ii) From Eqs. (i) and (ii), we get ⇒ $\left( \frac{I}{e} \right)^2 = e^{-4\lambda t} \Rightarrow e^{-2} = e^{-\lambda}$ Comparing the powers, we get $-2 = -4\lambda t \Rightarrow t = \frac{I}{2\lambda}$ Hence, the ratio of number of nuclei of A to those of B will be $\left( \frac{I}{e} \right)^2$ after a time of $\frac{I}{2\lambda}$ ∴ No option in the given question is correct