Q22 / 75 · JEE Main 2026 (24 January Shift 2) · Chemistry
The half-life of ${ }^{65} \mathrm{Zn}$ is 245 days. After $x$ days, $75 \%$ of original activity remained. The value of $x$ in days is $\_\_\_\_$. (Nearest integer)
(Given: $\log 3=0.4771$ and $\log 2=0.3010$)
(Given: $\log 3=0.4771$ and $\log 2=0.3010$)
Correct answer: 102
Solution
Using radioactive decay: $A = A_0 \left(\frac{1}{2}\right)^{t/t_{1/2}}$.
With 75% activity remaining: $0.75 = \left(\frac{1}{2}\right)^{x/245}$.
Taking log base 10: $\log(0.75) = \frac{x}{245}\log(0.5)$.
Calculate: $\log(0.75) = \log 3 - \log 4 = 0.4771 - 2(0.3010) = -0.1249$ and $\log(0.5) = -0.3010$.
Thus $-0.1249 = \frac{x}{245} \times (-0.3010)$, giving $x = \frac{0.1249 \times 245}{0.3010} = 101.66 \approx 102$ days (nearest integer).
With 75% activity remaining: $0.75 = \left(\frac{1}{2}\right)^{x/245}$.
Taking log base 10: $\log(0.75) = \frac{x}{245}\log(0.5)$.
Calculate: $\log(0.75) = \log 3 - \log 4 = 0.4771 - 2(0.3010) = -0.1249$ and $\log(0.5) = -0.3010$.
Thus $-0.1249 = \frac{x}{245} \times (-0.3010)$, giving $x = \frac{0.1249 \times 245}{0.3010} = 101.66 \approx 102$ days (nearest integer).
Hence, the correct answer is 102.