JEE Main 2026 — Jan 24, Shift 2 — Chemistry solution

JEE Main2026Jan 24, Shift 2Chemistry

The half-life of $^{65} Zn$ is 245 days. After x days, 75% of original activity remained. The value of x in days is ________. (Nearest integer)

(Given : $lo g 3 = 0.4771$ and $lo g 2 = 0.3010$ )

Your answer

Solution

Radioactive decay is first order: $t = \frac{2.303}{k} lo g \frac{a _{0}}{a _{t}}$ , with $k = \frac{ln 2}{t _{1/2}} = \frac{ln 2}{245}$ .

For 75% remaining, $\frac{a _{0}}{a _{t}} = \frac{4}{3}$ .

$t = \frac{245}{ln 2} \cdot ln \frac{4}{3} = 245 \cdot \frac{lo g ( 4/3 )}{lo g 2} = 245 \cdot \frac{2 lo g 2 - lo g 3}{lo g 2}$

$= 245 \cdot \frac{2 ( 0.3010 ) - 0.4771}{0.3010} = 245 \cdot \frac{0.1249}{0.3010} \approx 101.66$ days.

Nearest integer: $102$ .

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