JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Chemical Kinetics

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

)

Answer: 102

Solution

Radioactive decay is first order,

t = \frac{2.303}{λ} lo g \frac{a _{0}}{a _{t}}

with

λ = \frac{0.693}{t _{1/2}}

. With

75%

of the activity left,

\frac{a _{0}}{a _{t}} = \frac{100}{75} = \frac{4}{3}

, so

t = \frac{t _{1/2}}{l o g 2} lo g \frac{4}{3}

. Here

lo g \frac{4}{3} = 2 lo g 2 - lo g 3 = 2 (0.3010) - 0.4771 = 0.1249

, giving

t = 245 \times \frac{0.1249}{0.3010} = 101.7 \approx 102 days

JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Integer answer