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JEE Main 202624 JanuaryEvening Shift

Exam: JEE Main
Year: 2026
Date: 24 January
Shift: Evening Shift
Subject: Chemistry
Chapter: Chemical Kinetics

JEE Main 2026 · 24 January · Evening Shift · Chemistry · Chemical Kinetics

The half-life of ^ 65 Zn is 245 days. After x days, 75 \% of...

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

)

Correct answer

0

Step-by-step 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).

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