Q. In artificial radioactivity, $1.414\ \times \ 10^{6}$ nuclei are disintegrated into $10^{6}\ $nuclei in 10 min. The half-life in minutes must be

A) 5
B) 20
C) 15
D) 30
Explanation

(a) From Rutherford-Soddy law
$N = N_{0}\left( \frac{1}{2} \right)^{n}$
$\ \ \ \ \ \ \ \ \ \ \ n = \frac{t}{T}$
$\therefore\ 10^{6}\ = \ 1.414\ \times \ 10^{6}\left( \frac{1}{2} \right)^{t/T}$
$\Rightarrow \ \ \ \ \ \frac{1000}{1414} = \left( \frac{1}{2} \right)^{t/T}$
$\Rightarrow \ \left( \frac{1}{2} \right)^{2} = \left( \frac{10}{12} \right)^{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (Approximately)$
$\Rightarrow \ \ \ \ \ \ \ n = 2$
$\Rightarrow \ \ \ \ \ \ \ \ n = \frac{t}{T}$ =2
$\Rightarrow \ \ \ \ \ \ \ T = \ \frac{10}{2}\ = \ 5\ \min$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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