Q. The half-life of a sample of a radioactive substance is 1 hour. If $8 \times 10^{10}$ atoms are present at $t = 0$, then the number of atoms decayed in the duration $t = 2$ hour to $t = 4$ hour will be
A) $2 \times 10^{10}$
B) $1.5 \times 10^{10}$
C) Zero
D) Infinity
Explanation
(b) $N = N_{0}\left( \frac{1}{2} \right)^{\frac{t}{T_{1/2\ }}}$
No of atoms at $t = 2hr,N_{1} = 8 \times 10^{10}\left( \frac{1}{2} \right)^{\frac{2}{1}} = 2 \times 10^{10}$
No. of atoms at $t = 4hr,N_{2} = 8 \times 10^{10}\left( \frac{1}{2} \right)^{\frac{4}{1}} = \frac{1}{2} \times 10^{10}$
$\therefore$ No. of atoms decayed in given duration
$= \left( 2 - \frac{1}{2} \right) \times 10^{10} = 1.5 \times 10^{10}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.