Q. What is the half-life of Na-24 if$\ \ 2 \times 10^{- 4}\ g$ sample of it disintegrate at the rate of $7.0 \times 10^{12}$ atoms per s?
A) $4.97 \times 10^{5}\ s$
B) $4.97 \times 10^{4}\ s$
C) $4.97 \times 10^{6}\ s$
D) $4.97 \times 10^{2}\ s$
Explanation
(a)
$24\ g\ of\ Na\ has\ 6.02 \times 10^{23}$ atoms
$2 \times 10^{- 4}\$g of Na will have$\ = \frac{6.02 \times 10^{23}}{24} \times 2 \times 10^{- 4}\$atoms
$= 0.502 \times 10^{19}\ atoms$
$k = \frac{disintegration\ \left( in\ atom\ s^{- 1} \right)}{no.of\ atoms\ present}$
$= \frac{7.0 \times 10^{12}}{0.502 \times 10^{19}}$
$= 13.94 \times 10^{- 7}\ s^{- 1}$
Therefore, $t_{1/2} = \frac{0.693}{13.94\ \times 10^{- 7}} = 4.97 \times 10^{5}\ s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.