Question

The decay constant for an $\alpha$ -decay of $T{h}^{232}$ is $1.58\times {10}^{10}{s}^{-1}$. The number of $\alpha$-decays that occur from 1 g sample in 365 days are :

• A. $1.276\times {10}^{19}\alpha -atoms$
• B. $1.225\times {10}^{19}\alpha -atoms$
• C. $1.298\times {10}^{19}\alpha -atoms$
• D. none of these

Answer 1

The correct option is C $1.298\times {10}^{19}\alpha -atoms$

$t=2.303log\frac{N_0}{N_t}$
$365\times 24\times 60\times 60=\frac{2.303}{1.50\times {10}^{-10}}log\frac{1}{N}$
N = 0.955 g

Amount of $T{h}^{232}$ undergoing decay $=N_0-N$$=1-0.995=0.005g$
$\therefore$ 0.005 g Th on decay produces$=\frac{6.023\times {10}^{23}\times 0.005}{232}\alpha -atoms$$=1.298\times {10}^{19}\alpha -atoms$