Question

The half life of ${}_{92}{U}^{238}$ against $\alpha$ -decay is $4.5\times 10^9$ years. How many disintegrations occur per second in one gram of ${}_{92}{U}^{238}$ ?

• A. $2.34\times 10^4{s}^{-1}$
• B. $2.34\times 10^{-4}{s}^{-1}$
• C. $3.44\times 10^4{s}^{-1}$
• D. $3.44\times 10^{-4}{s}^{-1}$

Answer 1

Answer: (A)

$T=4.5\times 10^9$ years
$T=4.5\times 10^9\times 365\times 24\times 60\times 60$
$\lambda =\frac{0.693}{T}{s}^{-1}$
Number of atoms in one gram of ${}_{92}{U}^{238}$ is
$N=\frac{6.03\times 10^{23}}{238}$
$\frac{dN}{dt}=\lambda N$
$\Rightarrow \frac{dN}{dt}=\frac{0.693\times 6.03\times 10^{23}}{238\times 4.5\times 10^9\times 365\times 24\times 60\times 60}$
$\frac{dN}{dt}=2.34\times 10^4{s}^{-1}$