Question

The activity of radioactive sample is measured as $9750$ counts per minute at $t=0$ and as $975$ counts per minute at $t=5$ minutes, the decay constant is approximately

• A. $0.922$ per minute
• B. $0.270$ per minute
• C. $0.461$ per minute
• D. $0.39$ per minute

Answer 1

The correct option is C $0.461$ per minute

We know that
$\frac{dN}{dt}=\lambda N$
Now, $\frac{d{N}_0}{dt}=\lambda N_0;\frac{d{N}_t}{dt}=\lambda N_t$

$9750=\lambda N_0;975=\lambda N_t$
$\frac{N_0}{N_t}=\frac{9750}{975}=\frac{10}{1}\Rightarrow N_0=10N_t$
We know that
$N_t=N_0{e}^{-\lambda t}$
$\frac{N_t}{N_0}=e^{-\lambda t}\Rightarrow \frac{1}{10}=e^{-\lambda .5}$
${10}^{-1}=e^{-5\lambda }$
Taking log on both sides,
$-1=-5\lambda \times \frac{1}{2.303}$
$\lambda =\frac{1}{5}\times 2.303=0.461$ per minute.