Question

The activity of a radioactive sample is measured as $N_0$ count per minute at $t=0$ and $N_0/e$ counts per minute at $t=5min$. The time (in minutes) at which the activity reduces to half its value is

• A. ${\log }_e\frac{2}{5}$
• B. $\frac{5}{{\log }_e}2$
• C. $5{\log }_{{10}^{}}2$
• D. $5{\log }_{e}2$

Answer 1

The correct option is D $5{\log }_{e}2$

Radioactive material after time $t$ is

$A=A_0(\frac{1}{2}{)}^{t/T}$. . . . .(1)

After time $t=5min$, $A=\frac{N_0}{e}$ and $A=A_0$

$\frac{N_0}{e}=N_0(\frac{1}{2}{)}^{5min/T}$

$\frac{1}{e}=(\frac{1}{2}{)}^{5/T}$

Take log on both the sides,

$-loge=\frac{5}{T}log\left(1/2\right)$

$T=5lo{g}_{e}2$. . . . . . .(2)

Now, $N=\frac{N_0}{2}=N_0(\frac{1}{2}{)}^{t/T}$

$\left(\frac{1}{2}\right)=(\frac{1}{2}{)}^{t/T}$

$\frac{t}{T}=1$

From (2), $t=T=5lo{g}_{e}2$

The correct option is D.