Question

The activity of a radioactive sample is measured as N₀ counts per minute at t = 0 and N₀/e counts per minute at t = 5 minutes. The time (in minutes) at which the activity reduces to half its value is

• A. $5\mathrm{lo}{g}_{e}2$
• B. ${\log }_{e}2/5$
• C. $5/{\log }_{e}2$
• D. $5{\log }_{10}2$

Answer 1

The correct option is A

$5\mathrm{lo}{g}_{e}2$

Explanation for the correct option:

Option(A):

Step 1: formula applied:

We know, $N=\frac{N_o}{c}$

Therefore, the equation of nuclei left undecayed after time, t is given by,

$N=N_o{\left(\frac{1}{2}\right)}^{t/t_{1/2}}$

Step 2: calculation using the formula:

Here: t = $5min$

Hence, from the formula $N=\frac{N_o}{c}$

$\frac{N_o}{c}=N_o{\left(\frac{1}{2}\right)}^{t/t_{1/2}}{e}^{-1}=(\frac{1}{2}{)}^{t/t_{1/2}}$

Adding log on both the sides, we get:

$-\ln \left(e\right)=\frac{5}{t_{1/2}}\ln {\frac{1}{2}}^{}$

Step 4: Calculating $t_{1/2}$

$$\begin{gathered} t_{1/2}=-5\ln (1/2) \\ {t}_{1/2}=-5(-\ln 2) \\ {t}_{1/2}=5\ln 2 \\ {t}_{1/2}=5{\log }_{e}2 \end{gathered}$$

Hence The time (in minutes) at which the activity reduces to half its value is $5{\log }_{e}2$