Question

The activity of a radioactive sample is measured as No counts per minute at t=0 and No/e counts per minute at t=5 minutes. The time (in minutes) at which the activity reduces to half its value is

1-5loge2 2-loge2/5

3-5/loge2 4-5log10(5)

Answer 1

..
first option is correct !!

Answer 2

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

Explanation:

Activity of the radioactive sample after n half lives is,

$\bold{\frac{N}{N_{0}}=\left(\frac{1}{2}\right)^{n}=\left(\frac{1}{2}\right)^{t / T}}$

From question,

$\bold{N=\frac{N_{0}}{e}}$

On substituting above equation in half life equation, we get,

$\bold{\frac{N_{0}}{e N_{0}}=\left(\frac{1}{2}\right)^{5 / T}}$

$\bold{\Rightarrow \frac{1}{e}=\left(\frac{1}{2}\right)^{5 / T}}$

On taking log on both side, we get,

$\bold{\log 1-\log {e}=\frac{5}{T} \log \frac{1}{2}}$

$\bold{-1=\frac{5}{T}(-\log 2)}$

$\bold{T=\log _{e} 2}$

Now, time after the half life is:

$\bold{\left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)^{t / 5\log _{e} 2} }$

$\bold{\therefore t= 5 \log _{e} 2}$