Discussion Forum

The activity of a radioactive sample is measured as $N_{0}$ counts per minute at t=0 and $N_{0}/C$ counts per minute at t=5 min. The time , (in minute) at  which the activity reduces to half  its value is 

Answer:

Explanation:

After  n half -lives

$\frac{N}{N_{0}}=\left(\frac{1}{2}\right)^{n}=\left(\frac{1}{2}\right)^{t/7}$

$N=\frac{N_{0}}{e}=\frac{cN_{0}}{cN_{0}}=\left(\frac{1}{2}\right)^{5/7}$

$\Rightarrow$   $\frac{1}{e}$=$\left(\frac{1}{2}\right)^{5/7}$

Taking log on both sides , we get

$\log 1-\log e= \frac{5}{7} \log \frac{1}{2}$

$-1=\frac{5}{7}(-\log 2)$

T=$5 \log_{e}2$

Now, let  t be the time after which activity reduces to half 

$\left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)^{1'/5 log_{e}2}$

$t'=5 log_{e}2$