Answer:
201.8 days
Explanation:
The activity of a radioactive sample as a function of times is :
[tex]$R=R_0e^{\frac{0.693t}{T_{1/2}}}$[/tex]
Here, [tex]$R_0$[/tex] = the initial activity
[tex]$T_{1/2}$[/tex] = half life
t = elapsed time
Now rearranging the equation for time, t, we get:
[tex]$\frac{R}{R_0}=e^{-\frac{0.693t}{T_{1/2}}}$[/tex]
[tex]$\ln\left(\frac{R}{R_0}\right)=-\frac{0.693t}{T_{1/2}}$[/tex]
[tex]$t=\frac{-\ln\left(\frac{R}{R_0}\right)T_{1/2}}{0.693}$[/tex]
[tex]$t=\frac{-\ln\left(\frac{25}{350}\right)\times 53}{0.693}$[/tex]
= 201.8 days
Therefore, the required time is 201.8 days
