Question

If activity of a radioactive substance is R_(0) at time t=0 and (R_(0))/(e^(2)) at t=10 hours then the time (in hours) at which the activity reduces to half ((R_(0))/(2)) of its initial value is​

Answer 1

Given:

radioactive activity R_(0) at time t=0

radioactive activity $R_t = \frac{R_o}{e^2}$ at t=10

To find:

t =?

Explanation:

The emission of a radioactive substance is given by

λt = ln $\frac{R_o}{R_t}$            ....(1)

where

λ = decay constant

t = elapsed time

R0 = original no of radioactive atoms

Rt = no of radioactive atoms

at t = 10  $R_t = \frac{R_o}{e^2}$

Put given values in above eq (1)

λ (10) = ln$\displaystyle (\frac{R_o}{\frac{R_0}{e^2} })$

λ = 0.2

Now, we have to calculate the time  at which the activity reduces to half  of its initial value

i.e. R = $\frac{R_0}{2}$

from eq (1)

0.2 (t) = ln$\displaystyle (\frac{R_o}{\frac{R_0}{2} })$

solving above eq, we get,

t = 5 $\mathbf{ln^2}$