#### Question

The count rate from a radioactive sample falls from 4.0 × 10^{6} per second to 1.0 × 10^{6}per second in 20 hours. What will be the count rate 100 hours after the beginning?

#### Solution

Given:-

Initial count rate of radioactive sample, A_{0} = 4 × 10^{6 }disintegration/sec

Count rate of radioactive sample after 20 hours, A = 1 × 10^{6 }disintegration/sec

Time, t = 20 hours

Activity of radioactive sample (A) is given by

`A = A_0/2^(t/(T"/"2)`

Here , `T_"1/2"` = Half - life period

On substituting the values of `A_0` and `A` , we have

`2^(t/(T"/"2) = 2^2`

`⇒ t/(T"/"2) = 2`

`⇒ T"/"2 = t"/"2 = 20 "h""/"2 = 10 "h"`

100 hours after the beginning,

Count rate , A" `= A_0/2^(t/(T/2)`

⇒ A" `= (4 xx 10^6)/(2^(100/10))`

`= 0.390625 xx 10^4`

`= 3.9 xx 10^3 " disintegrations/sec"`

Is there an error in this question or solution?

Solution The Count Rate from a Radioactive Sample Falls from 4.0 × 106 per Second to 1.0 × 106per Second in 20 Hours. What Will Be the Count Rate 100 Hours After the Beginning? Concept: Radioactivity - Introduction of Radioactivity.