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Question
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?
Solution
Given:-
Initial count rate of radioactive sample, A0 = 4 × 106 disintegration/sec
Count rate of radioactive sample after 20 hours, A = 1 × 106 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"`
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