the count rate from a radioactive sample falls from `4.0xx10^6`per second to `1.0xx10^6` per second in 20 hours.What will be the count rate 100 hours after the begnning ?

Answer

A, C, D

Explanation

Given `A_0 = 4 xx 10^(6) disintegration//sec`
`A' = 1 xx 10^(6) disintegrations//sec`
`t = 20 hrs.`
`A' = A_0/2^(t//t_(1//2)) rArr 2^(t//t_(1//2)) = A_0/A'`
`rArr 2^(t//t_(1//2)) = 4 `
`rArr t//t^(1//2) = 2`
`t^(1/2) = t//2 = 20 hrs. //2 = 10 hrs.`
`A" = A_(0)/2^(t//t_(1/2))`
`A" = (4 xx 10^(6))/(2^(100//10))`
` = 0.00330625 xx 10^(4)`
`3.9xx 10^(3) disintegrations //sec.`