Question

A radioactive material of half-life T was kept in a nuclear reactor at two different instants. The quantity kept second time was twice of the kept first time. If now their present activities are `A_1` and `A_2` respectively, then their age difference equals
A. (a) `(T)/(1n2)1n(2A_1)/(A_2)`
B. (b) `T1n(A_1)/(A_2)`
C. (c) `(T)/(1n2)1n(A_2)/(2A_1)`
D. (d) `T1n(A_2)/(2A_1)`

Answer 1

Correct Answer - C
Activity `Aprop` Number of atoms`
`A_1=A_0e^(-lambdat_1)`
`:.` `t_1=1/lambda1n((A_0)/(A_1))=(T)/(1n2)1n(A_0)/(A_1)`
`A_2=2A_0e^(-lambda_(t_2))`
`t_2=(T)/(1n2)1n((2A_0)/(A_2))`
`t_1-t_2=(T)/(1n2)(A_0/A_1xx(A_2)/(2A_0))`
`=(T)/(1n2)1n((A_2)/(2A_1))`