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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 equalsA. (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)` |
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Answer» 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))` |
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