The ratio of molecular mass of two radioactive substances is `3//2` and the ratio of their decay cosntatnt is `4//3`. Then. The ratio of their initial activity per mole will be

Options

A. 2
B. `4//3`
C. `8//9`
D. `9//8`

Answer

A

Explanation

Activity, `A=lamdaN=(0.693)/(T_(1//2))N`, where `T_(1//2)` is the half-life of a radioactive sample.
`therefore (A_(1))/(A_(2))=(N_(1))/(T_(1))xx(T_(2))/(N_(2))`
`(T_(1))/(T_(2))=(A_(2))/(A_(1))xx(N_(1))/(N_(2))=(2A_(1))/(A_(1))xx(2N_(2))/(N_(2))=(4)/(1)`