A radioactive sample `S_(1)` having the activity `A_(1)` has twice the number of nuclei as another sample `S_(2)` of activity `A_(2)` If `A_(2)=2A_(1)`, then the ratio of half life of `S_(1)` to the life of `S_(2)` is

Options

A. 4
B. 2
C. `0.25`
D. 0.75`

Answer

A

Explanation

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