A fraction `f_(1)` of a radioactive sample decays in one mean lie and a fraction `f_(2)` decays in one half-life

Options

A. `f_(1) = f_(2)`
B. `f_(1) = (f_(2))/(2)`
C. `f_(1) lt f_(2)`
D. `f_(1) gt f_(2)`

Answer

C

Explanation

`N = N_(0) e^(- lambda t)`
`t = t_("mean") = (1)/(lambda) " " therefore f_(1) = (N)/(N_(0)) e^(-1) = (1)/(e)`.
`t = t_(("half")) = (ln2)/(lambda) " " f_(2) = (N)/(N_(0)) e^(-ln2) = (1)/(2)`
`therefore f_(2) gt f_(1)`.