`N` atoms of a radioactive element emit `n` alpha particles per second. The half-life of tge element is.

Options

A. `N/n sec`
B. `(0.693N)/(n) sec`
C. `(0.693n)/(n) sec`
D. `n/N sec`

Answer

B

Explanation

Generally we wire `(dN)/(dt) = -lamdaN`
But it is given that `(dN)/(dt) = n therefore n = - lamdaN therefore lamda = - (n)/(N)`
`therefore` Half life `= (0.693)/(lamda) = (0.693N)/(n)`
(-ve sign in to show the decay) .