Starting with a sample of pure `.^66 Cu, 7//8` of it decays into `Zn` in `15 min`. The corresponding half-life is.

Options

A. `10` min
B. `15` min
C. `5` min
D. `7(1)/(2)` min

Answer

C

Explanation

We know that, `(N)/(N_(0)) = ((1)/(2))^(n)` where, N are the radioactive nuclei left after n-half lives `N_(0)` are the initial number of nuclei and n is number of half-lives,
`(N_(0)//8)/(N_(0)) = (1)/(8) = ((1)/(2))^(3) = ((1)/(2))^(n) rArr n = 3`
Therefore `T_(1//2) = (15)/(3) = 5` min