Starting with a sample of pure `.^66Cu`, `3/4` of it decays into `Zn` in `15` minutes. The corresponding half-life is

Options

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

Answer

C

Explanation

(c ) `N = N_(0) (1 - e^(-lambda t))`
`implies (N_(0) - N)/(N_(0)) = e^(lambda t)`
`:. (1)/(8) = e^(- lambda t) implies 8 e^(lambda t) implies 3 1n 2 = lambda t implies lambda = (3 xx 0.693)/(15)`
Half-life period, `t_(1//2) = (0.693)/(3 xx 0.693) xx 15`
`t_(1//2) = 5 `min