Plutinium has atomic mass `210` and a decay constant equal to `5.8 xx 10^(-8)s^(-1)`. The number of `alpha`-particles emitted per second by `1 mg `plutonium is
(Avagadro's constant =` 6.0 xx 10^(23)`).

Options

A. `1.7 xx 10^9`
B. `1.7 xx 10^11`
C. `2.9 xx 10^11`
D. `3.4 xx 10^9`

Answer

B

Explanation

(b) Number of `alpha-`particles per second = activity `A`
`=(-dN//dt) = N lamda` is required, where
`N = (6.0 xx 10^23)/(210) xx 1 xx 10^-3`
And `lamda = 5.8 xx 10^-8 per sec`
So `A = N lamda`
`=(6.0 xx 10^23)/(210) xx 1 xx 10^-3 xx 5.8 xx 10^-8`
`=1.7 xx 10^11`.