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

Number of `alpha`particles per second =Activity
`=(-dN)/(dt)=N lambda`
Where
`N=6.0 xx(10^(23))/(210) xx 1 xx 10^(-3)`
`lambda =5.8 xx 10^(-8) s^(-1)`
So,
`A=N lambda `
`=6.0 xx(10^(23))/(210) xx 1 xx 10^(-3) xx 5.8 xx 10^(-8)`
`=1.7 xx 10^(11)` .