Let `N_(beta)` be the number of `beta` particles emitted by `1` gram of `Na^(24)` radioactive nuclei (half life `= 15` hrs) in `7.5` hours, `N_(beta)` is close to (Avogadro number `= 6.023 xx 10^(23) // "g. mole"`) :-

Options

A. `6. 2 xx 10^(21)`
B. `7.5 xx 10^(21)`
C. `1.25 xx 10^(22)`
D. `1. 75 xx 10^(22)`

Answer

B

Explanation

We know that `N_(beta) = N_(0) (1 - e^(-lambda t))`
`N_(beta) = (6.023 xx 10^(23))/(24) [1 - e (ln (2)/(15) xx 7.5)]`
on solving we get ,
`N_(beta) = 7.4 xx 10^(21)`