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

`tau_(1//2)=15` hours
`lambda=(ln(2))/tau_(1//2)=(ln(2))/15`
24 g of Na has `A_(0)` nuclei
1 g of Na has `A_(0)/24` nuclei `(therefore A_(0)=6.023 xx 10^(23))`
So, `N_(0)=A_(0)/24`
Let N be the remaining amount of nuclei at time t=7.5 hours.
`N=N_(0)e^(-lambdat) = N_(0)-N_(0)/sqrt(2)=N_(0)(1-1/sqrt(2))`
`N=N_(0)/sqrt(2)`
Number of `beta`-particle disintegrated at time t
`N_(beta)=N_(0)-N=N_(0)-N_(0)/sqrt(2)=N_(0)(1-1/sqrt(2))`
`A_(0)/24(1-1/sqrt(2))=(6.023 xx 10^(23))/24 xx 0.293`
`=0.0735 xx 10^(23) = 7.35 xx 10^(21) ~~ 7.5 xx 10^(21)`