A uranium nucleus `._(92)U^(238)` emits and `alpha`-particle and a `beta`-particle in succession. The atomic number and mass number of the final nucleus will be

Options

A. 90 and 233
B. 90 and 238
C. 91 and 238
D. 91 and 234

Answer

D

Explanation

Let `N_(0)` be number of 1g of `Na^(24)` nuclei at t=0
`=(N_(A))/("Atomic weight of Na")=(6.03xx10^(23))/(24)`
The decay constant is
`lamda=(0.693)/(T_(1//2))=(0.639)/(15)=("ln "2)/(15)`
So, the number of `beta`-particles emitted by 1 g of `Na^(24)` nuclei in time t=7.5 h is
`N_(beta)=(6.023xx10^(23))/(24)[1-e^(lamdat)]`
`=(6.023xx10^(23))/(24)[1-e^(((-ln2)/(15)xx7.5)))]`
`=(6.023xx10^(23))/(24)[1-e^((-ln2//2))]`
`implies = ((1-e^(-0.3465)xx6.023xx10^(23))/(24)`
`=(1-0.707106)xx(6.023xx10^(23))/(24)=0.0735039xx(10^(23))/(24)`
`=7.35xx10^(21)=7.5xx10^(21)`