Uranium `._(92)^(238)U` is an ustable nucleus. It decays to Thorium `._(92)^(238)Th`, which is again an unstable nucleus which further decays to `._(91)^(234)Pa`. Let `._(92)^(238)U` be called `A` of decay constant `lambda_(1)` and `._(90)^(234)Th` is called as `B` of decay constant `lambda_(2)` and stable nuclei `._(91)^(234)Pa` be called as `C`. Here `A` is called parent nucleus and `B` is called daughter nucleus of `A`. Any two adjacent nuclei may be consider parent or daughter nuclei `A, B` and `C` respectively at time `'t'`.
Then we can write `Aoverset(lambda_(1))rarrBoverset(lambda_(2))rarrC`
Rate of disintergration of `A=(dN_(1))/(dt)=lambda_(1)N_(1)`
Rate of disintergration of `B=(dN_(2))/(dt)=lambda_(1)N_(1)-lambda_(2)N_(2)`
Rate of formation of nuclei C is equal to
`(dN_(3))/(dt)=lambda_(2)N_(2)`
If at `t=0`, there are `N_(0)` number of nuclei of `A` where as nuclei `B` and `C` are absent in the sample Answer the following questions ltbgt Numeber of the nuclei of nuclei C at time `t` is

Options

A. `N_(0)[1+(lambda_(1))/(lambda_(2)-lambda_(1))e^(-lambda_(2)t)-(lambda_(2))/(lambda_(2)-lambda_(1))e^(-lambda_(1)t)]`
B. `N_(0)[(lambda_(1))/(lambda_(2)-lambda_(1))e^(-lambda_(2)t)-(lambda_(2))/(lambda_(2)-lambda_(1))e^(-lambda_(1)t)]`
C. `N_(0)(lambda_(1))/(lambda_(2)-lambda_(1))(e^(-lambda_(2)t)-e^(-lambda_(1)t))`
D. `N_(0)[(lambda_(1))/(lambda_(2)-lambda_(1))e^(-lambda_(2)t)-(lambda_(2))/(lambda_(2)-lambda_(1))e^(-lambda_(1)t)-1]`

Answer

A

Explanation

Successive disintegration