A radio nuclide consists of two isotopes. One of the isotopes decays by `alpha`-emission and other by `beta`-emission with half-lives `T_1=405s` and `T_2=1620s`, respectively. At `t=0`, probabilities of getting `alpha` and `beta`-particles from the radio nuclide are equal . Calculate their respective probabilities at `t=1620s`. If at `t=0`, total number of nuclei in the radio nuclide are `N_0`. Calculate the time t when total number of nuclei remained undecayed becomes equal to `N_0//2`.
`log_(10)2=0.3010`, `log_(10)5.94=0.7742` and `x^4+4x-2.5=0, `x=0.594`

Answer

A, B

Explanation

(i) At `t=0`, probabilities of getting `alpha` and `beta` particles are same. This implies that initial activity of both is equal. Say it is `R_0`.
Activity after `t=1620s`,
`R_1=R_0(1/2)^(1620//405)=R_0/16`
and `R_2=R_0(1/2)^(1620//1620)=R_0/2`
Total activity, `R=R_1+R_2=9/16R_0`
Probability of getting `alpha`-particles,
`=R_1/R=1/9`
and probability of getting `beta`-particles
`=R_2/R=8/9`
(ii) `R_(01)=R_(02)`
`:.` `(N_(01))/(T_1)=(N_(02))/(T_2)`
`:.` `(N_(01))/(N_(02))=1/4`
Let `N_0` be the total number of nuclei at `t=0`.
Then, `N_(01)=(N_0)/(5)` and `N_(02)=(4N_0)/(5)`
Given that `N_1+N_2=(N_0)/(2)`
or `(N_0)/(5)(1/2)^(t//405)+(4N_0)/(5)(1/2)^(t//1620)=(N_0)/(2)`
Let `(1/2)^(t//1620)=x`
Then, above equation becomes `x^4+4x-2.5=0`
`:.` `x=0.594` or `(1/2)^(t//1620)=0.594`
Solving, it we get `t=1215s`.