Question

A stable C can be obtained through unstable nuclides A nad B. the half life period for conversion $A\to C$ is $T$ and that for conversion $B\to C$ is $2T$. Initially a sample contains N nuclides of A, N nuclides of B and no nuclides of C. If the no. of nuclides of species C is equal to (27/16) N after time $t=nT$, then $n$ is

Answer 1

$A\stackrel{T}{\to }C$
$B\stackrel{2T}{\to }C$
$N_A=N{e}^{-\lambda t}$ : No. of decayed
$N_A^{\prime }=N-N_A=N(1-e^{-{\lambda }_{1}t})$
$N_B^{\prime }=N(1-e^{-{\lambda }_{2}t})$
$\therefore N_A^{\prime }+N_B^{\prime }=N[2-(e^{-{\lambda }_{1}t}+e^{-{\lambda }_{2}t}\left)\right]$
$\frac{27}{16}N=N[2-(e^{-{\lambda }_{1}t}+e^{-{\lambda }_{2}t}\left)\right]$
$\frac{5}{16}=e^{\frac{-\ln 2t}{T}}+e^{\frac{-\ln 2t}{T}}=$----(1)
If = y then $e^{\frac{-\ln 2t}{T}}=y^2$----(2)
From (1) and (2)
$\frac{5}{16}=y^2+y$ ; Solving $y=\frac{1}{4}$
$e^{\frac{-\ln 2t}{2T}}=\frac{1}{4}\Rightarrow \frac{-t\ln 2}{2T}=-2\ln 2\Rightarrow t=4T$