Question

A radioactive nucleus $X$ decays to a nucleus $Y$ with a decay constant ${\lambda }_x=0.1se{c}^{-1}$. $Y$ further decays to a stable nucleus $Z$ with a decay constant ${\lambda }_Y=1/30se{c}^{-1}$. Initially, there are only $X$ nuclei and their number is $N_0={10}^{20}$. Set up the rate equations for the populations of $X,Y$ and $Z$. The population of the $Y$ nucleus as a function of time is given by $N_Y\left(t\right)=\left\{N_0{\lambda }_X/\right({\lambda }_X-{\lambda }_Y\left)\right\}\left\{exp\right(-{\lambda }_{Y}t)-exp(-{\lambda }_{X}t\left)\right\}$. Find the time at which $N_Y$ is maximum and determine the population $X$ and $Z$ at that instant.

• A. $26.48,N_X=1.92\times {10}^{19},N_Z=2.302\times {10}^{19}$.
• B. $16.48,N_X=2.92\times {10}^{19},N_Z=2.302\times {10}^{19}$.
• C. $16.48,N_X=1.92\times {10}^{19},N_Z=3.302\times {10}^{19}$.
• D. $16.48,N_X=1.92\times {10}^{19},N_Z=2.302\times {10}^{19}$.

Answer 1

Answer: (D)

$16.48,N_X=1.92\times {10}^{19},N_Z=2.302\times {10}^{19}$.

$\begin{array}{c}{\lambda }_x=0.1=\frac{1}{10}{s}^{-1}\\ {\lambda }_y=\frac{1}{30}{s}^{-1}\\ N_0={10}^{20}\end{array}$

$N_y\left(t\right)=\left\{\frac{N_0{\lambda }_x}{{\lambda }_x-{\lambda }_y}\right\}\{e^{-{\lambda }_{y}t}-e^{-{\lambda }_{x}t}\}⟶\left(a\right)$

$\begin{array}{c}\text{For}{N}_y\text{maximum}\\ \begin{array}{c}\frac{dNy}{dt}=0\\ \frac{d{N}_y\left(t\right)}{dt}=\left\{\frac{N_0{\lambda }_n}{{\lambda }_n-{\lambda }_y}\right\}\end{array}\{-e^{-\lambda yt}\times {\lambda }_y+e^{-\lambda xt}\times {\lambda }_x\}=0\end{array}$

$\begin{array}{cc}\Rightarrow & {\lambda }_x{e}^{-\lambda xt}={\lambda }_y{e}^{-\lambda yt}\\ \Rightarrow \frac{e^{\frac{t}{10}}}{10}& =\frac{e^{-\frac{t}{30}}}{30}\\ 3e^{\frac{t}{30}}& =e^{t}10\\ \Rightarrow & 3=e^{t(\frac{1}{10}-\frac{1}{30})}\end{array}$

$\begin{array}{cc}\Rightarrow 3& =e^{\left(\frac{2t}{3}\right)}\\ 3& =e^{\left(\frac{t}{15}\right)}\\ \text{Taking}& \log \\ \ln 3& =\frac{t}{15}\\ t& =15\times \ln 3\\ & =16.489\end{array}$

$\text{Putting}t=16.48\text{sec in}e{q}^n\text{a}$

$\Rightarrow Ny=5.778\times {10}^{19}$

$\begin{array}{c}X\stackrel{A_x}{⟶}Y\stackrel{\lambda y}{⟶}Z\\ -\frac{d{N}_x}{dt}={\lambda }_x{N}_x⟶\left(i\right)\end{array}$

$\begin{array}{c}-\frac{d{N}_y}{dt}={\lambda }_y{N}_y-{\lambda }_x{N}_x⟶\left(ii\right)\\ +\frac{d{N}_z}{dt}={\lambda }_{y}Ny⟶\left(iii\right)\end{array}$

$\begin{array}{c}\text{Integrating (i)}\\ N_x=N_0{e}^{-\lambda t}\\ \text{Putting}t=16.48s\text{and}{N}_0={10}^{20}\\ N_x=1.92\times {10}^{19}\end{array}$

$\begin{array}{cc}\text{At (t=16.48 sec),}& N_0=N_x+N_y+N_z\\ \Rightarrow 10\times {10}^{19}& =1.92\times {10}^{19}+5.778\times {10}^{19}+N_2\\ \Rightarrow & N_z=2.302\times {10}^{19}\end{array}$