Modern Physics Ques 137
- A radioactive nucleus $X$ decays to a nucleus $Y$ with a decay constant $\lambda _x=0.1 $ $s^{-1}, Y$ further decays to a stable nucleus $Z$ with a decay constant $\lambda _y=1 / 30 $ $s^{-1}$. 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 $Y$ nucleus as a function of time is given by $N _y(t)=\{N _0 \lambda _x /\left(\lambda _x-\lambda _y\right) \}\left[\exp \left(-\lambda _y t\right)-\exp \left(-\lambda _x t\right)\right]$. Find the time at which $N _Y$ is maximum and determine the populations $X$ and $Z$ at that instant.
(2001, 5M)
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Answer:
Correct Answer: 137.(a) $ \left(\frac{d N _X}{d t}\right) =-\lambda _X N _X , \left(\frac{d N _Y}{d t}\right) =\lambda _X N _X-\lambda _Y N _Y ,\left(\frac{d N _Z}{d t}\right) =\lambda _Y N _Y $
(b) $16.48 $ $s$
(c) $N_X= 1.92 \times 10^{19} ,N _Z =2.32 \times 10^{19}$
Solution:
Formula:
- (a) Let at time $t=t$, number of nuclei of $Y$ and $Z$ are $N _Y$ and $N _Z$. Then,
Rate equations of the populations of $X, Y$ and $Z$ are
$ \begin{aligned} \left(\frac{d N _X}{d t}\right) & =-\lambda _X N _X \quad …….(i)\\ \left(\frac{d N _Y}{d t}\right) & =\lambda _X N _X-\lambda _Y N _Y \quad …….(ii)\\ \text { and } \quad \left(\frac{d N _Z}{d t}\right) & =\lambda _Y N _Y \quad …….(iii) \end{aligned} $
(b) Given $N _Y(t)=\frac{N _0 \lambda _X}{\lambda _X-\lambda _Y}\left[e^{-\lambda _Y t}-e^{-\lambda _X t}\right]$
For $N _Y$ to be maximum
$\frac{d N _Y(t)}{d t}=0 $
$\text { i.e } \quad \lambda _X N _X=\lambda _Y N _Y \quad …….(iv) \quad \text { [from Eq. (ii)] } $
$\text { or } \lambda _X\left(N _0 e^{-\lambda _X t}\right)=\lambda _Y \frac{N _0 \lambda _X}{\lambda _X-\lambda _Y}\left[e^{-\lambda _Y t}-e^{-\lambda _X t}\right] $
$\text { or } \quad \frac{\lambda _X-\lambda _Y}{\lambda _Y}=\frac{e^{-\lambda _Y t}}{e^{-\lambda _X t}}-1 $
$\frac{\lambda _X}{\lambda _Y}=e^{\left(\lambda _X-\lambda _Y\right) t} $
$\text { or }\left(\lambda _X-\lambda _y\right) t \ln (e)=\ln (\frac{\lambda _X}{\lambda _Y}) $
$\text { or } t=\frac{1}{\lambda _X-\lambda _Y} \ln (\frac{\lambda _X}{\lambda _Y})$
Substituting the values of $\lambda _X$ and $\lambda _Y$, we have
$ \begin{aligned} & t=\frac{1}{(0.1-1 / 30)} \ln (\frac{0.1}{1 / 30})=15 \ln (3) \\ & t=16.48 s \end{aligned} $
(c) The population of $X$ at this moment,
$ \begin{aligned} N _X & =N _0 e^{-\lambda _X t}=(10)^{20} e^{-(0.1)(16.48)} \\ N _X & =1.92 \times 10^{19} \\ N _Y & =\frac{N _X \lambda _X}{\lambda _Y} \text { [From Eq. (iv) ]} \\ & =\left(1.92 \times 10^{19}\right) \frac{0.1}{(1 / 30)}=5.76 \times 10^{19}\\ N _Z & =N _0-N _X-N _Y \\ & =10^{20}-1.92 \times 10^{19}-5.76 \times 10^{19} \\ \text { or } \quad N _Z & =2.32 \times 10^{19} \end{aligned} $
BETA
