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  • A radioactive nucleus X decays to a nucleusYwith a decay constant\lambda <img alt="\mathrm{x=0.1\sec ^{-1}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%3D0.1%5Csec

A radioactive nucleus X decays to a nucleusYwith a decay constant\lambda \mathrm{x=0.1\sec ^{-1}}, yfurther decays to a stable nucleus Z with a decay constant \mathrm{\lambda_y=1 / 30\sec ^{-1}}. Initially, there are only X nuclei and their number is \mathrm{N_0=10^{20}}. The population of Y nucleus as a function of time is given by\mathrm{ N_y(t)=\left\{N_0 \lambda_x /\left(\lambda_x-\right.\right.\left.\left.\lambda_y\right)\right\}\left[\exp \left(-\lambda_y t\right)-\exp \left(-\lambda_x t\right)\right]. }The time at which \mathrm{N_y} is maximum is:

 

Option: 1

16.48 sec


Option: 2

16.95 sec


Option: 3

17.25 sec


Option: 4

17.85sec


Answers (1)

best_answer

Let at time t = t, number of nuclei of Y and Z are \mathrm{N_y \text { and } N_z \text {. }} Then
Rate equations of the populations of X, Y and Z are

\mathrm{\begin{aligned} \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 \\ \text { and } \quad\left(\frac{d N_Z}{d t}\right) & =\lambda_Y N_Y \end{aligned}}

\mathrm{\text { Given } N_Y(t)=\frac{N_0 \lambda_X}{\lambda_X-\lambda_Y}\left[e^{-\lambda_Y t}-e^{-\lambda_X t}\right]}

\mathrm{\text { For } N_Y \text { to be maximum } }

\mathrm{\frac{d N_Y(t)}{d t}=0 }

\mathrm{\text { i.e., } \lambda_X N_X=\lambda_Y N_Y \ldots \text { (4) (from Eq. (2)) } }

\mathrm{\begin{aligned} & \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 } \frac{\lambda_X-\lambda_Y}{\lambda_Y}=\frac{e^{-\lambda_Y t}}{e^{-\lambda_Y t}}-1 \\ & \frac{\lambda_X}{\lambda_Y}=e^{\left(\lambda_Y-\lambda_X\right) t} \\ & \operatorname{or}\left(\lambda_X-\lambda_y\right) t \ln (e) \quad=\ln \left(\frac{\lambda_X}{\lambda_Y}\right) \end{aligned} }

or    \mathrm{t=\frac{1}{\lambda_X-\lambda_Y} \ln \left(\frac{\lambda_X}{\lambda_Y}\right) }

\mathrm{\text { Substiuting the values of } \lambda_x \text { and } \lambda_r \text { we have } }

\mathrm{\begin{aligned} t & =\frac{1}{(0.1-1 / 30)} \ln \left(\frac{0.1}{1 / 30}\right) \\ & =15 \ln (3) \\ t & =16.48 \mathrm{~s} . \end{aligned} }

 

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Rishi

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