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At time t = 0, a material is composed of two radioactive atoms A and B ,where N_{A}\left ( 0 \right )= 2N_{B}\left ( 0 \right ). The decay constant of both kind of radioactive atoms is \lambda, However,A disintegrates to B and B disintegrates to C. Which of the following figures represents  the evolution of N_{B}\left ( t \right )/N_{B}\left ( 0 \right ) with respect to time t ?
\begin{bmatrix} N_{A} \left ( 0 \right )= No.Of \, A\, atoms\, at\, t = 0& \\ N_{B} \left ( 0 \right )= No.Of \, B\, atoms\, at\, t = 0& \end{bmatrix}
Option: 1
Option: 2
Option: 3
Option: 4

Answers (1)

best_answer

At t = 0
,N_{A}\left ( 0 \right )= 2N_{B}\left ( 0 \right )\left ( Given \right )
\lambda _{A}=\lambda _{B}= \lambda \left ( Given \right )
A\overset{\lambda }{\rightarrow}B\overset{\lambda }{\rightarrow}C
\frac{dN_{A}}{dt}= -\lambda N_{A}
\frac{dN_{B}}{dt}= \lambda N_{A}-\lambda N_{B}
N_{A}= \left ( N_{A} \right )_{0}e^{-\lambda t}= \left ( 2\left ( N_{B} \right ) _{0}\right )e^{-\lambda t}
\frac{dN_{B}}{dt}= \lambda \left (2 N_{B} \right )_{0}e^{-\lambda t}-\lambda \left ( N_{B} \right ) \\ e^{+\lambda t}\left ( \frac{dN_{B}}{dt}+\lambda N_{B} \right )= 2\left ( N_{B} \right )_{0}\lambda e^{0}
\frac{d}{dt}\left ( N_{B}e^{+\lambda t} \right )= 2\lambda N_{B_{0}}
N_{B}e^{\lambda t}= 2\lambda t\left ( N_{B} \right )_{0}+C
at t = 0 , N_{B}= N_{B_{0}}
C= N_{B_{0}}
\therefore N_{B}e^{\lambda t}= 2\lambda t\left ( N_{B} \right )_{0}+N_{B_{0}}
N_{B}= N_{B_{0}}\left [ 1+2\lambda t \right ]e^{-\lambda t}
\frac{dN_{B}}{dt}= 0
-\lambda \left [ 1+2\lambda t \right ] e^{-\lambda t}+2\lambda e^{-\lambda t}= 0
t= \frac{1}{2\lambda }
\therefore \left ( N_{B} \right )_{max}\: at\: t= \frac{1}{2\lambda }
The correct option is (3)

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vishal kumar

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