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  • A given sample contains two types of atoms A and B in the ratio 3 : 1. Atoms of type A undergo -decay with a half life of 30 days to form ‘Br

A given sample contains two types of atoms A and B in the ratio 3 : 1. Atoms of type A undergo \alpha-decay with a half life of 30 days to form ‘B’ while ‘atoms of type B’ undergo \alpha-decay with a half life of 45 days to form ‘C’, which is stable.The time after which the activities of A and that of B are in the ratio 9:22 is

 

Option: 1

30 days


Option: 2

60 days


Option: 3

90 days


Option: 4

120 days


Answers (1)

best_answer

The radioactive decay series is given
\begin{aligned} & A \stackrel{T_{\frac{1}{2}}=\text { 3odays }}{\longrightarrow} B \stackrel{T_{\frac{1}{2}}^2=45 \text { days }}{\longrightarrow} C \\ & \text { Initially } \mathrm{N}_{\mathrm{A}}(0): \mathrm{N}_{\mathrm{B}}(0)=3: 1 \\ & \frac{d N_A}{d t}+\lambda_A N_A=0 \\ & \frac{d N_B}{d t}=\lambda_A N_A-\lambda_B N_B \\ & \frac{\mathrm{dN}_{\mathrm{C}}}{\mathrm{dt}}=\lambda_{\mathrm{B}} \mathrm{N}_{\mathrm{B}} \\ \end{aligned}\begin{aligned} & \mathrm{N}_{\mathrm{A}}=\mathrm{N}_{\mathrm{A}}(0) e^{-\lambda_A t} \\ & \mathrm{~N}_{\mathrm{B}}=\mathrm{c}_1 e^{-\lambda_B t}+\frac{\lambda_A N_A(0) e^{-\lambda_A t}}{-\lambda_A+\lambda_B} \\ \end{aligned}
Then we get, \mathrm{c}_1=\frac{5}{2} \mathrm{~N}_0
\therefore \mathrm{N}_{\mathrm{A}}(\mathrm{t})=\frac{3}{4} \mathrm{~N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\frac{t}{T_1}}=\frac{3}{4} \mathrm{~N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\frac{t}{30 * \text { days }}}\\ and, \mathrm{N}_{\mathrm{B}}(\mathrm{t})=\left[\frac{5}{2} N_o\left(\frac{1}{2}\right)^{\frac{t}{45 \text { days }}}-\frac{9}{4} N_o\left(\frac{1}{2}\right)^{\frac{t}{30 \text { days }}}\right]\\ Now, \frac{\lambda_{\mathrm{A}} \mathrm{N}_{\mathrm{A}}}{\lambda_{\mathrm{B}} \mathrm{N}_{\mathrm{B}}}=\frac{9}{22} i.e. \frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\frac{3}{11}\\ or, \left(\frac{1}{2}\right)^{-\frac{t}{90}}=2\\ or, \mathrm{t}=90 days

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himanshu.meshram

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