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

\mathrm{ \mathrm{A} \stackrel{\mathrm{T}_{\mathrm{y} 2}=30 \text { days }}{\longrightarrow} \mathrm{B} \stackrel{\mathrm{T}_{12}=45 \text { days }}{\longrightarrow} \mathrm{C} }

 

Option: 1

40 Days
 


Option: 2

60 Days

 


Option: 3

80 Days
 


Option: 4

90 days


Answers (1)

best_answer

The radioactive decay series is given

Initially \mathrm{N}_{\mathrm{A}}(0): \mathrm{N}_{\mathrm{B}}(0)=3: 1

\mathrm{ \frac{d N_A}{d t}+\lambda_A N_A=0 }

\mathrm{ \frac{d N_B}{d t}=\lambda_A N_A-\lambda_B N_B }

\mathrm{ \frac{d N_C}{d t}=\lambda_B N_B }

\mathrm{ N_A=N_A(0) e^{-\lambda_A t} }

\mathrm{ N_B=c_1 e^{-\lambda_B t}+\frac{\lambda_A N_A(0) e^{-\lambda_A t}}{-\lambda_A+\lambda_B}}

Then we get, \mathrm{ c_1=\frac{5}{2} N_0}

\mathrm{ \therefore \quad N_A(t)=\frac{3}{4} N_{\circ}\left(\frac{1}{2}\right)^{\frac{t}{T_1 / 2}}=\frac{3}{4} N_{\circ}\left(\frac{1}{2}\right)^{\frac{t}{30 \text { days }}} }

\mathrm{ and, N_B(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] }

\mathrm{ Now, \frac{\lambda_A N_A}{\lambda_B N_B}=\frac{9}{22} i.e. \frac{N_A}{N_B}=\frac{3}{11} }

\mathrm{ or, \left(\frac{1}{2}\right)^{-t / 90}=2 }

\mathrm{ or, \quad t=90 days }

Hence option 4 is correct.




 


 

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Nehul

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