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Two radioactive substances \mathrm{A} and \mathrm{B} have decay constant 5 \lambda and \lambda respectively. At \mathrm{t}=0, they have the same number of nuclei. The ratio of number of nuclei of \mathrm{A} to that of \mathrm{B} will be (1 / e)^2 after a time interval of:

Option: 1

\frac{1}{\lambda}


Option: 2

\frac{1}{2 \lambda}


Option: 3

\frac{1}{3 \lambda}


Option: 4

\frac{1}{4 \lambda}


Answers (1)

best_answer

According to radioactive decay, \mathrm{N}=\mathrm{N}_0 \mathrm{e}^{-\lambda \mathrm{t}}

Where, \mathrm{N}_0= Number of radioactive nuclei present in the sample at \mathrm{t}=0

\mathrm{N}= Number of radioactive nuclei left undecayed after time \mathrm{t}

\lambda=\ decay constant 

\text { For } \mathrm{A}, \mathrm{N}_{\mathrm{A}}=\left(\mathrm{N}_0\right) \mathrm{Ae} \mathrm{}^{-5 \lambda t}                       (i)

\text { For } B, N_B=\left(N_0\right) \mathrm{Be}^{-\lambda t}                          (ii)

As per question \left(\mathrm{N}_0\right)_{\mathrm{A}}=\left(\mathrm{N}_0\right)_{\mathrm{B}}  (Given) 

Dividing (i) by (ii), we get

\frac{N_A}{N_B}=\frac{e^{-5 \lambda t}}{e^{-\lambda t}}=e^{-4 \lambda t} \quad \text { or }\left(\frac{1}{e}\right)^2=e^{-4 \lambda t} \quad \text { or } \quad \frac{1}{e^2}=\frac{1}{e^{4 \lambda t}}

\mathrm{e}^{4 \lambda \mathrm{t}}=\mathrm{e}^2 \quad                               \text { or } \quad 4 \lambda \mathrm{t}=2 \quad               \text { or } \quad \mathrm{t}=\frac{2}{4 \lambda}=\frac{1}{2 \lambda}

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Rishabh

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