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  • Some nuclei of radioactive material are undergoing radioactive decay. The time gap between the instances when a quarter of the nuclei have decoyed and when half of the nuclei have decayed is give b

Some nuclei of radioactive material are undergoing radioactive decay. The time gap between the instances when a quarter of the nuclei have decoyed and when half of the nuclei have decayed is give by _ (\mathrm{\lambda } is decay constant)

Option: 1

\mathrm{ \frac{1}{2} \frac{\ln (2)}{\lambda} }


Option: 2

\mathrm{\frac{1}{\lambda} \ln (2)}


Option: 3

\frac{2 \ln (2)}{\lambda}


Option: 4

\frac{\ln (3 / 2)}{\lambda}


Answers (1)

best_answer

\mathrm{use N=N o e^{-\lambda t} Case I N=\frac{N_0}{4}, \frac{N_0}{4} }
\mathrm{\rightarrow decayed Undecayed =N_0-\frac{N_0}{4}=3 \frac{N_0}{4} \frac{3 N_0}{4}=N_0 e^{-\lambda t} }
\mathrm{Cos \mathbb{I} N=\frac{N_0}{2}, \quad \frac{N_0}{2}=N_0 e^{-\lambda t_2} Divide O \perp(0) \frac{3}{4} \times \frac{2}{1}=e^{-\lambda\left(t_1-t_2\right)}=e^{\lambda\left(t_2-t_1\right)} }\mathrm{but \left(t_2-t_1\right)=\Delta t= time gap. \begin{aligned} \frac{3}{2} & =e^{\lambda \Delta t} \\ \ln \frac{3}{2} & =\ln e^{\lambda \Delta t}=\lambda \Delta t \ln e \\ \Delta t & =\frac{1}{\lambda} \ln (3 / 2) \end{aligned} }

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Gaurav

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