The half life of a radioactive nucleus is 50 | vedclass

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Question

According to radioactive decay law

$N=N_{e} e^{-\lambda t}$

where $N_{0}=$ Number of radioactive nuclei at time $t=0$

$N=$ Number of radioact

Vedclass · Accepted Answer

$50$

Vedclass · Answer

According to radioactive decay law

$N=N_{e} e^{-\lambda t}$

where $N_{0}=$ Number of radioactive nuclei at time $t=0$

$N=$ Number of radioactive nuclei left undecayed at any time $t$

$\lambda=$ decay constant

At time $t_{2}, \frac{2}{3}$ of the sample had decayed

$	herefore N=\frac{1}{3} N_{0}$

$	herefore \frac{1}{3} N_{0}=N_{0} e^{-\lambda t_{2}}$  ...... $(i)$

At time $t_{1}, \frac{1}{3}$ of the sample had decayed,

$	herefore N=\frac{2}{3} N_{0}$

$	herefore \frac{2}{3} N_{0}=N_{0} e^{-\lambda t_{1}}$  ...... $(ii)$

Divide $(i)$ by $(ii)$, we get

$\frac{1}{2} = \frac{{{e^{ - \lambda {t_2}}}}}{{{e^{ - \lambda {t_1}}}}}$ $ \Rightarrow \frac{1}{2} = {e^{ - \lambda \left( {{t_2} - {t_1}} ight)}}$

${\lambda\left(t_{2}-t_{1}ight)=\ln 2} $

${t_2} - {t_1} = \frac{{\ln 2}}{\lambda } = \frac{{\ln 2}}{{\left( {\frac{{\ln 2}}{{{T_{1/2}}}}} ight)}}$ $\left( {\because \lambda  = \frac{{\ln 2}}{{{T_{1/2}}}}} ight)$

${=T_{10}=50 	ext { days }}$

The half life of a radioactive nucleus is $50$ days. The time interval $\left( t _2-t_1\right)$ between the time $t _2$ when $\frac{2}{3}$ ot it has decayed and the time $t_1$, when $\frac{1}{3}$ of it had decayed is ......days

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