કોઇ રેડિયો એકિટવ નમુનાની એકિટવિટી t=0 સમયે N | vedclass

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Question

According to activity law

$R=R_{0} e^{-\lambda t}$  ..... $(i)$

According to given problem,

$R_{0}=N_{0}$ counts per minute

$R=\frac{

Vedclass · Accepted Answer

$5$$\log_e2$

Vedclass · Answer

According to activity law

$R=R_{0} e^{-\lambda t}$  ..... $(i)$

According to given problem,

$R_{0}=N_{0}$ counts per minute

$R=\frac{N_{0}}{e}$ counts per minute

$t=5\,minutes$

Substituting these values in equation $(i)$, we get

${\frac{N_{0}}{e}=N_{0} e^{-5 \lambda}}$

${e^{-1}=e^{-5 \lambda}}$

$5 \lambda=1$ or $\lambda=\frac{1}{5}$ per minute

At $t=T_{1/2},$ the activity $R$ reduces to $\frac{R_{0}}{2}$

where $T_{1 / 2}=$ half life of a radioactive sample From equation $(i)$, we get

$\frac{R_{0}}{2}=R_{0} e^{-\lambda T_{1 / 2}}$

$e^{\lambda T_{1 / 2}}=2$

Taking natural logarithms of both sides of above equation, we get

$\lambda T_{1 / 2}=\log _{e} 2$

or $T_{1 / 2}=\frac{\log _{e} 2}{\lambda}=\frac{\log _{e} 2}{\left(\frac{1}{5}\right)}=5 \log _{e} 2$ minutes

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