The graph shows the log of activity log R of | vedclass

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Question

$(b)$ Activity of a radioactive sample is given by

$R=-\frac{d N}{d t}=-\frac{d}{d t} N_{0} e^{-\lambda t}=\lambda N_{0} \cdot e^{-\lambda t}$

S

Vedclass · Accepted Answer

$3.0$

Vedclass · Answer

$(b)$ Activity of a radioactive sample is given by

$R=-\frac{d N}{d t}=-\frac{d}{d t} N_{0} e^{-\lambda t}=\lambda N_{0} \cdot e^{-\lambda t}$

So, $\log R=\log \left(\lambda N_{0}\right)+\log \left(e^{-\lambda t}\right)$

$\Rightarrow \log R=-\lambda t+\log \left(\lambda N_{0}\right)$

This equation is form of $y=m x+C$

So, absolute value of slope of $\log R$ versus

$t$ graph gives decay constant $\lambda$.

Now, from graph,

We get, slope $=\frac{8-6}{8-16} \mid=\frac{1}{4}=\lambda$

So, half-life time period of sample is

$T_{1 / 2}=\frac{\log 2}{\lambda}=\frac{0.693}{1 / 4} \approx 3.0 \,min$

The graph shows the $\log$ of activity $\log R$ of a radioactive material as a function of time $t$ in minutes.The half-life (in minute) for the decay is closest to

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