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To determine the half-life of a radioactive element, a student plots a graph of $\ln \left|\frac{d N(t)}{d t}\right|$ versust. Here $\frac{d N(t)}{d t}$ is the rate of radioactive decay at time $t$. If the number of radioactive nuclei of this element decreases by a factor of $p$ after $4.16 \mathrm{yr}$, the value of $p$ is
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Verified Answer
The correct answer is:
8
$\left|\frac{d N}{d t}\right|=\mid$ Activity of radioactive substance|
$$
=\lambda N=\lambda N_0 e^{-\lambda t}
$$
Taking log both sides
$$
\ln \left|\frac{d N}{d t}\right|=\ln \left(\lambda N_0\right)-\lambda t
$$
Hence, $\ln \left|\frac{d N}{d t}\right|$ versus $t$ graph is a straight line with slope $-\lambda$
From the graph we can see that,
$$
\lambda=\frac{1}{2}=0.5 \mathrm{yr}^{-1}
$$
Now applying the equation,
$$
\begin{aligned}
N & =N_0 e^{-\lambda t} \\
& =N_0 e^{-0.5 \times 4.16} \\
& =N_0 e^{-2.08}=0.125 N_0 \\
& =\frac{N_0}{8}
\end{aligned}
$$
ie, nuclei decreases by a factor of 8 . Hence the answer is 8 .
$$
=\lambda N=\lambda N_0 e^{-\lambda t}
$$
Taking log both sides
$$
\ln \left|\frac{d N}{d t}\right|=\ln \left(\lambda N_0\right)-\lambda t
$$
Hence, $\ln \left|\frac{d N}{d t}\right|$ versus $t$ graph is a straight line with slope $-\lambda$
From the graph we can see that,
$$
\lambda=\frac{1}{2}=0.5 \mathrm{yr}^{-1}
$$
Now applying the equation,
$$
\begin{aligned}
N & =N_0 e^{-\lambda t} \\
& =N_0 e^{-0.5 \times 4.16} \\
& =N_0 e^{-2.08}=0.125 N_0 \\
& =\frac{N_0}{8}
\end{aligned}
$$
ie, nuclei decreases by a factor of 8 . Hence the answer is 8 .
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