JEE Advanced · Physics · 28. Nuclear Physics
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

- A 2
- B 4
- C 6
- D 8
Answer & Solution
Correct Answer
(D) 8
Step-by-step Solution
Detailed explanation
\(\left|\frac{d N}{d t}\right|=\mid\) 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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