KCET · Physics · Nuclear Physics
The natural logarithm of the activity \(R\) of a radioactive sample varies with time \(t\) as shown. At \(t=0\), there are \(N_0\) undecayed nuclei. Then, \(N_0\) is equal to [Take \(e^2=7.5\) ]
- A \(7500\)
- B \(3500\)
- C \(75000\)
- D \(150000\)
Answer & Solution
Correct Answer
(C) \(75000\)
Step-by-step Solution
Detailed explanation
From the diagram at \(t=0\)
\(\log _e R_0=2\)
\(\Rightarrow \quad R_0=e^2=7.5\)
Since, \(R_0=\lambda N_0\) ....(i)
Since, \(\lambda=\) Slope of graph
\(=\frac{1}{10 \times 10^3}=10^{-4} \mathrm{~s}^{-1}\)
\(\therefore\) From Eq. (i), \(N_0=\frac{R_0}{\lambda}=\frac{7.5}{10^{-4}}=75000\)
\(\log _e R_0=2\)
\(\Rightarrow \quad R_0=e^2=7.5\)
Since, \(R_0=\lambda N_0\) ....(i)
Since, \(\lambda=\) Slope of graph
\(=\frac{1}{10 \times 10^3}=10^{-4} \mathrm{~s}^{-1}\)
\(\therefore\) From Eq. (i), \(N_0=\frac{R_0}{\lambda}=\frac{7.5}{10^{-4}}=75000\)
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