JEE Mains · Physics · STD 12 - Nuclei
- JEE Mains 2017
- Hard
- MCQ
A radioactive nucleus \(A\) with a half life \(T\), decays into a nucleus \(B.\) At \(t = 0\), there is no nucleus \(B\). At sometime \(t\), the ratio of the number of \(B\) to that of \(A\) is \(0.3\). Then, \(t\) is given by
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(t = T\;\frac{{\log 1.3}}{{\log 2}}\)
Step-by-step Solution
Detailed explanation
Let initially there are total \(\mathrm{N}_{0}\) number of nuclei
At time \(\mathrm{t} \frac{N_{B}}{N_{A}}=0.3(\text { given })\)
\(\Rightarrow \quad N_{B}=0.3 N_{A}\)
\(\mathrm{N}_{0}=N_{A}+N_{B}=N_{A}+0.3 N_{A}\)
\(\therefore \quad N_{A}=\frac{\mathrm{N}_{0}}{1.3}\)
As we know \(N_{t}=N_{0} e^{-\lambda t}\)
or, \(\frac{\mathrm{N}_{0}}{1.3}=N_{0} e^{-\lambda t}\)
\(\frac{1}{1.3}=e^{-\lambda t} \Rightarrow \ln (1.3)=\lambda t\)
or, \(t=\frac{\ln (1.3)}{\lambda} \Rightarrow t=\frac{\ln (1.3)}{\frac{\ln (2)}{T}}=\frac{\ln (1.3)}{\ln (2)} T\)
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