JEE Mains · Physics · STD 12 - 13. Nuclei
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
- A \(t = \frac{T}{2}\;\frac{{\log 2}}{{\log 1.3}}\)
- B \(t = T\;\frac{{\log 1.3}}{{\log 2}}\)
- C \(t=T \log(1.3)\)
- D \(t = \frac{T}{{{{log}}\left( {1.3} \right)}}\)
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…
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