TS EAMCET · Physics · Nuclear Physics
Two radioactive materials \(R_1\) and \(R_2\) have decay constants \(6 \lambda\) and \(\lambda\), respectively. The half life of \(R_2\) is \(1.4 \times 10^{17} \mathrm{~s}\). Initially they contain some number of nuclei. The time at which the ratio of the remaining nuclei of \(R_2\) to that of \(R_1\) will be \(e\) is \((\) Let \(\ln 2=0.7)\)
- A \(2 \times 10^{16} \mathrm{~s}\)
- B \(4 \times 10^{16} \mathrm{~s}\)
- C \(3 \times 10^{16} \mathrm{~s}\)
- D \(5 \times 10^{16} \mathrm{~s}\)
Answer & Solution
Correct Answer
(B) \(4 \times 10^{16} \mathrm{~s}\)
Step-by-step Solution
Detailed explanation
For two radioactive elements \(R_1\) and \(R_2\), decay constants, \(\lambda_1=6 \lambda\) and \(\lambda_2=\lambda\) Half-life of \(R_2,\left(t_{1 / 2}\right)_2=1.4 \times 10^{17} \mathrm{~s}\) If \(N_0\) be the initial number of nuclei at \(t=0\), in both \(R_1\) and \(R_2\) or…
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