JEE Mains · Physics · STD 12 - 13. Nuclei
The decay constant for a radioactive nuclide is \(1.5 \times 10^{-5} s ^{-1}\). Atomic of the substance is \(60\,g\) mole \(^{-1},\left( N _{ A }=6 \times 10^{23}\right)\). The activity of \(1.0\,\mu g\) of the substance is \(.......\,\times 10^{10}\,Bq\)
- A \(14\)
- B \(13\)
- C \(12\)
- D \(15\)
Answer & Solution
Correct Answer
(D) \(15\)
Step-by-step Solution
Detailed explanation
\(\lambda=1.5 \times 10^{-5} s ^{-1}\) No. of mole \(=\frac{1 \times 10^{-6}}{60}=\frac{10^{-7}}{6}\) No. of atoms \(=\) no. of moles \(\times N _{ A }\) \(=\frac{10^{-7}}{6} \times 6 \times 10^{23}=10^{16}\) \(A=N_0 \lambda e^{-\lambda t}\) For, \(t =0, A = A _0= N _0 \lambda\)…
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