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JEE Mains · Physics · STD 12 - Nuclei
  • JEE Mains 2023
  • Hard
  • Numerical

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 \(.......\,\times 10^{10}\,Bq\)

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Answer & Solution

Correct Answer

(D) \(15\)

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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, \(t =0, A = A _0= N _0 \lambda\)

\(=1.5 \times 10^{-5} \times 10^{16}=15 \times 10^{10} Bq .\)

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