AP EAMCET · PHYSICS · Nuclear Physics
The half-life of \(\mathrm{Ra}^{226}\) is 1620 years. Then the number of atoms decay in one second in \(1 \mathrm{~g}\) of radium (Avogadro number \(=6.023 \times 10^{23}\) )
- A \(4.23 \times 10^9\)
- B \(3.16 \times 10^{10}\)
- C \(3.61 \times 10^{10}\)
- D \(2.16 \times 10^{10}\)
Answer & Solution
Correct Answer
(C) \(3.61 \times 10^{10}\)
Step-by-step Solution
Detailed explanation
The number of atoms decay in one second \(\frac{d N}{d t}=\lambda N\) \(\begin{aligned} & =\frac{0.693}{1620 \times 365 \times 86 \times 400} \times \frac{6.023 \times 10^{23}}{226} \\ & =3.61 \times 10^{10}\end{aligned}\)
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