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TS EAMCET · Physics · Nuclear Physics

A certain radioactive element disintegrates with a decay constant of \(7.9 \times 10^{-10} / \mathrm{s}\). At a given instant of time, if the activity of the sample is equal to \(55.3 \times 10^{11}\) disintegration/second, then number of nuclei at that instant of time is

  1. A \(7.0 \times 10^{21}\)
  2. B \(4.27 \times 10^{13}\)
  3. C \(4.27 \times 10^3\)
  4. D \(6 \times 10^{23}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(7.0 \times 10^{21}\)

Step-by-step Solution

Detailed explanation

Decay constant, \(\lambda=7.9 \times 10^{-10} /\) second Activity, \(\quad A=55.3 \times 10^{11} \frac{\text { disintegration }}{\text { second }}\) Let the number of nuclei at that instant be \(N\). We know activity, \(A=\lambda N\)…
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