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TS EAMCET · Physics · Nuclear Physics

An \(\alpha\)-particle of energy ' \(E\) ' is liberated during the decay of a nucleus of mass number 236. The total energy released in this process is

  1. A \(58 E\)
  2. B \(59 E\)
  3. C \(\frac{58 E}{59}\)
  4. D \(\frac{59 E}{58}\)
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Answer & Solution

Correct Answer

(D) \(\frac{59 E}{58}\)

Step-by-step Solution

Detailed explanation

In decay of nucleus, \((\mathrm{KE})_{\infty}=\mathrm{E}\) As, \(\mathrm{KE}=\frac{\mathrm{P}^2}{2 \mathrm{~m}} \Rightarrow \mathrm{KE} \propto \frac{1}{\mathrm{~m}}\)…
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