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
- A \(58 E\)
- B \(59 E\)
- C \(\frac{58 E}{59}\)
- D \(\frac{59 E}{58}\)
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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