KCET · Physics · Nuclear Physics
A radioactive nucleus has specific binding energy \(\mathrm{E}_{1}\). It emits an \(\alpha\)-particle. The resulting nucleus has specific binding energy \(\mathrm{E}_{2}\). Then
- A \(\mathrm{E}_{2}=0\)
- B \(\mathrm{E}_{2}=\mathrm{E}_{1}\)
- C \(\mathrm{E}_{2} < \mathrm{E}_{1}\)
- D \(\mathrm{E}_{2}>\mathrm{E}_{1}\)
Answer & Solution
Correct Answer
(D) \(\mathrm{E}_{2}>\mathrm{E}_{1}\)
Step-by-step Solution
Detailed explanation
The nucleus become more stable on emission of \(\alpha\)-particle, hence \(\mathrm{E}_{2}>\mathrm{E}_{1}\).
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