JEE Mains · Physics · STD 12 - 13. Nuclei
A nucleus with mass number \(184\) initially at rest emits an \(\alpha\)-particle. If the \(Q\) value of the reaction is \(5.5\, {MeV}\), calculate the kinetic energy of the \(\alpha-\) particle. (In \({MeV}\))
- A \(5.0\)
- B \(5.5\)
- C \(0.12\)
- D \(5.38\)
Answer & Solution
Correct Answer
(D) \(5.38\)
Step-by-step Solution
Detailed explanation
\(\frac{1}{2}(4\, m) v^{2}+\frac{1}{2}(180\, m)\left(\frac{4\, v}{180}\right)^{2}=5.5\, {MeV}\) \(\Rightarrow \frac{1}{2} 4 m v^{2}\left[1+45\left(\frac{4}{180}\right)^{2}\right]=5.5\, {MeV}\) \({K} . E._{a}=\frac{5.5}{1+45 \cdot\left(\frac{4}{180}\right)^{2}}\, {MeV}\)…
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