GATE 2023 PH – Question 62
At rest $^{241}_{94}\mathrm{Am}\to{}^{237}_{92}\mathrm U+\alpha$. The respective rest masses are 224.544, 220.811 and 3.728 GeV/c squared. Find the alpha kinetic energy (source units MeV/c squared), to two decimal places.
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Correct answer: 4.90 to 4.94
Explanation
In the decay at rest, $\text{Am}\to\text{U}+\alpha$, energy and momentum are conserved. The momenta are equal and opposite, $p_U=p_\alpha=p$.
**Energy conservation** (units $c=1$, masses in GeV):
$$M=\sqrt{p^2+m_U^2}+\sqrt{p^2+m_\alpha^2}.$$
For two-body decay, the energy of the alpha particle is
$$E_\alpha=\frac{M^2+m_\alpha^2-m_U^2}{2M}.$$
**Numbers** (GeV): $M=224.544$, $m_U=220.811$, $m_\alpha=3.728$:
- $M^2=224.544^2=50\,420.008$,
- $m_U^2=220.811^2=48\,757.498$,
- $m_\alpha^2=3.728^2=13.898$.
$$E_\alpha=\frac{50\,420.008+13.898-48\,757.498}{2\times224.544}=\frac{1676.408}{449.088}=3.732917\ \text{GeV}.$$
**Kinetic energy:**
$$T_\alpha=E_\alpha-m_\alpha=3.732917-3.728=0.004917\text{ GeV}=\mathbf{4.92\ MeV}.$$