Question:
A particle of mass \(200\ \mathrm{MeV/c^2}\) collides with a hydrogen atom at rest. Soon after the collision the particle comes to rest, and the atom recoils and goes to its first excited state. The initial kinetic energy of the particle (in eV) is \(\dfrac{N}{4}\). The value of \(N\) is :
(Given the mass of the hydrogen atom to be \(1\ \mathrm{GeV/c^2}\)) ______ .
A particle of mass \(200\ \mathrm{MeV/c^2}\) collides with a hydrogen atom at rest. Soon after the collision the particle comes to rest, and the atom recoils and goes to its first excited state. The initial kinetic energy of the particle (in eV) is \(\dfrac{N}{4}\). The value of \(N\) is :
(Given the mass of the hydrogen atom to be \(1\ \mathrm{GeV/c^2}\)) ______ .
(Given the mass of the hydrogen atom to be \(1\ \mathrm{GeV/c^2}\)) ______ .
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Conserve momentum (particle stops, atom takes all momentum) and conserve energy with the 10.2 eV excitation energy (n = 1 to n = 2).
Updated On: Nov 5, 2026
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Correct Answer: 51
Solution and Explanation
Let the particle have mass \(m = 200\) MeV/c\(^2\), KE \(K\), momentum \(p\); atom mass \(M = 1000\) MeV/c\(^2\).
Momentum conservation (particle stops): atom momentum \(= p\).
Atom KE after collision \(= \dfrac{p^2}{2M}\), and \(K = \dfrac{p^2}{2m}\), so atom KE \(= K\dfrac{m}{M} = 0.2K\).
Energy conservation: \(K = 0.2K + \Delta E\), where \(\Delta E = 13.6 - 3.4 = 10.2\) eV (n = 1 to n = 2).
\(0.8K = 10.2 \Rightarrow K = 12.75\) eV \(= \dfrac{51}{4}\) eV.
So \(N = 51\).
Answer: 51
Momentum conservation (particle stops): atom momentum \(= p\).
Atom KE after collision \(= \dfrac{p^2}{2M}\), and \(K = \dfrac{p^2}{2m}\), so atom KE \(= K\dfrac{m}{M} = 0.2K\).
Energy conservation: \(K = 0.2K + \Delta E\), where \(\Delta E = 13.6 - 3.4 = 10.2\) eV (n = 1 to n = 2).
\(0.8K = 10.2 \Rightarrow K = 12.75\) eV \(= \dfrac{51}{4}\) eV.
So \(N = 51\).
Answer: 51
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