A particle of mass 200 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 kine

Question

A particle of mass 200 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 N 4 . The value of N is: (Given the mass of the hydrogen atom to be 1 GeV c - 2 ).........

TestHub · Accepted Answer

Given, M = 200 MeV c - 2 and m = 1 GeV c - 2 , now use the concept of conservation of linear momentum, M v 0 = m v ⇒ v = M v 0 m . . . 1 . Now use the conservation of energy, 1 2 M v 0 2 = 1 2 m v 2 + ∆ E . . 2 , here ∆ E = 13 . 6 eV , for the hydrogen atom in the ground state, now from equation, 1 & 2 , 1 2 M v 0 2 = 1 4 3 ∆ E × 10 8 e V = N 4 , ⇒ N 4 = 51 4 e V ⇒ N = 51 .

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