A deuteron is a nucleus with one proton and one neutron, and to good approximati
ID: 1571074 • Letter: A
Question
A deuteron is a nucleus with one proton and one neutron, and to good approximation, has the mass of two protons. A deuteron with KE = 36 eV collides with a stationary hydrogen atom. The first excited state of hydrogen lies 10.2 eV above the ground state, while its second excited state lies 12.1 eV above the ground state. What photon energies could you observe as the excited hydrogen atom de-excites? To a good approximation, the mass of the deuteron is twice that of the hydrogen atom.
A) Only 1.9 eV
B) 1.9 eV, 10.2 eV, 12.1 eV
C) Only 10.2 eV
D) Only 1.9 eV, 10.2 eV
E) Only 12.1 eV
Explanation / Answer
Let us consider that the collision is perfectly inelastic therefore there would be the maximum loss of energy.
So applying the momentum conservation
Initially hydrogen atom is at rest therfore velocity would be zero .
Let us consider that the mass of the hydrogen atom is = m
mass of deutron = 2m
Initial Velocity of deutron = V
Intial momentum of the system = (2m*V) +(m*0) = 2mV
Final momentum of the system = (2m+m)v
where v is the final velocity , they both have same velocity because they are sticked after collision becasue
we have considered inelastic collision.
Applying conservation of momentum
Initial momentum = final momentum
2mV = 3mv
v = (2/3)V
Initial energy(K) = (1/2)(2m)V2
Final energy = (1/2)(3m)v2 = (1/2)(3m)((2/3)*V)2 = (2/3)K
Loss of energy = K -(2/3)K = (1/3)K = 36/3= 12 eV
Hence the energy absorbed should be less than 12 eV
Which is onl 10.2 eV
Therfore the correct option is C.