The electron in a hydrogen atom at rest makes a transition from `n = 2` energy state to the `n = 1` ground state.
Assuming that all of the energy corresponding to transition from `n = 2 to n = 1` is carried off by the photon. By setting the momentum of the system (atom + photon) equal to zero after the emission and assuming that the recoil energy of the atom is smaller compared with the `n = 2 to n = 1` energy level separation , find the energy of the recoiling hydrogen atom.

Options

A. `2.75 xx 10^(-7) eV`
B. `5.54 xx 10^(-8) eV`
C. `8.11 xx 10^(-8) eV`
D. `10.36 xx 10^(-7) eV`

Answer

B

Explanation

From conservation of momentum, as the total momentum before emission is zero. The photon and atom therefore move off in oposite direction , with `mv = E_(photon)//C`
where `m and v` are the mass and recoil speed of hydrogen atom `E_(photon)` is the actual energy of the photon (less than `10.2 eV)` and `c` is the speed of light. The energy difference between the `n = 2 to n = 1` levels . `E` is the source of both the photon energy and the recoil kinetic energy energy of the atom.
From energy conservation , we have
`E = E_(photon) + (1)/(2) mv^(2)`
Atom being massive, we can assume that its recoil speed `v` and kinrtic energy are so small that `E = E_(photon)`. Substituting `E_(photon) = 10.2 eV` into the experession for `mV` yields.
`mv = 10.2 eVc`
The recoil kinetic energy of the hydrogen atom can now be calculated.
`K = (1)/(2) mv^(2) = (1)/(2) ((mv)^(2))/(m) = (0.5) ((10.2)^(2))/(m c^(2))`
`= ((0.5) (10.2)^(2))/(938.8 xx 10^(6)) = 5.54 xx 10^(-8) eV`
Thus , the fraction of the energy difference between the `n = 3 to n = 1` level that goes into atomic recoil energy is very small.
`(K)/(E) = (5.54 xx 10^(-8))/(10.2) = 5.43 xx 10^(-9)`
That is why by equanting the photon's energy to the atomic energy level separation yields accurate answer because little energy is needed to conserve momentum.