An electron of a stationary hydrogen atom passes from the fifth energy level to the fundamental state. What velocity did the atom acquire as the result of photon emission? What is the recoil energy?

Answer

`4.2m//s;9.1xx10^(-6)eV`.

Explanation

The energy of transition from the excited to the ground state is shared by the photon and the atom: `epsi=hv+D`, where D = `p^(2)//2M` is the atom.s recoil energy, and p is the momentum due to the emission of a photon. In accordance with the law of conservation of momentum, `p=p_(ph)=hv//c`. Hence `D=h^(2)v^(2)//2Mc^(2)` and the transition energy is
`epsi=hv(1+(hv)/(2Mc^(2)))`
Solving this quadratic equation, we obtain the expression for the energy of the photon:
`hv=(2epsi)/(1+sqrt(1+2epsi//Mc^(2)))`
Since the transition energy in a hydrogen atom is below 13.6 eV, and its rest energy is 1 GeV, it follows that `2epsi//Mc^(2)~~10^(-8)`, and so this term in the denominator may be left out without appreciable loss in accuracy. Since, according to the statement of the problem, the transition is from the fifth to the first level, it follows that
`epsi=hcR(1/1^(2)-1/5^(2))=24/25hcR`
Hence `hv=(24hcR)/25`, the recoil energy is `D=(24^(2)h^(2)R^(2))/(2xx25^(2)M)`, the velocity of the atom is v = `(24hR)/(25M)`.