A hydrogen atom is at its first excited state. When it de-excites it emits a photon. What is the energy of the photon and the kinetic energy of the atom?


Is it correct to assume that after the emission they acquire the same momentum, just oppositely directed? Then it's easy to calculate the kinetic energy of the atom, since the energy of the photon is its momentum times the speed of light in vacuum. While the energy of the photon is also the change in the atoms energy states, -3,4 keV - (-13,6 keV). (The energy of the atom would then be the square of the momentum divided by two times the mass of the hydrogen atom).

This seems to, so I assume that I've been to careless in my calculations :)

up vote 2 down vote accepted

Yes, the conservation of momentum is valid. The photon and the Hydrogen atom acquire equal and opposite momenta when the photon is emitted. However you must be a little careful when you are calculating the momentum as the total energy of the photon and the hydrogen atom is 10.2 eV. This means your equation will be $$ p^2/2m + pc = 10.2 eV$$ You must conserve total energy before and after the event. So the potential energy stored by the electron when it was in the 1st excited level is now converted into both the kinetic energy of the Hydrogen atom as well as the energy of the photon.

  • Gosh you 're right ! Thank you! – A. Fågel May 11 '16 at 21:41

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