The vessel contains rarefied atomic hydrogen. The hydrogen atom in the ground state (E1 = -13.6 eV) absorbs a photon with a frequency of 3.7 * 10 ^ 15 Hz. With what speed V moves away from the nucleus the electron ejected from the atom as a result of ionization? The energy of the thermal motion of hydrogen atoms should be neglected.
Since the energy of the thermal motion of hydrogen atoms can be neglected, all the energy of the photon goes to ionization of the electron (this requires transferring 13.6 eV to the atom in order to transfer the electron from a bound state with energy to a free state with an energy of 0 eV) and to communicate the kinetic energy that it will have when moving to infinity (when the interaction with the hydrogen ion can be neglected):
Efot = 13.6 eV + Ekin.
Thus, the frequency of the photon is:
v = Efot / h = (13.6 * 1.6 * 10 ^ -19 J + 9.1 + 10 ^ -31 kg * (10 ^ 6 m / s) ^ 2/2) / 6.6 * 10 ^ -34 J * s
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