Stationary States

In the Bohr model of the hydrogen atom, the angular momentum for the orbital motion of the electron of mass m at distance r is quantized in integral units of Planck’s constant h that is, assuming the proton position to be fixed, mvr = nh/2π, where n is an integer and v the electron velocity. Using mv = h/λ, de Broglie was able to derive Bohr’s quantization rule and nλ = 2πr. If f₁ and f₂ are the frequencies of the Bohr orbital motion of the electron in energy states E₁ and E₂, then if an electron jumps down from state 2 to state 1, why isn’t the energy of the emitted photon the difference energy hf₁ – hf₂?

Answer

In the Bohr model of the hydrogen atom, one would calculate the frequency f = 2πr/v of the electron’s orbital motion. The virial theorem states that twice the kinetic energy plus the potential energy add to zero, so mv² = ke² r, from which the electron’s frequency of orbit is f = n³h³/(4π²me⁴). The actual Bohr energy E = –2π²me⁴/(n²h²) is clearly a different quantity, and for an electron jump between two energy states, E₂ – E₁ ≠ hf₂ – hf₁.