# Quantized energy levels of Bohr hydrogen atom

When considering the quantized velocity of the electron in the Bohr model of the hydrogen atom at the various energy levels we have: $$v_{n} = \bigg(\frac{e^2}{4 \pi \epsilon_0}\bigg)\frac{1}{n \hbar}$$ thus the total energy (kinetic energy plue potential energy) for each energy level is given as $$E_{n} = \frac{1}{2}m_{e}v_{n}^2 - \frac{1}{4 \pi \epsilon_0}\frac{e^2}{r}.$$

Question: As the expression of the energy above shows, the binding energy $E_{n}$ decreases as the velocity $v_{n}$ increases (example is $E_{1} = 13.6 eV$, if we remove the kinetic energy we would obtain $E_{1} < -13.6 eV$, hence higher binding energy). Therefore if the electron was stationary as opposed to having velocity $v_{n}$, it would have a higher binding energy. Why this the case? The velocity referred to seems to be the tangential velocity of the orbit of the electron, why does it decrease the binding energy in this way?

Thanks.

• Since the Bohr model is not correct, why worry about it too much? – Jon Custer Feb 2 '17 at 20:12
• Just to explain my d/v, Bohr's model is a picture in everybody's head, but a question about epicycles would be sent to HistorySE, and this is something similar, imo, sorry. – user140606 Feb 2 '17 at 21:14
• Velocity is NOT quantized. – ZeroTheHero Feb 2 '17 at 22:50
• Ask your self about the velocity and binding energy of the planets in the solar system. Bohr's atom (with it's many faults and few virtues) is just a quantization assumption patched onto a atom-as-solar-system model, so before blaming a feature on quantization you might check to see if the non-quantized system shares it. Suggesting reading topic: Virial theorem as applied to 1/r^2 force laws. – dmckee Feb 3 '17 at 0:19