# Physics with Calculus/Modern/The Bohr Model

If we have an electron orbiting the nucleus, the electron is accelerating towards the nucleus. It is well known that an accelerating charge produces radiation, which contains energy. Therefore, as the electron orbits, it loses energy and consequently will eventually fall into the nucleus. The classical model of electrons orbiting a nucleus must therefore be wrong because not only do we not observe the radiation, but the electrons do NOT fall into the nucleus.

Bohr proposed that the angular momentum of the electrons could only take on discrete values, say, ${\displaystyle n\hbar }$. The energy of the electrons is also quantized as a result, so unless an electron gets enough energy to jump energy levels, it cannot emit a photon; that is, it will not radiate.

We will work through a very simple case of the Bohr model.

Assume we have a single electron of mass m orbiting a single proton in a circular orbit, and the angular momentum is ${\displaystyle n\hbar }$ where n is any integer greater than zero. Also assume that the mass of an electron is negligible compared to that of a proton.

${\displaystyle L=mvr=n\hbar }$

${\displaystyle F={\frac {ke^{2}}{r^{2}}}={\frac {mv^{2}}{r}}}$

${\displaystyle v^{2}={\frac {ke^{2}}{mr}}}$

${\displaystyle v={\frac {ke^{2}}{mvr}}={\frac {ke^{2}}{n\hbar }}}$

${\displaystyle r={\frac {n\hbar }{mv}}=n^{2}{\frac {\hbar ^{2}}{mke^{2}}}}$

${\displaystyle E={\frac {1}{2}}mv^{2}-{\frac {ke^{2}}{r}}=-{\frac {1}{2}}{\frac {ke^{2}}{r}}=-{\frac {1}{2}}{\frac {m(ke^{2})^{2}}{n^{2}\hbar ^{2}}}}$.

The Bohr model, surprisingly, makes very good predictions about the energy of the photons emitted from an electron jumping levels. We could continue our model and calculate such things as the magnetic moment of hydrogen and figure out what we expect the magnetic properties to be.

### Flaws of the Bohr Model

Only works for hydrogenic(one-electron) atoms.

The assumption of discrete angular momentum values is unexplained.