This Quantum World/Serious illnesses/Bohr

Bohr

In 1913 Niels Bohr postulated that the angular momentum $L$ of an orbiting atomic electron was quantized: its "allowed" values are integral multiples of $\hbar$:

$L=n\hbar$ where $n=1,2,3,\dots$

Why quantize angular momentum, rather than any other quantity?

• Radiation energy of a given frequency is quantized in multiples of Planck's constant.
• Planck's constant is measured in the same units as angular momentum.

Bohr's postulate explained not only the stability of atoms but also why the emission and absorption of electromagnetic radiation by atoms is discrete. In addition it enabled him to calculate with remarkable accuracy the spectrum of atomic hydrogen — the frequencies at which it is able to emit and absorb light (visible as well as infrared and ultraviolet). The following image shows the visible emission spectrum of atomic hydrogen, which contains four lines of the Balmer series.

Visible emission spectrum of atomic hydrogen, containing four lines of the Balmer series.

Apart from his quantization postulate, Bohr's reasoning at this point remained completely classical. Let's assume with Bohr that the electron's orbit is a circle of radius $r.$ The speed of the electron is then given by $v=r\,d\beta /dt,$ and the magnitude of its acceleration by $a=dv/dt=v\,d\beta /dt.$ Eliminating $d\beta /dt$ yields $a=v^{2}/r.$ In the cgs system of units, the magnitude of the Coulomb force is simply $F=e^{2}/r^{2},$ where $e$ is the magnitude of the charge of both the electron and the proton. Via Newton's $F=ma$ the last two equations yield $m_{e}v^{2}=e^{2}/r,$ where $m_{e}$ is the electron's mass. If we take the proton to be at rest, we obtain $T=m_{e}v^{2}/2=e^{2}/2r$ for the electron's kinetic energy.

If the electron's potential energy at infinity is set to 0, then its potential energy $V$ at a distance $r$ from the proton is minus the work required to move it from $r$ to infinity,

$V=-\int _{r}^{\infty }F(r')\,dr'=-\int _{r}^{\infty }\!{e^{2} \over (r')^{2}}\,dr'=+\left[{e^{2} \over r'}\right]_{r}^{\infty }=0-{e^{2} \over r}.$

The total energy of the electron thus is

$E=T+V=e^{2}/2r-e^{2}/r=-e^{2}/2r.$

We want to express this in terms of the electron's angular momentum $L=m_{e}vr.$ Remembering that $m_{e}v^{2}=e^{2}/r,$ and hence $rm_{e}^{2}v^{2}=m_{e}e^{2},$ and multiplying the numerator $e^{2}\,$ by $m_{e}e^{2}$ and the denominator $2r$ by $rm_{e}^{2}v^{2},$ we obtain

$E=-{e^{2} \over 2r}=-{m_{e}e^{4} \over 2m_{e}^{2}v^{2}r^{2}}=-{m_{e}e^{4} \over 2L^{2}}.$

Now comes Bohr's break with classical physics: he simply replaced $L$ by $n\hbar$. The "allowed" values for the angular momentum define a series of allowed values for the atom's energy:

$E_{n}=-{1 \over n^{2}}\left({m_{e}e^{4} \over 2\hbar ^{2}}\right),\quad n=1,2,3,\dots$

As a result, the atom can emit or absorb energy only by amounts equal to the absolute values of the differences

$\Delta E_{nm}=E_{n}-E_{m}=\left({1 \over n^{2}}-{1 \over m^{2}}\right)\,{\hbox{Ry}},$

one Rydberg (Ry) being equal to $m_{e}e^{4}/2\hbar ^{2}=13.6056923(12)\,{\hbox{eV.}}$ This is also the ionization energy $\Delta E_{1\infty }$ of atomic hydrogen — the energy needed to completely remove the electron from the proton. Bohr's predicted value was found to be in excellent agreement with the measured value.

Using two of the above expressions for the atom's energy and solving for $r,$ we obtain $r=n^{2}\hbar ^{2}/m_{e}e^{2}.$ For the ground state $(n=1)$ this is the Bohr radius of the hydrogen atom, which equals $\hbar ^{2}/m_{e}e^{2}=5.291772108(18)\times 10^{-11}m.$ The mature theory yields the same figure but interprets it as the most likely distance from the proton at which the electron would be found if its distance from the proton were measured.