BELIEVE   ME   NOT!    - -     A   SKEPTICs   GUIDE  

The Bohr Radius

We can play more games with Bohr's hydrogen atom if we like, using just Eq. (8) to relate r_n and p_n. Suppose we ask, ``What is keeping the electron in its orbit?'' The answer is, of course, ``The Coulomb force of attraction between the positive nucleus and the negative electron!'' This centripetal force has the value (in SI units)

$$F(r) \; = \; {1 \over 4 \pi \epsilon_\circ} {e^2 \over r^2}$$ (24.10)

where $e = 1.60217733 \times 10^{-19}$ C is the magnitude of the charge on either an electron (-e) or a proton (+e) and the ugly mess out front is the legacy of SI units - a constant stuck in to make it come out right. The corresponding electrostatic potential energy is

$$V(r) \; = \; - {1 \over 4 \pi \epsilon_\circ} {e^2 \over r}$$ (24.11)

(relative to $V \to 0$ at $r \to \infty$). We'll need that momentarily.

A simple application of NEWTON'S SECOND LAW gives

$$m {v^2 \over r} \; = \; {p^2 \over m r} \; = \; {1 \over 4 \pi \epsilon_\circ} {e^2 \over r^2}$$

where m is the mass of the electron. Cancelling one r and rearranging gives

$$p^2 \; = \; {1 \over 4 \pi \epsilon_\circ} {m e^2 \over r} .$$ (24.12)

Substituting Eq. (8) into Eq. (12) gives

$$\left( n \hbar \over r_n \right)^2 \; = \; {1 \over 4 \pi \epsilon_\circ} {m e^2 \over r_n}$$

or (after some shuffling)

$$r_n \; = \; { 4 \pi \epsilon_\circ n^2 \hbar^2 \over m e^2 }$$ (24.13)

for the radius of the $n^{\rm th}$ Bohr orbit of the H atom. The lowest orbit (n = 1) has a special name and symbol: the BOHR RADIUS,

$$a_\circ \; = \; { 4 \pi \epsilon_\circ \hbar^2 \over m e^2 } \; = \; 0.529189379 \hbox{\rm\AA}$$ (24.14)

where 1 Å = 10^-10 m.