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Niels Bohr's Analysis
of the Quantum States
of an Electron in an Atom

Niels Bohr's Analysis of the Electron in an Atom

The analysis of Niels Bohr that is referred to here is not the highly simplified version that is usually associated with his name but a more sophisticated analysis that preceded that planetary model. The version of Bohr's analysis that is utilized is the one presented by David Bohm in his Quantum Theory. It is this analysis that gives a basis for the quantization of the angular momentum of an electron in an atom.

The electron of charge −e is attracted to the point charge of the nucleus of charge +Ze. The nucleus is so massive compared to the electron that it can be consider fixed in position while the electron revolves around it. The potential energy of an electron at a distance r from the nucleus is −αZe²/r, where α is the constant for the Coulomb force. The kinetic energy of the electron moving at a velocity of v is in classical physics ½mv², where m is the mass of the electron. Thus the total energy of the electron is

E = ½mv² −αZe²/r

In a circular orbit the balancing of the Coulombic attraction with the centrifugal force gives:

αZe²/r² = mv²/r
and thus
mv²r = αZe²
and hence
the potential energy
αZe²/r = mv²

This latter condition means that

E = ½mv² − mv² = −½mv²

The angular momentum pθ is equal to mvr; i.e. the linear momentum mv times the distance to the center of revolution. From the previous condition it follows that

vpθ = αZe²
and thus
v = αZe²/pθ

This means that the total energy is E = −½m(αZe²)²/pθ²

If the electron goes from angular momentum pθ1 to angular momentum pθ2 there will be a change in energy ΔE which will go into a photon of frequency ν where

ΔE = hν = −½m(αZe²)²[1/pθ1² − 1/pθ2²]

For a change in angular momentum which is small the change in energy may be approximated by

ΔE = dE = (dE/dpθ)Δpθ
= m[(αZe²)²/pθ³]Δpθ

According to classical physics the angular frequency ω associated with a charge revolving in a circular orbit of radius r with a tangential velocity v is ω=v/r. The regular frequency f is related to the angular frequency by the relation f=ω/(2π). This should be the frequency associated with the escape of an electron from the hydrogen atom. Therefore

(h/2π)(v/r) = = m[(αZe²)²/pθ³]Δpθ

The expression h/2π is usually denoted as h and called h bar.

Since pθ=mvr the above equation reduces to

h(v/r) = [(αZe²)²/(m²v³r³)]Δpθ

Previously the relation αZe²/r = mv² was derived. Squaring both sides of the equation gives

(αZe²)²/r² = m²v4
and dividing both sides
by m²v³r gives
(αZe²)²/(m²v³r³) = v/r

Comparing this equation with the equation

h(v/r) = [(αZe²)²/(m²v³r³)]Δpθ

reveals the stunning result that

Δpθ = h

In words, the angular momentum of an electron in an atom must change by an increment equal to Planck's constant divided by 2π. Or, stated differently the angular momentum of an electron is quantized.

The condition on the incremental change means that angular momentum must be of the form

pθ = lh + k

Imposing the condition that the physical relationships must be the same for left-handed and right-handed coordinate system requires that k must be either 0 or ½. This means that an increment in angular momentum has to be a multiple of h.

A humorous note: David Bohm in his presentation has h in his derivation but he knows the final result should involve h-bar so he shamelessly changes from h in one equation to h-bar in the next equation without a word of justification. The original h's were a mistake stemming from treating angular frequency as being the regular frequency.

(To be continued.)

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