San José State University
Thayer Watkins
Silicon Valley
& Tornado Alley

The Old Quantum Physics of Niels Bohr
and the Spectrum of Helium:
A Modified Version of
the Bohr Model

Niels Bohr's model of the atom provided a wonderfully accurate explanation of the spectrum of hydrogen, but when it was applied to the spectrum of helium it failed. Werner Heisenberg developed a modification of Bohr's analysis but it involved half-integral values for the quantum numbers. Bohr considered non-integral values of quantum numbers nonsensical in terms of the logic of quantum theory. There was a crisis in quantum theory that ultimately led to the creation in the 1920's of the new quantum theory which became known as quantum mechanics.

This is an attempt to apply the Old Quantum Physics to explain the spectrum of helium. The energy level of an electron in an atom in which the effective charge is Z is −RZ²/n² where R is the Rydberg constant and n is the quantum number of the electron (its number of angular momentum units). The quantum number n is necessarily a positive integer, but Z can be either an integer or half-integer such as {1/2, 3/2, 5/2, … }.

To see how a half-integer value for Z can arise consider a unit negative charge distributed throughout a thin spherical shell of radius r. From a point within the shell; i.e., a point at a distance from the center of the shell less than r, the charge cancels out and the effect of the charge on the shell is zero. From a point whose distance for the center of the shell is greater than r the effect of the charge is the same as if it were concentrated at the center of the shell. Thus as distance from the center of the shell goes from a value greater than r to one less than r the effect charge goes from −1 to zero. Thus at a distance of exactly r the effective charge has to −½. Thus for an electron in a shell the effect of the other electrons in the shell is the same as if one half of the charge of the other electrons were concentrated at the center of the shell.


For the two electrons of a helium atom, if they have different quantum numbers and hence are in different shells the outer electron experiences an effective charge in the nucleus of 2−1=+1 and the inner electron experiences an effective charge of 2−0=+2. But if the electrons have the same quantum number they both experience an effective charge in the nucleus of 2−½=+3/2.

Energy Levels

Let n1 and n2 be the quantum numbers of the two electrons in a helium atom. For simplicity assume that n1≥n2. If n1>n2 then the energy levels of the electrons are

thus the total is
−R[1/n1² + 4/n2²]

If n1=n2 then the energy levels are both

so the total is
−2R(9/4)/n1² = −(9/2)/n1².

Energy Level Changes

For the case in which n1>n2 and n1 is reduced to n3 but n3>n2 the energy change is

ΔE = R[1/n3² − 1/n1²]

This portion of the spectrum would be the same as for hydrogen.

For n1>n2 and n2 is reduced to n4 the energy change is

ΔE = R[4/n4² − 4/n1²] = 4R[1/n4² − 1/n1²]

The transition of an outer electron from quantum number n1>n2 to n2 involves an energy change of

−R(9/2)/n2² − [ −R(1/n1² + 4/n2²)]
which is equal to
R[(4−9/2)/n2² + 1/n1²)] = R[1/n1² − ½/n2²]

The transition of an electron from quantum number n2=n1 to n4<n1 involves an energy change of

−R[1/n1² + 4/n4²] − [− R(9/2)/n1²]
or, equivalently
R[(9/2−1)/n1² − 4/n4²] = R[(7/2)/n1² − 4/n4²]

Finally there is the case in which the quantum number of the outer electron changes from n1>n2 to n3 which is less than n2. The energy change is

ΔE = −R(1/n2+4/n2²) − [−R(1/n1² + 4/n2²)]
R[(4−1)/n2² + (1/n1² − 4/n3²)]
R[3/n2² + 1/n1² − 4/n3²]

There are thus five cases. If the helium model is valid then for any spectral line of helium there should exist integer solutions for at least one of the five cases.

The Helium Spectrum

There are three spectral lines for helium, {438.793 nm, 443.755 nm, 447.148 nm} for which the best explanation is a transition of an electron from quantum number 5 to quantum number 2 while the other electron remains at quantum number 1. The wave length is given by the equation

ΔE = hc/λ = R(1/n3² − 1/n1²)
1/λ = (R/hc)(1/n3² − 1/n1²)

The value of the coefficient (R/hc) is 1.097373×107 m−1. If n1=5 and n3=2 then the wavelength for the transition should be 433.937 nm. For the spectral line at 438.793 this is in error by about 1.1 percent. For the other two lines the computed value is in error by 2.3 and 3.0 percent, respectively.

For the 471.314 nm, 492.193 nm and 501.5675 nm spectral lines the closest fit is for n1=4 and n3=2 with n2=1. For this transition the wavelength should be 486.009 nm. For the 471.314 nm line the computed value is 3.1 percent too high. For the 492.193 nm and 501.5675 nm lines the computed values are 1.3 percent and 3.2 percent too low.

Another line of the helium spectrum is at 667.815 nm. The best explanation for this line is a transition from n1=3 to n3=2. The computed wavelength for this transition is 656.112 nm, an error of 1.8 percent.

Here are these results in tabulated form:

Comparison of Measured Helium Spectrum Lines
with Values Computed from a Modified Version
of the Bohr Model
438.793 nm433.937 nm-1.1%
471.314 nm486.009 nm+3.1%
492.193 nm 486.009 nm-1.3%
501.5675 nm486.009 nm-3.2%
667.815 nm656.112 nm-1.8%

Unfortunately there is no readily available computations of the Helium spectrum using the New Quantum Theory (Quantum Mechanics) for comparison.

HOME PAGE OF applet-magic
HOME PAGE OF Thayer Watkins