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Class 27: The Bohr model for the atom
Hydrogen line spectra
When excited by an electric discharge, a hydrogen gas emits light at a set of discrete wavelengths. Four
strong lines are seen in the visible part of the spectrum at wavelengths 656.3, 486.1, 434.0 and 410.2 nm.
Balmer in 1885 discovered an empirical formula for these wavelengths plus some strong UV lines
observed in hot stars
2
2 364.705 nm, 3, 4,5,. 4
n n n
λ = = −
(Note here the wavelength is for measurements made in vacuum.) By writing the Balmer formula in
terms of wave number, rather than wavelength, Rydberg proposed in 1888 a more general formula
2 2 0 0 0 0
RH , n n 1, n 2, n 3, ,
λ n n
where RH is a constant, now known as the Rydberg constant for hydrogen. The numerical value is
7 1 RH 1.09678 10 m.
− = × The corresponding wavelength is 91.1763 nm.
Over the following 25 years or so, series of lines corresponding to different values of n 0 were discovered.
n 0 Series name Discovered Longest wavelength in
the series in nm
1 Lyman 1906 - 1914 121.6 (UV)
2 Balmer 656.
3 Paschen 1908 1875 (IR)
4 Brackett 1922 4050 (IR)
5 Pfund 1924 7460 (IR)
6 Humphreys 1953 12400 (IR)
At the time of Rydberg’s discovery, there was no physical explanation for his formula.
The Bohr model
In the Rutherford model for the atom, the electrons reside outside the positively charge nucleus, and
hence are accelerated by the Coulomb force. According to classical physics, the electrons must radiate
electromagnetic waves. The energy loss would result in the electrons spiraling into the nucleus.
At this time Niels Bohr was contemplating how Planck’s quantum nature of radiation ideas could be
applied to atomic spectra, and in particular the Rydberg formula for hydrogen. Bohr pictured the electron
in hydrogen orbiting the central atomic nucleus. In 1915 he postulated a number of general assumptions:
a) Atoms have ‘stationary states’ of definite total energy. In these stationary states, the electrons do
not radiate.
b) Emission or absorption of electromagnetic radiation occurs only in transitions from one stationary
state to another. The frequency of the electromagnetic radiation is proportional to the energy
difference of the two states, and the constant of proportionality is Planck’s constant.
c) Classical physics describes the dynamical equilibrium of the atom in a stationary state but does
not describe transitions between stationary states.
d) The mean value of the kinetic energy of the electron – nucleus system is quantized. For a circular
orbit, Bohr pointed out that the quantization of kinetic energy was equivalent to the angular
momentum of the system being an integer multiple of ℏ.
To derive the Rydberg formula for hydrogen from Bohr’s assumptions, we will assume that the central
nucleus is fixed. For an electron of mass m moving with speed v in a circular orbit of radius r , its angular
momentum about the nucleus is
L = mvr. (27.2)
According to assumption d), L = mvr = n ℏ, where the integer n is called the principal quantum number.
According to assumption c), we can apply Newton’s second law to the motion of the electron. The
Coulomb force provides the centripetal acceleration. Hence
2 2
2
e v k m r r
The orbital velocity is then given by
2 2 .
e v k mr
Using the quantization of angular momentum to eliminate v , we find that only certain values of the orbital
radius are allowed:
2 2 2 r n (^) 2 n a 0 , kme
(27.5)
where
2
a 0 (^) (^2) kme
(27.6)
is called the Bohr radius. Its numerical value is a 0 (^) =0.0529 nm.
The kinetic and potential energies are
r = r 1 (^) − r 2 (^). (27.15)
Hence
( ) ( ) ( )
2 2 2 1 2 2 2 2 1 2 1 2 1 2 1 2
d d d
dt dt dt m m m m
r r r F r r F r r F r (27.16)
This can be considered as the equation of motion of a single particle of mass μ and position r acted on
by a force whose magnitude depends only on the magnitude of the particle’s position vector. The mass is
the reduced mass
1 2
1 2
m m
m m
(27.17)
Now further suppose that the positions of the two particles are measured in the center of mass frame.
Then
m 1 1 r (^) + m 2 r 2 (^) =0. (27.18)
Since this holds for all time, we also have that
1 2 1 2 0.
d d m m dt dt
r r (27.19)
In terms of r , we now have
2 1 1 2 1 2 1 2
m m
m m m m
r r r r (27.20)
with similar expressions for the particle velocities.
The angular momentum of the system about the center of mass is
1 2 1 1 2 2
2 2 1 1 1 2 1 2 1 2 1 2 1 2
1 2
1 2
d d m m dt dt
m m d m m d m m m m m m dt m m m m dt
m m d
m m dt
d
dt
= × + ×
= × + ×
= ×
= ×
r r L r r
r r r r
r r
r r
(27.21)
This is the same as the angular momentum of a single particle of mass equal to the reduced mass at
position r.
We now see that to generalize the Bohr model to a finite mass nucleus the electron mass must be replaced
by the reduced mass in the quantization of angular momentum (see equation (27.2)) and in the Newton’s
second law equation (27.3).
De Broglie relation
It is Bohr’s postulate (d) that lead de Brogle to formulate his relation between momentum and
wavelength. If the electron in the hydrogen atom is to be represented by a wave, then the
circumference of the orbit must be an integer number of wavelengths, i.e.
n λ = 2 π r. (27.22)
Equation (27.2) can be written in terms of the electron momentum and radius of the orbit
L = rp. (27.23)
Bohr’s quantization condition gives that
rp = n ℏ. (27.24)
Eliminating r from equations (27.22) and (27.24), we find
n nh h p
r π r λ
which is the de Broglie relation.
Ionization potential
From equation (27.9), we see that the most tightly bound orbit occurs for n = 1. The minimum energy
input needed to remove the electron from this ground state is called the ionization potential. The
ionization potential for hydrogen is
2 4
2
13.5984 eV. 2
H H
k e I
μ = = ℏ
(27.26)
Single electron ions
For a single electron in an ion of nuclear charge Ze , the analysis is the same as for the Bohr model of the
hydrogen atom, except that e 2 must be replaced by Ze 2 , and the appropriate reduced mass must used. In
particular the Rydberg constant is
2 4 2 2 3
H H
Z e R k Z R c
(27.27)
