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21.2Quantisation of angular momentum
Niels Bohr (1885â1962), Figure 21.9a, was a Danish physicist who studied the hydrogen atom. This is the simplest atom, consisting of a nucleus of a single proton and a single electron orbiting it, Figure 21.9b.

Figure 21.9: a Niels Bohr. b An electron orbiting a proton. The force on the electron is the electric force.

EXAM TIP
There is a lot of algebra in this derivation that must be learned carefully.
Bohr objected to the Rutherford model because an electron in orbit experiences centripetal acceleration. It was a prediction of the classical theory of electromagnetism that accelerated charges emit electromagnetic radiation and so the electron would lose energy and plunge into the nucleus. Bohr assumed that the electron could only exist in certain special orbits satisfying the condition:
mvr=nh2Ïn=1,2,3,â¦
In these orbits the electron would not radiate, in other words, the classical laws of electromagnetism would not apply to these states. This was a revolutionary step! Notice that the quantity L = mvr is the angular momentum of the electron. The condition states that the angular momentum is quantised; in other words, it is an integral multiple of a basic unit (in this case, h2Ï).
What is the consequence of this condition?
The total energy ET of the orbiting electron is (k is the Coulomb constant):
ET=12mv2îîîîîîîîîîîîîîkinetic energyÂ +(âke2r)îîîîîîîîîîîîîîîîîîîîîîelectric potential energyÂ 
This is because the proton has charge e and the electron charge âe. But the electron is acted upon by the electric force, and so:
ke2r2=mv2r
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From this we deduce that mv2=ke2r, and so the total energy becomes:
ET=12ke2râke2r
ET=â12ke2r
Squaring both sides of the Bohr condition equation:
m2v2r2=n2h24Ï2
and so mv2=n2h24Ï2mr2
But earlier we found that mv2=ke2r, so substituting for mv2 we get:
ke2r=n2h24Ï2mr2
This gives the extraordinary result that the orbital radius cannot be anything we wish:
it equals r=h24Ï2ke2mÃn2.
We define a0=h24Ï2ke2m to be the Bohr radius. Numerically,
a0=(6.626Ã10â34)24Ï2Ã8.988Ã109Ã(1.602Ã10â19)2Ã9.109Ã10â31=5.29Ã10â11Â mÂ so thatÂ r=a0Ãn2.
Figure 21.10 shows the radii corresponding to n = 1, 2 and 3.

Figure 21.10: The lowest three allowed orbits of the electron in the Bohr model of hydrogen.

If we substitute this value of the orbit radius in the formula for the total energy of the electron, ET=â12ke2r, we get:
ET=â12ke2a0Ã1n2.
Putting in accurate values of the constants we finally get:
E=â13.6n2eV
In other words, the theory predicts that the electron in the hydrogen atom has discrete or quantised energy. As we saw in SectionÂ 21.1, this explains the emission and absorption spectra of hydrogen.
Links
Much of what you have learned in this chapter will be familiar from the work we did on gravitation in SectionsÂ 17.2 and 17.3 showing the combined use of conservation of energy and Newtonâs second law.
WORKED EXAMPLE 21.3
In gravitation the period of revolution T of a planet in a circular orbit of radius R around the sun obeys T2 â R3. Deduce the corresponding relation in the Bohr hydrogen model for an electron.
Answer
We know that v2=ke2mr and so (2ÏrT)2=ke2mr, leading to T2=4Ï2mke2r3 i.e. T2 â r3, just as in gravitation.
THEORY OF KNOWLEDGE
Where did the Bohr angular momentum quantisation condition come from? And why did the scientific community accept it even though no explanation was given for it? The answer is that this condition led to an explanation of a long-standing unsolved problem, namely atomic spectra. The fact that such a convincing explanation was found implied that most likely the assumptions behind the theory were correct or almost correct; that was the beginning of a new era in physics and everyone suspected that with more developments and better understanding of what was going on, a plausible argument for this condition would be found. Sure enough, the SchrÃ¶dinger model provided the missing explanation quite a few years later through the introduction of the wavefunction. (But try the last end-of-chapter problem for Bohrâs own explanation.)
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TEST YOUR UNDERSTANDING
14Calculate, for an electron in the n = 2 state of hydrogen,
athe orbit radius
bthe linear momentum
cthe angular momentum
dthe kinetic energy
ethe potential energy.
15aShow that the speed of an electron in the nth state of hydrogen is given by v=2Ïke2nh.
bHence calculate the ratio of the speed of the electron in the state n = 2 to that in the state n = 1.
16aShow that the relation between the period of revolution Tn of an electron in the nth state of hydrogen is related to the period in the n = 1 state T1 by Tn = n3T1.
bFind an expression for T1 in terms of fundamental constants.
cEvaluate T1 in seconds and 1T1 in Hz.
dUse the result in a and the formula for the radius of the nth orbit to deduce Keplerâs third law for the hydrogen electron.
17The Balmer series consists of transitions from a state n down to the state n = 2.
aShow that the wavelengths, in nanometers, of the Balmer series are given by Î»=365Ãn2n2â4.
bEvaluate this wavelength for the transition from n = 3.
cCalculate the frequency of revolution of the electron in the states n = 3 and n = 2 and compare with the frequency of the photon in the transition from n = 3 to n = 2.
18A planet (mass m) circling the sun (mass M) in a circular orbit has its angular momentum quantised as in the Bohr model.
aShow that the allowed radii of the orbit are given by r=n2h24Ï2GMm2.
bApply this to the earth whose orbit radius is 1.5 Ã 1011 m. What value of n corresponds to this radius?
cIs the assumption of a quantised angular momentum meaningful in this case?
19A particle of electric charge q moving in a circle in a region of magnetic field B has its angular momentum quantised as in the Bohr model. Show that the allowed radii are given by r=nh2ÏqBââââââ.
20The Bohr model can also be applied to the singly ionised atom of helium. By retracing the derivations in the Bohr model for hydrogen, determine for the n = 1 state of ionised helium:
athe radius in terms of the Bohr radius a0,
bthe energy in eV.
21Show that when n is very large, in the transition from the state n + 1 to the state n the frequency of the emitted photon equals the frequency of revolution of the electron in the state n.
22For the really interested: Bohr introduced his correspondence principle which states that as the quantum number of a system becomes very large, the quantum system behaves like the corresponding classical system. In the case of the hydrogen atom, the corresponding classical system would be an orbiting electron radiating EM waves with a frequency equal to the frequency of revolution of the electron around the proton, exactly as predicted by Maxwellâs classical theory. We can use this to show why angular momentum has to be quantised!
Consider an electron that orbits a proton with angular frequency Ï.
aShow that the angular momentum can be written as L = mÏr2.
bApplying Coulombâs law to the orbiting electron, show that L3=mk2e4Ï.
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cShow that r=L2mke2.
dHence show that the total energy of the electron is given by E=âmk2e42L2 and therefore dEdL=mk2e4L3.
eUsing your answer to part b, deduce that dEdL=Ï.
fTake dE to be the energy of the photon emitted when an electron makes a transition from a state n + 1 to a state n where n is large. According to the correspondence principle this photon must have a frequency equal to the frequency of revolution of the electron, i.e. dE=hf=hÏ2Ï.
Hence deduce that hÏ2Ï=ÏdL i.e. dL=h2Ï.
This says that the angular momentum increases in steps of h2Ïâ², i.e. it is quantised.
