A repair by new rules
In 1913 the Danish physicist Niels Bohr faced the failure of the last lesson. He kept Rutherford's nucleus and the circling electron. Then he did something bold. He did not derive the electron's behaviour from the old laws. Instead he laid down three new rules, his postulates, and let the results judge them.
First, the electron moves round the nucleus only on certain circular orbits, called stationary orbits. On them it does not radiate. Second, on a stationary orbit its angular momentum is a whole number times \(h/2\pi\): \(mvr = nh/2\pi\), with \(n = 1, 2, 3\) and so on. Third, the atom emits or absorbs light only when the electron jumps between stationary orbits. Each jump gives out or takes in one photon, of energy \(h\nu = E_i - E_f\).
The first postulate: orbits without radiation
A stationary orbit is an orbit on which the electron keeps a fixed energy. The word stationary describes the orbit and its energy, not the electron. The electron moves fast on it, at over two thousand kilometres per second in hydrogen. The first postulate simply says that on these orbits the classical rule of radiation does not apply. So the atom does not collapse.
The second postulate: angular momentum in steps
Which orbits are stationary? Bohr's answer uses the angular momentum \(L = mvr\) of the electron, where \(m\) is its mass, \(v\) its speed and \(r\) the radius of the orbit. Only orbits with \(L\) equal to a whole number times \(h/2\pi\) are allowed. Here \(h = 6.63 \times 10^{-34}\,\text{J s}\) is Planck's constant, from the chapter on Photons and the Photoelectric Effect.
The whole number \(n\) is the principal quantum number of the orbit. The smallest is \(n = 1\), so the angular momentum can never be less than \(h/2\pi\), and there is a smallest orbit. Between \(n = 1\) and \(n = 2\) no orbit exists at all. The angular momentum is quantised: it comes only in steps.
The third postulate: one jump, one photon
The electron can change its energy only by jumping from one stationary orbit to another. It jumps down by giving out one photon, and up by taking in one photon. The photon carries exactly the difference of the two energies. With \(E_i\) the energy before and \(E_f\) after, the photon has frequency \(\nu\) given by \(h\nu = E_i - E_f\).
In Fig 2.1 the orbits give no light while the electron circles on them. Light appears only at the jump, as one photon. Since the orbits have fixed energies, the jumps have fixed energies too, and so the photons can have only certain frequencies. That is why hydrogen gives sharp lines. The later lessons of this chapter turn this into numbers.
Find the angular momentum of the electron of hydrogen on the orbits \(n = 1\) and \(n = 3\). By how much does it change when the electron jumps from \(n = 3\) to \(n = 1\)?
\(h/2\pi = 6.63 \times 10^{-34}/(2\pi) = 1.055 \times 10^{-34}\,\text{J s}\).
For \(n = 1\), \(L = 1.055 \times 10^{-34}\,\text{J s}\). For \(n = 3\), \(L = 3 \times 1.055 \times 10^{-34} = 3.17 \times 10^{-34}\,\text{J s}\).
The jump from 3 to 1 lowers \(L\) by \(2h/2\pi = 2.11 \times 10^{-34}\,\text{J s}\). The angular momentum can change only in whole steps of \(h/2\pi\).
Values used: \(h = 6.63 \times 10^{-34}\,\text{J s}\).
According to Bohr, why does an electron on a stationary orbit not radiate?
Which of these can be the angular momentum of the electron of a hydrogen atom in Bohr's model?
An electron of hydrogen jumps from \(n = 4\) straight down to \(n = 2\). How many photons does this single jump give out?
Bohr had worked in Rutherford's laboratory in Manchester in 1912, and he believed in the nucleus. Back in Copenhagen he tried to make the electrons stable, without much success. Early in 1913 a colleague asked him about Balmer's formula for the lines of hydrogen, which Bohr had not studied. Bohr said later that as soon as he saw the formula, the whole thing became clear to him. His paper on the hydrogen atom appeared that July.
Where the second postulate comes from
Bohr's second rule looked like a guess. Eleven years later Louis de Broglie gave a reason for it. In the chapter on Matter Waves you meet his idea: a moving electron behaves as a wave of wavelength \(\lambda = h/mv\), its de Broglie wavelength. On a circular orbit this wave runs round the circle and meets itself.
A wave on a closed loop lasts only if it comes back in step with itself. Then every trip adds to the last, and a steady standing wave forms. It is like the standing waves of the chapter on Standing Waves on a String. If it comes back out of step, successive trips cancel. So the circumference must hold a whole number of wavelengths.
\[ 2\pi r = n\lambda \]
Why. a whole number \(n\) of wavelengths fits round the orbit, as in Fig 2.2.
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Your turn. put in the de Broglie wavelength. Which line is it?
\[ mvr = \dfrac{nh}{2\pi} \]
Read it. multiply both sides by \(mv/2\pi\). This is Bohr's second postulate.
So Bohr's rule is a condition for standing waves. The orbit \(n = 1\) holds one wavelength, the orbit \(n = 2\) holds two, and so on. An orbit that would hold three and a half wavelengths cannot exist, because its wave would cancel itself. The whole number \(n\) is simply the number of waves round the orbit.
The lowest orbit of hydrogen has radius \(r_1 = 0.529\,\text{Å}\) (\(1\,\text{Å} = 10^{-10}\,\text{m}\)), and the electron moves on it at \(v_1 = 2.18 \times 10^6\,\text{m/s}\). The next lesson derives both. Find the de Broglie wavelength of the electron in two ways, and compare them.
From the orbit: one wavelength fits round it, so \(\lambda = 2\pi r_1 = 2\pi \times 0.529 = 3.32\,\text{Å}\).
From the speed: \(\lambda = h/mv_1 = 6.63 \times 10^{-34}/(9.1 \times 10^{-31} \times 2.18 \times 10^6) = 3.34 \times 10^{-10}\,\text{m} = 3.34\,\text{Å}\).
The two agree to within the rounding of the constants, about half a percent. The wave on the lowest orbit is about one third of a nanometre long.
How many de Broglie wavelengths fit round the orbit \(n = 3\) of hydrogen?
Put the steps of the standing-wave argument for Bohr's rule in order.
Bohr's rule can be applied to other circular motions, and the ladder it gives depends on the force. For a particle held by a spring-like force, \(U = \tfrac12 m\omega^2 r^2\), the force balance gives \(v = \omega r\). The rule then gives \(r^2 \propto n\) and energies \(E_n = nh\omega/2\pi\), equally spaced. The atom's ladder, as the next lessons show, crowds together instead.
- Reading "stationary orbit" as an electron at rest.The orbit and its energy are fixed; the electron moves fast on it.
- Saying that \(n\) counts the radius.\(n\) counts the wavelengths round the orbit and the steps of \(h/2\pi\). The radius grows as \(n^2\).
- Thinking a jump from \(n = 4\) to \(n = 2\) gives two photons.One jump gives one photon. Two photons need two jumps, 4 to 3 and 3 to 2.