The photon's push
A photon carries momentum as well as energy. In the chapter on Photons and the Photoelectric Effect you found that a photon of energy \(E\) has momentum \(p = E/c\). So when an atom at rest gives out a photon, momentum must still add up to zero. The atom moves off the other way with momentum \(E/c\), like a gun recoiling from its bullet.
The recoiling atom takes some kinetic energy, \(K_R = p^2/2M\), where \(M\) is its mass. With \(p = E/c\), this is \(K_R = E^2/2Mc^2\). The energy \(\Delta E\) released by the jump is shared between the photon and the recoil. So the photon gets a little less than the gap: \(E = \Delta E - K_R\).
\[ \dfrac{E}{c} = Mv \]
Why. the atom starts at rest, so the photon's momentum and the atom's momentum are equal and opposite.
\[ K_R = \dfrac{(Mv)^2}{2M} = \dfrac{E^2}{2Mc^2} \]
Why. kinetic energy is \(p^2/2M\), with \(p = E/c\).
\[ \Delta E = E + \dfrac{E^2}{2Mc^2} \]
Why. energy is conserved: the gap pays for the photon and the recoil.
\[ E \approx \Delta E - \dfrac{\Delta E^2}{2Mc^2} = \Delta E\left(1 - \dfrac{\Delta E}{2Mc^2}\right) \]
Read it. the recoil is tiny, so put \(E \approx \Delta E\) inside the small term.
A hydrogen atom at rest, of mass \(1.67 \times 10^{-27}\,\text{kg}\), falls from \(n = 2\) to \(n = 1\) and emits a photon of \(10.2\,\text{eV}\). Find the momentum of the photon, the recoil speed of the atom, and its recoil kinetic energy.
Momentum is conserved: the atom's momentum equals the photon's, \(p = E/c\), in the opposite direction.
\(p = E/c\), \(v = p/M\), \(K_R = \tfrac12 Mv^2\), or \(K_R = E^2/2Mc^2\).
\(p = 10.2 \times 1.6 \times 10^{-19}/3 \times 10^8 = 5.44 \times 10^{-27}\,\text{kg m/s}\). \(v = 5.44 \times 10^{-27}/1.67 \times 10^{-27} = 3.3\,\text{m/s}\). \(K_R = (10.2)^2/(2 \times 939 \times 10^6) = 5.5 \times 10^{-8}\,\text{eV}\), or \(8.9 \times 10^{-27}\,\text{J}\).
The atom recoils at about \(3.3\,\text{m/s}\), walking pace, and takes only five parts in a thousand million of the energy. Values used: \(e = 1.6 \times 10^{-19}\,\text{C}\), \(c = 3 \times 10^8\,\text{m/s}\), mass of the hydrogen atom \(1.67 \times 10^{-27}\,\text{kg}\).
In absorption the recoil works the other way. A photon absorbed by an atom at rest pushes the atom along its own direction. The photon must therefore supply the gap \(\Delta E\) and the recoil energy as well: \(E = \Delta E + K_R\), slightly more than the gap. For hydrogen the extra is again only \(5.5 \times 10^{-8}\,\text{eV}\).
An atom at rest drops through an energy gap \(\Delta E\) and emits a photon of energy \(E\). How do \(E\) and \(\Delta E\) compare?
A hydrogen atom at rest emits the red Balmer photon of \(1.89\,\text{eV}\). Find its recoil speed in m/s. Take the mass of the atom as \(1.67 \times 10^{-27}\,\text{kg}\), \(e = 1.6 \times 10^{-19}\,\text{C}\), \(c = 3 \times 10^8\,\text{m/s}\).
The nucleus moves too
So far the nucleus has stayed fixed. A real nucleus of mass \(M\) feels the same pull as the electron, in the opposite direction, so it moves too. Both circle their common centre of mass, always on opposite sides of it. The two stars of a binary do the same, in the chapter on Gravitational Energy and Orbits.
The centre of mass stays at rest, so \(m r_e = M r_N\), where \(r_e\) and \(r_N\) are the radii of the two circles. The heavier body moves on the smaller circle. The two-body problem then becomes a one-body problem. The electron's mass \(m\) is replaced by the reduced mass, \(\mu = mM/(m + M)\), and every formula of Bohr's model holds as before.
The energy of a level is proportional to the mass in Bohr's formula, and the radius to one over the mass. Since \(\mu\) is a little less than \(m\), every level becomes slightly shallower and every orbit slightly larger. For hydrogen, \(M = 1836\,m\), so \(\mu = (1836/1837)\,m = 0.99946\,m\). The ground state moves from \(-13.60\) to \(-13.59\,\text{eV}\), a change of \(0.054\,\%\).
A spectrometer measures wavelengths far better than one part in two thousand, so this tiny shift matters. Deuterium, heavy hydrogen, has a nucleus twice as heavy, \(3672\,m\). Its reduced mass is a little larger, so its levels are a little deeper and its lines a little shorter. Its red line lies \(0.18\,\text{nm}\) below hydrogen's. In 1932 Harold Urey found deuterium by seeing exactly this faint second line.
(a) Positronium is an electron bound to a positron, a particle of the same mass and opposite charge. Find its reduced mass, ground-state energy and ionisation energy. Find also its ground-state separation and the wavelength of its line from 2 to 1. (b) A muon, of mass \(207\,m\) and charge \(-e\), replaces the electron of hydrogen (proton mass \(1836\,m\)). Find the reduced mass, the ground-state energy and the radius.
(a) \(\mu = m \cdot m/2m = m/2\). Every energy halves: \(E_1 = -6.8\,\text{eV}\), so the ionisation energy is \(6.8\,\text{eV}\). Every radius doubles: \(r_1 = 2 \times 0.529 = 1.06\,\text{Å}\).
The 2 to 1 photon has half of hydrogen's \(10.2\,\text{eV}\), \(5.1\,\text{eV}\), so \(\lambda = 1240/5.1 = 243\,\text{nm}\), twice hydrogen's \(121.6\,\text{nm}\).
(b) \(\mu = 207 \times 1836/(207 + 1836)\,m = 186\,m\). So \(E_1 = -13.6 \times 186 = -2530\,\text{eV} = -2.53\,\text{keV}\), and \(r_1 = 0.529/186 = 2.84 \times 10^{-3}\,\text{Å}\), or \(284\,\text{fm}\).
The muon is bound 186 times more tightly and sits 186 times closer, well inside the electron's orbit but still outside the nucleus.
Find the wavelength, in nm, of the line from \(n = 3\) to \(n = 2\) in positronium. Hydrogen's line for this jump is at \(656.5\,\text{nm}\).
The red line of deuterium is at a slightly shorter wavelength than the red line of ordinary hydrogen. Why?
Exciting an atom by a collision
An atom can also be excited by a blow. A moving particle strikes it, and some kinetic energy goes into lifting its electron. But an atom cannot take energy in any amount. Its internal energy changes only by a gap of its ladder: \(10.2\,\text{eV}\), \(12.09\,\text{eV}\) and so on for hydrogen. If the blow cannot supply a whole gap, the atom takes nothing, and the collision is elastic.
How much of the kinetic energy can a collision hand over? Momentum decides. The pair cannot stop dead after a head-on blow, because the momentum brought in must still be there. The most kinetic energy is lost when the two move on together, as in the chapter on Collisions. So only part of \(K\) is available for the atom.
\[ p = \sqrt{2mK} \]
Why. the particle's momentum, from \(K = p^2/2m\). The atom starts at rest.
\[ K' = \dfrac{p^2}{2(m + M)} = \dfrac{m}{m + M}\,K \]
Why. at the least kinetic energy after the blow, the two move together with the same total momentum \(p\).
\[ K - K' = \dfrac{M}{m + M}\,K \]
Why. this is the most energy the collision can hand over to the atom.
\[ \dfrac{M}{m + M}\,K_{\min} = \Delta E \;\Rightarrow\; K_{\min} = \Delta E\,\dfrac{m + M}{M} \]
Read it. at the threshold the energy available just equals the gap.
For a neutron or a hydrogen atom striking a hydrogen atom, \(m = M\). Then only half of \(K\) is available, and excitation to \(n = 2\) needs \(K \ge 2 \times 10.2 = 20.4\,\text{eV}\). The threshold energy is the least kinetic energy that can do it. At exactly the threshold the two move on together with the other \(10.2\,\text{eV}\) as kinetic energy.
Find the least kinetic energy in four cases. (a) A neutron ionises a hydrogen atom at rest. (b) An electron excites a hydrogen atom at rest to \(n = 2\). (c) A helium atom (\(4\,\text{u}\)) excites a hydrogen atom (\(1\,\text{u}\)) at rest to \(n = 2\). (d) A hydrogen atom excites a \(\text{He}^+\) ion (\(4\,\text{u}\)) at rest to \(n = 2\).
Use \(K_{\min} = \Delta E(m + M)/M\), with \(m\) the moving particle and \(M\) the target at rest.
(a) \(m = M\): \(2 \times 13.6\). (b) \(m/M = 1/1836\): \(10.2 \times 1837/1836\). (c) \(5/1 \times 10.2\). (d) \(5/4 \times 40.8\).
(a) \(27.2\,\text{eV}\). (b) \(10.2\,\text{eV}\), more exactly \(10.21\,\text{eV}\). (c) \(51\,\text{eV}\). (d) \(51\,\text{eV}\).
An electron needs little more than the gap, because the heavy atom barely recoils. A heavy projectile on a light target must bring five times the gap. Values used: hydrogen levels \(-13.6/n^2\,\text{eV}\), \(M = 1836\,m\) for hydrogen.
Two more cases complete the picture. If two hydrogen atoms move towards each other with equal speeds, their total momentum is zero. They can then stop dead, and all their kinetic energy is available: each needs only \(10.2/2 = 5.1\,\text{eV}\). And a photon, unlike a particle, is absorbed whole and disappears. So it must fit the gap exactly, while a particle may bring any energy above the threshold and carry the surplus away.
A hydrogen atom with kinetic energy \(25\,\text{eV}\) strikes a hydrogen atom at rest, head-on. What is the most energy, in eV, that can go into exciting the struck atom?
A neutron of \(16\,\text{eV}\) strikes a hydrogen atom at rest head-on. \(16\,\text{eV}\) is more than the \(10.2\,\text{eV}\) gap. Why can it still not excite the atom?
A photon can excite a hydrogen atom from \(n = 1\) to \(n = 2\) only if its energy is \(10.2\,\text{eV}\). An electron can do it with any energy above the threshold. Explain the difference.
A photon is absorbed whole and disappears, so all of its energy must go into the atom; it must match the gap exactly, apart from the tiny recoil. A particle survives the collision. It can give the atom exactly the gap and carry the rest away as its own kinetic energy, so any energy at or above the threshold will do.
In 1914 James Franck and Gustav Hertz fired electrons through mercury vapour and measured how much energy they kept. Below \(4.9\,\text{eV}\) the electrons bounced off the atoms without loss. At \(4.9\,\text{eV}\) they suddenly lost almost all their energy, and the vapour gave out ultraviolet light of \(253\,\text{nm}\), which is \(1240/4.9\). Atoms took energy only in one fixed step. It was the first direct proof of Bohr's levels, and it won them the Nobel Prize of 1925.
- Comparing the whole kinetic energy of the projectile with the gap.Only \(K M/(m + M)\) is available. For equal masses that is half of \(K\).
- Writing the reduced mass as larger than \(m\).\(\mu = m/(1 + m/M)\) is always less than \(m\). For positronium it is \(m/2\).
- Taking the photon energy equal to the gap in a recoil question.In emission \(E = \Delta E - K_R\); in absorption \(E = \Delta E + K_R\).