7.P — Worked Problems: Quantum Mechanics

Fourteen problems across Part 7. Every solution shows the arithmetic. Attempt each before opening it.

Problem 1 — Photons in a laser pointer

A 5.0 mW red laser pointer emits at 650 nm. How many photons per second, and what force does the beam exert on a mirror?

Solution

Photon energy:

E = \frac{hc}{\lambda} = \frac{(6.626\times10^{-34})(3\times10^{8})}{650\times10^{-9}} = \frac{1.988\times10^{-25}}{6.5\times10^{-7}} = 3.058\times10^{-19}\ \text{J}

= 1.91\ \text{eV}

Photons per second:

N = \frac{P}{E} = \frac{5.0\times10^{-3}}{3.058\times10^{-19}} = 1.64\times10^{16}\ \text{s}^{-1}

Force on a mirror. Each photon carries p = E/c, and reflection reverses it, so the momentum transferred is 2E/c:

F = \frac{2P}{c} = \frac{2(5.0\times10^{-3})}{3\times10^{8}} = 3.3\times10^{-11}\ \text{N}

What to notice. Sixteen thousand million million photons a second, which is why light looks continuous — the granularity is invisible at any ordinary intensity. And the force is 33 piconewtons, about the weight of a bacterium. That is nonetheless enough to trap and move single cells with a focused beam, which is what optical tweezers do (Chapter 4.7).

Problem 2 — Photoelectric threshold

Light of 300 nm ejects electrons from a metal with maximum kinetic energy 1.2 eV. Find the work function, the threshold wavelength, and the stopping voltage.

Solution

Photon energy:

E = \frac{1240\ \text{eV nm}}{300\ \text{nm}} = 4.13\ \text{eV}

using the very useful shortcut hc = 1240 eV·nm.

Work function:

\phi = E - KE_{\max} = 4.13-1.20 = 2.93\ \text{eV}

Threshold wavelength:

\lambda_0 = \frac{1240}{2.93} = 423\ \text{nm}

Stopping voltage — the reverse voltage that just stops the fastest electrons:

eV_s = KE_{\max} \quad\Longrightarrow\quad V_s = 1.20\ \text{V}

What to notice. Memorise hc = 1240 eV·nm; it converts wavelength to photon energy in one step and appears constantly. Note also that 423 nm is violet, so this metal responds to violet and ultraviolet and is blind to everything from blue to red, however bright. That is the threshold behaviour classical physics cannot produce.

Problem 3 — De Broglie wavelength of a thermal neutron

Find the wavelength of a neutron in thermal equilibrium at 300 K, and comment on its usefulness for crystallography.

Solution

A thermal neutron has E = \frac{3}{2}k_BT:

E = 1.5(1.381\times10^{-23})(300) = 6.21\times10^{-21}\ \text{J} = 0.0388\ \text{eV}

p = \sqrt{2mE} = \sqrt{2(1.675\times10^{-27})(6.21\times10^{-21})} = \sqrt{2.081\times10^{-47}} = 4.56\times10^{-24}

\lambda = \frac{h}{p} = \frac{6.626\times10^{-34}}{4.56\times10^{-24}} = 1.45\times10^{-10}\ \text{m} = 1.45\ \text{Å}

What to notice. 1.45 Å is almost exactly an interatomic spacing, which is why neutron diffraction works so beautifully and why research reactors are built for it. The coincidence is not accidental: room temperature is set by the same chemistry that sets bond lengths, both being governed by electron-volt-scale energies.

Neutrons have two advantages over X-rays. They scatter off nuclei rather than electron clouds, so light elements like hydrogen — invisible to X-rays — show up strongly, which is essential for studying water, polymers and proteins. And they carry a magnetic moment, so they reveal magnetic ordering directly.

Problem 4 — Electron in a nanowire

An electron is confined to a one-dimensional wire 20 nm long. Find the first three energy levels and the wavelength of light emitted in the n=2 \to n=1 transition.

Solution

E_n = \frac{n^2h^2}{8mL^2}

E_1 = \frac{(6.626\times10^{-34})^2}{8(9.109\times10^{-31})(2.0\times10^{-8})^2} = \frac{4.390\times10^{-67}}{8(9.109\times10^{-31})(4\times10^{-16})}

= \frac{4.390\times10^{-67}}{2.915\times10^{-45}} = 1.506\times10^{-22}\ \text{J} = 9.40\times10^{-4}\ \text{eV}

E_2 = 4E_1 = 3.76\ \text{meV}, \qquad E_3 = 9E_1 = 8.46\ \text{meV}

Transition energy:

\Delta E = 3E_1 = 2.82\ \text{meV}

\lambda = \frac{1240\ \text{eV nm}}{2.82\times10^{-3}\ \text{eV}} = 4.4\times10^{5}\ \text{nm} = 0.44\ \text{mm}

Far infrared, almost microwave.

What to notice. Compare with k_BT at room temperature, 25.9 meV. The level spacing is ten times smaller than thermal energy, so at room temperature the electron is smeared across many levels and the quantisation is washed out. To see discrete levels in a 20 nm wire you must cool it below about 4 K — which is exactly why quantum transport experiments run in dilution refrigerators. Shrink the wire to 2 nm and the spacing rises by a factor of 100 to 0.28 eV, comfortably above thermal energy at room temperature, which is why quantum dots work in a television and quantum wires do not.

Problem 5 — Tunnelling through an STM gap

An STM tip is 0.5 nm from a surface with an effective barrier height of 4.0 eV. Find the transmission probability, and how much the current changes if the gap increases by 0.1 nm.

Solution

\kappa = \frac{\sqrt{2m_e V_0}}{\hbar}

taking E \ll V_0:

\kappa = \frac{\sqrt{2(9.109\times10^{-31})(4.0\times1.602\times10^{-19})}}{1.055\times10^{-34}} = \frac{\sqrt{5.836\times10^{-48}}}{1.055\times10^{-34}}

= \frac{2.416\times10^{-24}}{1.055\times10^{-34}} = 2.29\times10^{10}\ \text{m}^{-1}

At 0.5 nm:

2\kappa a = 2(2.29\times10^{10})(5\times10^{-10}) = 22.9

T \sim e^{-22.9} = 1.13\times10^{-10}

At 0.6 nm:

2\kappa a = 27.5, \qquad T \sim e^{-27.5} = 1.14\times10^{-12}

\frac{T(0.5)}{T(0.6)} = e^{4.58} = 97.5

What to notice. Moving the tip 0.1 nm — less than one atomic diameter — changes the current by a factor of a hundred. That is the sensitivity that lets an STM image individual atoms, and it also means essentially all the current flows through the single atom at the very apex of the tip, since any atom even slightly further back contributes a hundred times less. The microscope's resolution comes from an exponential, not from a lens.

Problem 6 — Zero-point energy of a bond

The carbon–hydrogen bond in methane vibrates at 9.0\times10^{13} Hz. Find its zero-point energy, and the zero-point energy of a C–D bond in deuterated methane. Estimate the effect on reaction rate at 300 K.

Solution

C–H:

E_0 = \frac{1}{2}hf = \frac{1}{2}(6.626\times10^{-34})(9.0\times10^{13}) = 2.98\times10^{-20}\ \text{J} = 0.186\ \text{eV}

C–D. The frequency goes as \sqrt{k/\mu} with \mu the reduced mass. For C–H, \mu \approx 1\times12/13 = 0.923 u; for C–D, \mu \approx 2\times12/14 = 1.714 u.

\frac{f_{CD}}{f_{CH}} = \sqrt{\frac{0.923}{1.714}} = \sqrt{0.5385} = 0.734

E_0(CD) = 0.734\times0.186 = 0.137\ \text{eV}

Difference:

\Delta E_0 = 0.186-0.137 = 0.049\ \text{eV} = 4.7\ \text{kJ/mol}

Since the C–D bond starts lower in the well, it needs 0.049 eV more to break. The rate ratio, from the Arrhenius factor (Chapter 10.6):

\frac{k_{CH}}{k_{CD}} = e^{\Delta E_0/k_BT} = e^{0.049/0.0259} = e^{1.89} = 6.6

What to notice. Replacing one hydrogen with deuterium — a change in nothing but the number of neutrons — slows the reaction sevenfold. This is a purely quantum effect with no classical counterpart, and it is one of the most useful diagnostic tools in physical organic chemistry: measure the ratio, and if it is around 6–7 you know the rate-limiting step involves breaking that bond.

It is also used in medicine. Deuterated drugs, where a key C–H is replaced by C–D, are metabolised more slowly by the liver, so they last longer in the body. The first deuterated drug was approved in 2017.

Problem 7 — Uncertainty in a nucleus

Estimate the kinetic energy of a proton confined to a nucleus of radius 5.0 fm, and compare with typical nuclear binding energies.

Solution

\Delta x \approx 5.0\times10^{-15}\ \text{m}

\Delta p \approx \frac{\hbar}{2\Delta x} = \frac{1.055\times10^{-34}}{10^{-14}} = 1.055\times10^{-20}\ \text{kg m/s}

KE \approx \frac{(\Delta p)^2}{2m_p} = \frac{(1.055\times10^{-20})^2}{2(1.673\times10^{-27})} = \frac{1.113\times10^{-40}}{3.346\times10^{-27}} = 3.33\times10^{-14}\ \text{J}

= 0.21\ \text{MeV}

Using \Delta p \approx \hbar/\Delta x instead gives about 0.83 MeV, and a more careful treatment gives a few MeV.

What to notice. A few MeV, which is the same order as the nuclear binding energy of about 8 MeV per nucleon. The nucleus is barely bound, in the sense that the confinement energy is a substantial fraction of the binding — which is why nuclear structure is difficult and why small changes in the strong force parameters would make most nuclei unstable.

Compare with Chapter 7.5's electron calculation, which gave 39 MeV for an electron in the same volume, far exceeding the binding. The proton stays because it is 1836 times heavier and KE \propto 1/m. The mass ratio is what decides whether a particle can be confined in a nucleus.

Problem 8 — Hydrogen transitions

Find the wavelength of the n=3 \to n=2 transition in hydrogen, and the shortest wavelength in the Lyman series.

Solution

E_n = -\frac{13.606}{n^2}\ \text{eV}

n=3 \to n=2:

\Delta E = 13.606\left(\frac{1}{4}-\frac{1}{9}\right) = 13.606\times\frac{9-4}{36} = 13.606\times0.13889 = 1.890\ \text{eV}

\lambda = \frac{1240}{1.890} = 656.2\ \text{nm}

Red — the H-alpha line, which is the colour of every emission nebula in every astronomical photograph.

Lyman series limit is n=\infty \to n=1:

\Delta E = 13.606\ \text{eV} \quad\Longrightarrow\quad \lambda = \frac{1240}{13.606} = 91.1\ \text{nm}

Far ultraviolet.

What to notice. Every Lyman photon has enough energy to ionise another hydrogen atom, since the series limit is the ionisation energy. So Lyman radiation cannot travel far through neutral hydrogen — it is absorbed almost immediately. This is why the early universe was opaque before recombination and why intergalactic hydrogen leaves a forest of absorption lines in quasar spectra (Chapter 12.5).

Note also the accuracy: 656.2 nm against Balmer's measured 656.3. The small discrepancy is the finite proton mass, corrected by using the reduced mass instead of the electron mass, which shifts everything by a factor of 1/(1+m_e/m_p) = 0.99946.

Problem 9 — Counting states in a shell

How many electrons fit in the n = 4 shell, and how many in the 4d subshell specifically?

Solution

For n = 4: \ell runs from 0 to 3.

\ell	Label	m_\ell values	Orbitals	Electrons
0	4s	1	1	2
1	4p	3	3	6
2	4d	5	5	10
3	4f	7	7	14

\text{Total} = 2+6+10+14 = 32 = 2n^2 = 2(16)$$ ✔ **The $4d$ subshell holds 10.** **What to notice.** The $2n^2$ formula and the subshell capacities 2, 6, 10, 14 are the entire numerical skeleton of the periodic table. The ten in the d subshell is why there are exactly ten transition metals in each row, and the fourteen in f is why the lanthanides and actinides each have fourteen members. **But the shells do not fill in order of $n$**, because of screening (Chapter 7.6): 4s fills before 3d, which is why the fourth row of the periodic table has 18 elements rather than 32. Chapter 9.3 works through the filling order and its exceptions.

Problem 10 — A Bell test with real numbers

In a CHSH experiment the measured correlations are E(a,b) = 0.68, E(a,b') = -0.66, E(a',b) = 0.70, E(a',b') = 0.69. Is local realism violated?

Solution

S = E(a,b)-E(a,b')+E(a',b)+E(a',b')

using the standard sign arrangement:

S = 0.68-(-0.66)+0.70+0.69 = 0.68+0.66+0.70+0.69 = 2.73

Local realism requires |S| \leq 2. Measured: 2.73.

Violated, and by a wide margin. The quantum maximum is 2\sqrt{2} = 2.828, so this experiment reaches 97 % of the theoretical limit.

How significant? With a typical statistical error of 0.01 on each correlation, the error on S is about 0.01\times2 = 0.02, so:

\frac{2.73-2.00}{0.02} = 36\ \text{standard deviations}

What to notice. Real experiments routinely reach tens or hundreds of standard deviations, which is why this is not a marginal result. For comparison, particle physics calls 5 standard deviations a discovery.

The gap between 2.73 and 2.828 is due to imperfect detectors, imperfect entanglement and imperfect alignment. No experiment has ever exceeded 2\sqrt{2}, which is itself interesting: quantum mechanics violates local realism but does not violate it as much as pure logic would allow.

Problem 11 — Casimir force in a MEMS device

Two parallel gold plates of area 100 μm × 100 μm are separated by 100 nm in a MEMS device. Find the Casimir force and compare with the weight of the upper plate if it is 1 μm thick.

Solution

\frac{F}{A} = \frac{\pi^2\hbar c}{240d^4} = \frac{(9.8696)(1.055\times10^{-34})(3\times10^{8})}{240(10^{-7})^4}

= \frac{3.124\times10^{-25}}{240\times10^{-28}} = \frac{3.124\times10^{-25}}{2.4\times10^{-26}} = 13.0\ \text{Pa}

A = (10^{-4})^2 = 10^{-8}\ \text{m}^2

F = 13.0\times10^{-8} = 1.30\times10^{-7}\ \text{N}

Weight of the plate. Gold density 19,300 kg/m³, volume 10^{-8}\times10^{-6} = 10^{-14} m³:

m = 19300\times10^{-14} = 1.93\times10^{-10}\ \text{kg}

W = mg = 1.89\times10^{-9}\ \text{N}

\frac{F_{\text{Casimir}}}{W} = \frac{1.30\times10^{-7}}{1.89\times10^{-9}} = 69

What to notice. The Casimir force is seventy times the plate's own weight. In micro-mechanical devices the vacuum is a dominant force, and it is attractive, so it pulls parts together and holds them there. This "stiction" is a genuine and well-documented failure mode in MEMS manufacture, and designers avoid sub-100 nm gaps between parallel surfaces specifically because of it.

Note the d^{-4}: double the gap to 200 nm and the force falls by sixteen, to about four times the weight. Go to 1 μm and it is 10^{-4} of the weight and irrelevant.

Problem 12 — Range of a force

The pion has mass 139.6 MeV/c². Find the range of the force it mediates. Then find what mass a carrier would need for a force of range 1 mm.

Solution

R = \frac{\hbar}{mc} = \frac{\hbar c}{mc^2} = \frac{197\ \text{MeV fm}}{139.6\ \text{MeV}} = 1.41\ \text{fm}

Which is the observed range of the nuclear force, and the basis of Yukawa's 1935 prediction.

For R = 1 mm = 10^{12} fm:

mc^2 = \frac{197\ \text{MeV fm}}{10^{12}\ \text{fm}} = 1.97\times10^{-10}\ \text{MeV} = 1.97\times10^{-4}\ \text{eV}

About 0.2 milli-electron-volts.

What to notice. Searches for a "fifth force" at millimetre ranges are searches for a boson of about this mass, and the constraint is severe: torsion-balance experiments have shown that gravity behaves normally down to about 50 μm, ruling out any new force of comparable strength above that scale.

The same arithmetic constrains the photon's mass. Since electromagnetism has been verified over astronomical distances, the photon mass must be below about 10^{-18} eV, and measurements of the solar wind's magnetic field push it below 10^{-27} eV. The photon is massless to about fifty decimal places of any reasonable comparison.

Problem 13 — Making a proton–antiproton pair

What minimum kinetic energy must a proton have, striking a stationary proton, to produce a proton–antiproton pair? The reaction is p+p \to p+p+p+\bar{p}.

Solution

Use the invariant mass method from Chapter 6.5. The threshold is when all four final particles move together with no relative motion, so the invariant mass of the system equals 4m_pc^2.

Final state invariant mass:

Mc^2 = 4(938.3) = 3753\ \text{MeV}

Initial state. Beam proton with total energy E, target at rest with energy m_pc^2:

(Mc^2)^2 = (E+m_pc^2)^2 - (pc)^2

Using (pc)^2 = E^2-(m_pc^2)^2:

(Mc^2)^2 = E^2+2Em_pc^2+(m_pc^2)^2-E^2+(m_pc^2)^2 = 2Em_pc^2+2(m_pc^2)^2

Solve for E:

E = \frac{(Mc^2)^2-2(m_pc^2)^2}{2m_pc^2} = \frac{(3753)^2-2(938.3)^2}{2(938.3)}

= \frac{1.4085\times10^{7}-1.7608\times10^{6}}{1876.6} = \frac{1.2324\times10^{7}}{1876.6} = 6567\ \text{MeV}

Kinetic energy:

KE = 6567-938 = 5629\ \text{MeV} = 5.63\ \text{GeV}

What to notice. Making a pair whose combined rest energy is 1.88 GeV requires a 5.63 GeV beam — three times as much — because most of the energy goes into the centre-of-mass motion, as Chapter 6.P Problem 5 showed.

This number determined the design of a machine. The Bevatron at Berkeley was built in 1954 with a 6.2 GeV proton beam, chosen deliberately to exceed this threshold with margin. It found the antiproton in 1955, and Segrè and Chamberlain received the Nobel Prize in 1959. An accelerator was designed around one line of relativistic kinematics.

Problem 14 — Compton shift and the electron's recoil

A 0.100 nm X-ray photon scatters at 60° from a free electron. Find the scattered wavelength, the energy given to the electron, and the electron's speed.

Solution

Wavelength shift:

\Delta\lambda = \lambda_C(1-\cos 60°) = (2.426\times10^{-12})(1-0.5) = 1.213\times10^{-12}\ \text{m}

\lambda' = 0.100\ \text{nm}+0.001213\ \text{nm} = 0.101213\ \text{nm}

Photon energies:

E = \frac{1240}{0.100} = 12{,}400\ \text{eV}, \qquad E' = \frac{1240}{0.101213} = 12{,}251\ \text{eV}

Energy to the electron:

KE = 12400-12251 = 149\ \text{eV}

Electron speed. Since 149 eV is far below m_ec^2 = 511 keV, use the non-relativistic form:

v = \sqrt{\frac{2KE}{m_e}} = \sqrt{\frac{2(149)(1.602\times10^{-19})}{9.109\times10^{-31}}} = \sqrt{5.24\times10^{13}} = 7.24\times10^{6}\ \text{m/s}

About 2.4 % of light speed — relativistic corrections would change this by about 0.03 %, so the non-relativistic answer is fine.

What to notice. The photon lost only 1.2 % of its energy, because the shift is a fixed 1.2 pm out of 100 pm. To transfer a large fraction you need a photon whose wavelength is comparable to \lambda_C = 2.4 pm, meaning an energy near m_ec^2 = 511 keV — a gamma ray.

This is why Compton scattering dominates gamma-ray attenuation in the 100 keV to 10 MeV range and matters enormously in radiation shielding and radiotherapy dose calculations. Below that range the photoelectric effect dominates; above about 10 MeV, pair production takes over.