Problems 1 · Quantum Mechanics (Phillips)

The instructive answers here are the ones that come out wrong by a factor of ten thousand rather than by a factor of two.

Twelve problems, each with Phillips’ own hint from Hints to selected problems (pp. 249–261) and a full worked solution. Both are collapsed — try the problem first; the boxes are there for when you are stuck or want to check.

Every numerical answer below was obtained by running the code shown with it, so the numbers on this page and the numbers your own script prints should agree.

What this problem set is actually training

Look at the twelve as a group and a pattern appears. Only two (1 and 5) are derivations. Two more (3 and 4) are dimensional analysis. The remaining eight are all the same exercise: work out a de Broglie wavelength , then compare it with the size of something.

That comparison is the entire skill chapter 1 is teaching. Quantum behaviour shows up when $\lambda$ is comparable to the apparatus and vanishes when it is not — so “compute $\lambda$ , then compare” answers why electrons diffract off crystals but cricket balls do not, and it answers it for every case in the same three lines. Problems 8 through 12 are that question asked five times with different numbers.

1 Derive the Compton shift derivation

A photon of energy $\epsilon_i$ and momentum $p_i$ strikes a stationary electron and is scattered through an angle $\theta$ . For a relativistic electron $\epsilon^2 - P^2c^2 = m_e^2c^4$ , and for a photon $\epsilon = pc$ .

Show, by conserving momentum and energy separately and then combining them, that

\Delta\lambda = \frac{h}{m_ec}(1 - \cos\theta)

💡 Phillips' own hint

Rewrite the two conservation laws as $|\mathbf{p}_i - \mathbf{p}_f| = |\mathbf{P}_f - \mathbf{P}_i|$ and $\epsilon_i - \epsilon_f = E_f - E_i$ , then use the energy–momentum relation for each particle. Finish with $\epsilon = hc/\lambda$ .

✓ Worked solution

This is the nine-step derivation stepped through on §1.1 — open the Compton scattering widget there and work along with it rather than reading a static wall of algebra.

The single move that makes it work: square both conservation laws, then subtract. Squaring turns the vector momentum equation into a scalar one containing $\cos\theta$ (the cosine rule), and the subtraction is exactly what eliminates the two unknowns you do not care about — the recoiling electron’s energy and momentum — because they appear only in the combination $E_f^2 - P_f^2c^2$ , which you already know equals $m_e^2c^4$ .

The result contains nothing about the incoming photon. That is the physical content: the shift depends only on the scattering angle and the mass of what you hit.

2 The photoelectric threshold numerical

Electromagnetic radiation on a metal surface may eject electrons — the photoelectric effect — but only above a threshold frequency. Show this follows if the mechanism is a photon–electron interaction and there is a minimum energy needed to free an electron.

Magnesium needs at least $3.68\ \mathrm{eV}$ . Show that light below $8.89\times10^{14}\ \mathrm{Hz}$ produces no photoelectrons, however intense.

💡 Phillips' own hint

If the minimum energy to eject an electron is $W$ , the photon must have frequency at least $W/h$ .

✓ Worked solution

One photon is absorbed by one electron, all or nothing. If the photon’s energy $h\nu$ is less than the work function $W$ , no single absorption can free an electron — and turning up the intensity only sends more photons, each still too feeble. Hence a sharp threshold at $\nu_{\min} = W/h$ :

\nu_{\min} = \frac{W}{h} = \frac{3.68 \times 1.602\times10^{-19}\ \mathrm{J}}{6.626\times10^{-34}\ \mathrm{J\,s}} = 8.90\times10^{14}\ \mathrm{Hz}

That is $\lambda = c/\nu = 337$ nm — near ultraviolet. Magnesium is blind to all visible light, however bright.

Why this killed the wave picture. A classical wave deposits energy continuously, so a dim beam should just take longer to accumulate $W$ — a delay, never a threshold. And a brighter beam of any colour should work. Both predictions are wrong, and the disagreement is qualitative rather than numerical.

Reproduce it in NumPy

h, c, eV = 6.626e-34, 2.998e8, 1.602e-19
W = 3.68 * eV
nu = W / h
print(f"nu_min = {nu:.3e} Hz   ->  lambda_max = {c/nu*1e9:.0f} nm  (near ultraviolet)")
E500 = h*c/500e-9/eV
print(f"a 500 nm photon carries {E500:.2f} eV  <  3.68 eV  ->  no ejection, at any intensity")

printsnu_min = 8.897e+14 Hz -> lambda_max = 337 nm (near ultraviolet) a 500 nm photon carries 2.48 eV < 3.68 eV -> no ejection, at any intensity

3 Stefan–Boltzmann by dimensional analysis dimensional

Black-body radiation can be treated as a gas of photons moving at $c$ with typical energies of order $kT$ , so the emitted intensity $I$ (in $\mathrm{J\,s^{-1}\,m^{-2}}$ ) is a function of $h$ , $c$ and $kT$ only.

Use dimensional analysis to confirm $I \propto T^4$ , and find how $\sigma$ depends on $h$ and $c$ .

💡 Phillips' own hint

$kT$ is an energy and $hc$ is an energy × length, so $(kT)^4/(hc)^3$ is an energy per unit volume, and $c(kT)^4/(hc)^3$ is an energy per second per unit area.

✓ Worked solution

Follow the hint’s construction:

$kT$ — an energy.
$hc$ — an energy × length.
$\dfrac{(kT)^4}{(hc)^3}$ — energy⁴ / (energy³ × length³) = energy per unit volume.
Multiply by a speed and you get energy per second per unit area:

I = f\,\frac{c\,(kT)^4}{(hc)^3} = f\,\frac{k^4}{h^3c^2}\,T^4

with $f$ a pure number dimensional analysis cannot supply. So $I \propto T^4$ falls straight out, and

\sigma = f\,\frac{k^4}{h^3c^2}

Check it against the measured $\sigma$ . Computing $k^4/h^3c^2$ gives $1.391\times10^{-9}$ , and $\sigma = 5.670\times10^{-8}$ , so $f = 40.8$ . The exact value from the full Planck calculation is $2\pi^5/15 = 40.80$ .

Dimensional analysis got everything except a factor of $2\pi^5/15$ — and, as §0.2 put it, that is exactly what it promises: the scale, never the pure number.

Reproduce it in NumPy

import numpy as np
h, c, k = 6.626e-34, 2.998e8, 1.381e-23
combo = k**4 / (h**3 * c**2)
sigma = 5.670e-8
print(f"k^4/(h^3 c^2)     = {combo:.4e}")
print(f"measured sigma    = {sigma:.4e}")
print(f"=> f = sigma / that = {sigma/combo:.2f}")
print(f"exact 2 pi^5 / 15   = {2*np.pi**5/15:.2f}")

printsk^4/(h^3 c^2) = 1.3911e-09 measured sigma = 5.6700e-08 => f = sigma / that = 40.76 exact 2 pi^5 / 15 = 40.80

4 Why the electron must be a wave, not a fast particle dimensional

§1.3 built an atomic size from $\hbar$ , $m_e$ and $e^2/4\pi\epsilon_0$ . Now try it the classical relativistic way instead: build a length from $m_e$ , $e^2/4\pi\epsilon_0$ and $c$ . Show it is possible — and far too small.

💡 Phillips' own hint

$m_ec^2$ is a rest energy and $e^2/4\pi\epsilon_0R$ is a potential energy if $R$ is a length, so $R = e^2/4\pi\epsilon_0 m_ec^2$ . Show this equals $\alpha^2a_0$ , where $\alpha = e^2/4\pi\epsilon_0\hbar c \approx 1/137$ .

✓ Worked solution

Setting the two energies equal gives the only available length:

R = \frac{e^2}{4\pi\epsilon_0 m_ec^2} = 2.82\times10^{-15}\ \mathrm{m}

— the classical electron radius, about the size of a nucleus, roughly 19 000 times smaller than an atom.

The relationship to $a_0$ is worth seeing, because it says exactly where the factor came from:

R = \alpha^2 a_0, \qquad \alpha = \frac{e^2}{4\pi\epsilon_0\hbar c} \approx \frac{1}{137}

Two factors of $\alpha$ , each $1/137$ , giving $1/18\,800$ .

What this really says — the two dimensional arguments differ by four orders of magnitude, and only one is right

This is the sharpest result in the problem set. Both calculations are legitimate dimensional analysis; both produce a length; they differ by four orders of magnitude, and only one is right.

The difference is which constant you brought. Using $c$ says the electron is a fast classical particle and relativity sets its scale. Using $\hbar$ says the electron is a wave and its wavelength sets its scale. Nature picked $\hbar$ , and the evidence is that the answer comes out at $10^{-10}$ m, which is where atoms actually are.

$\alpha$ is precisely the ratio between the two answers — which is why §0.2 could describe it as the number measuring how non-relativistic atomic physics is.

Reproduce it in NumPy

import numpy as np
e, eps0, me, c = 1.602e-19, 8.854e-12, 9.109e-31, 2.998e8
a0, alpha = 0.5292e-10, 1/137.036

R = e**2 / (4*np.pi*eps0*me*c**2)
print(f"R = e^2 / (4 pi eps0 me c^2) = {R:.4e} m")
print(f"a0                          = {a0:.4e} m")
print(f"a0 / R                      = {a0/R:.0f}   (about 4 orders of magnitude)")
print(f"alpha^2 * a0                = {alpha**2*a0:.4e} m   <- same as R")

printsR = e^2 / (4 pi eps0 me c^2) = 2.8174e-15 m a0 = 5.2920e-11 m a0 / R = 18784 (about 4 orders of magnitude) alpha^2 * a0 = 2.8181e-15 m <- same as R

5 When is a classical orbit legitimate? derivation

An electron in a circular orbit about a proton can be described classically if its angular momentum $L \gg \hbar$ . Show this holds when the orbit radius satisfies $r \gg a_0 = 4\pi\epsilon_0\hbar^2/e^2m_e$ .

💡 Phillips' own hint

The Coulomb force $e^2/4\pi\epsilon_0r^2$ supplies the centripetal acceleration $m_ev^2/r$ , and the orbital angular momentum is $L = m_evr$ .

✓ Worked solution

Balance the forces:

\frac{e^2}{4\pi\epsilon_0 r^2} = \frac{m_ev^2}{r} \quad\Longrightarrow\quad v^2 = \frac{e^2}{4\pi\epsilon_0 m_e r}

Then

L = m_evr = m_e r\sqrt{\frac{e^2}{4\pi\epsilon_0m_er}} = \sqrt{\frac{e^2m_er}{4\pi\epsilon_0}}

Requiring $L \gg \hbar$ and squaring:

\frac{e^2m_er}{4\pi\epsilon_0} \gg \hbar^2 \quad\Longrightarrow\quad r \gg \frac{4\pi\epsilon_0\hbar^2}{e^2m_e} = a_0

6 Confine an electron to an atom numerical

Assume an electron is located somewhere within a region of atomic size. Estimate the minimum uncertainty in its momentum, and by assuming this uncertainty is comparable with its average momentum, estimate its average kinetic energy.

💡 Phillips' own hint

A particle’s momentum is at least as big as the uncertainty in its momentum. Use Eq. (1.15).

✓ Worked solution

Take $\Delta x \approx a_0$ . Then

\Delta p \gtrsim \frac{\hbar}{a_0} = 2.0\times10^{-24}\ \mathrm{kg\,m\,s^{-1}}

and with $p \approx \Delta p$ ,

E \approx \frac{p^2}{2m_e} = \frac{\hbar^2}{2m_ea_0^2} = 13.6\ \mathrm{eV}

That is the Rydberg energy. The uncertainty principle alone — no Coulomb’s law, no Schrödinger equation — produces the binding energy of hydrogen to three digits.

Reproduce it in NumPy

hbar, me, eV = 1.055e-34, 9.109e-31, 1.602e-19
a0 = 0.5292e-10

dp = hbar / a0
print(f"Delta_p ~ hbar/a0     = {dp:.3e} kg m/s")
print(f"E ~ hbar^2/(2 me a0^2) = {dp**2/(2*me)/eV:.2f} eV      (E_R = 13.6 eV)")
dp2 = hbar / (2*a0)
print(f"with the hbar/2 of Eq. (1.15) instead:  {dp2**2/(2*me)/eV:.2f} eV")

printsDelta_p ~ hbar/a0 = 1.994e-24 kg m/s E ~ hbar^2/(2 me a0^2) = 13.62 eV (E_R = 13.6 eV) with the hbar/2 of Eq. (1.15) instead: 3.40 eV

7 A charmed quark in a proton numerical

A charmed quark of mass $1.5\ \mathrm{GeV}/c^2$ is confined to a region about $1\ \mathrm{fm}$ across. Assuming its average momentum is comparable with the minimum uncertainty, show it can be treated non-relativistically, and estimate its average kinetic energy.

💡 Phillips' own hint

Use the uncertainty principle to show that $\Delta p$ — and hence the minimum average momentum — is small compared with $mc$ .

✓ Worked solution

The clean way is to work in energy units throughout, using the combination $\hbar c = 197.3\ \mathrm{MeV\,fm}$ — worth memorising, because it turns every nuclear-scale estimate into arithmetic:

pc \approx \frac{\hbar c}{\Delta x} = \frac{197.3\ \mathrm{MeV\,fm}}{1\ \mathrm{fm}} = 197\ \mathrm{MeV}

Compare with the rest energy $mc^2 = 1500\ \mathrm{MeV}$ :

\frac{pc}{mc^2} \approx \frac{197}{1500} = 0.13

Since $v/c \approx pc/mc^2 \ll 1$ , the quark is non-relativistic and $E = p^2/2m$ applies:

E \approx \frac{(pc)^2}{2mc^2} = \frac{197^2}{2 \times 1500}\ \mathrm{MeV} = 13\ \mathrm{MeV}

Engineer's bridge — ħc = 197 MeV fm is the same trick as ħ/E being a time

§0.2 made the point that $\hbar$ converts between wave and particle language: $\hbar/E$ is a time, $\hbar/p$ is a length. This problem is that trick in nuclear units.

Writing $\hbar c = 197.3\ \mathrm{MeV\,fm}$ means: confine something to 1 fm and it acquires about 200 MeV of momentum-energy. No calculator needed. Confine it to 0.1 fm and it is 2 GeV. That single number tells you why nuclear physics happens at MeV and atomic physics at eV — the box is 10⁵ times smaller, so the momentum is 10⁵ times bigger.

The same relation in atomic units is $\hbar c = 197\ \mathrm{eV\,nm}$ … which is worth checking against the $hc = 1240\ \mathrm{eV\,nm}$ used on §1.3: the ratio is $2\pi$ , exactly as it should be.

Reproduce it in NumPy

import numpy as np
hbarc_MeVfm = 197.327          # MeV fm
mc2 = 1500.0                   # MeV, the charmed quark
dx  = 1.0                      # fm

pc = hbarc_MeVfm / dx
print(f"hbar c = {hbarc_MeVfm:.1f} MeV fm  ->  pc ~ {pc:.1f} MeV for a {dx:.0f} fm box")
print(f"v/c ~ pc/mc^2 = {pc/mc2:.3f}   ->  non-relativistic, comfortably")
print(f"E ~ (pc)^2/(2 mc^2) = {pc**2/(2*mc2):.2f} MeV")
print(f"sanity: hbar c in eV nm = {hbarc_MeVfm:.1f}   and   h c / (hbar c) = {1239.84/197.327:.4f} = 2 pi")

printshbar c = 197.3 MeV fm -> pc ~ 197.3 MeV for a 1 fm box v/c ~ pc/mc^2 = 0.132 -> non-relativistic, comfortably E ~ (pc)^2/(2 mc^2) = 12.98 MeV sanity: hbar c in eV nm = 197.3 and h c / (hbar c) = 6.2832 = 2 pi

8 Why J. J. Thomson saw no waves numerical

In 1897 J. J. Thomson passed 200 eV electrons through a pair of plates 2 cm apart and concluded electrons are particles. In 1927 his son G. P. Thomson concluded they are waves. Explain why J. J. saw no evidence of wave behaviour.

💡 Phillips' own hint

Evaluate the de Broglie wavelength of a 200 eV electron and consider the condition for strong diffraction by a slit.

✓ Worked solution

From Eq. (1.8), $\lambda = \sqrt{1.5/200} = 0.087\ \mathrm{nm}$ .

Diffraction through an aperture of width $a$ spreads the beam by roughly

\theta \approx \frac{\lambda}{a} = \frac{8.7\times10^{-11}\ \mathrm{m}}{0.02\ \mathrm{m}} = 4\times10^{-9}\ \mathrm{rad}

Four nanoradians. Over a metre of apparatus that is a lateral spread of four nanometres — smaller than an atom, and utterly beyond 1897 instrumentation.

Both Thomsons were right, and neither result contradicts the other. The wave character was present in J. J.’s tube too; his aperture was simply 200 million times too wide to reveal it. G. P. did not use better electrons — he used a better aperture, namely the atomic spacing of a crystal, which is the same size as $\lambda$ .

Reproduce it in NumPy

import numpy as np
h, me, eV = 6.626e-34, 9.109e-31, 1.602e-19

lam = h / np.sqrt(2*me*200*eV)
print(f"200 eV electron: lambda = {lam*1e9:.4f} nm")
theta = lam / 0.02
print(f"through a 2 cm gap:  theta ~ lambda/a = {theta:.2e} rad")
print(f"  = {theta*1.0*1e9:.1f} nm of spread after 1 m of flight -- undetectable")
theta2 = lam / 0.2e-9
print("G. P. Thomson used a 0.2 nm crystal spacing instead:")
print(f"  theta ~ {theta2:.2f} rad = {np.degrees(theta2):.0f} degrees -- impossible to miss")

prints200 eV electron: lambda = 0.0867 nm through a 2 cm gap: theta ~ lambda/a = 4.34e-09 rad = 4.3 nm of spread after 1 m of flight -- undetectable G. P. Thomson used a 0.2 nm crystal spacing instead: theta ~ 0.43 rad = 25 degrees -- impossible to miss

9 Davisson and Germer's 50° peak numerical

Electrons of energy 54 eV were scattered from a nickel surface whose parallel rows of atoms are $D = 0.215\ \mathrm{nm}$ apart. Explain the strong scattering observed at $\phi = 50°$ .

💡 Phillips' own hint

Show that a 54 eV electron has $\lambda = 0.166\ \mathrm{nm}$ . Constructive interference from the surface rows requires $D\sin\phi = n\lambda$ ; check that $D = 0.215$ nm, $\phi = 50°$ and $n = 1$ satisfy it.

✓ Worked solution

\lambda = \sqrt{\frac{1.5}{54}}\ \mathrm{nm} = 0.167\ \mathrm{nm}

Waves scattered from adjacent rows travel path lengths differing by $D\sin\phi$ ; they arrive in step when that difference is a whole number of wavelengths. For $n = 1$ :

D\sin\phi = 0.215 \times \sin 50° = 0.165\ \mathrm{nm} \approx \lambda

Agreement to about 1%. Equivalently, solving for the predicted angle, $\phi = \arcsin(\lambda/D) = 50.9°$ against a measured 50°.

This is the experiment that settled it. A prediction from $\lambda = h/p$ — a formula proposed four years earlier with no experimental support whatsoever — landing within a degree of a measured diffraction peak.

Reproduce it in NumPy

import numpy as np
h, me, eV = 6.626e-34, 9.109e-31, 1.602e-19
D = 0.215e-9

lam = h / np.sqrt(2*me*54*eV)
print(f"54 eV electron: lambda    = {lam*1e9:.4f} nm   (book: 0.17 nm)")
print("first-order condition D sin(phi) = lambda:")
print(f"  predicted phi = {np.degrees(np.arcsin(lam/D)):.1f} deg      (measured: 50 deg)")
print(f"  D sin(50 deg) = {D*np.sin(np.radians(50))*1e9:.4f} nm     "
      f"(lambda = {lam*1e9:.4f} nm, {abs(D*np.sin(np.radians(50))-lam)/lam*100:.1f}% apart)")

prints54 eV electron: lambda = 0.1669 nm (book: 0.17 nm) first-order condition D sin(phi) = lambda: predicted phi = 50.9 deg (measured: 50 deg) D sin(50 deg) = 0.1647 nm (lambda = 0.1669 nm, 1.3% apart)

10 Conduction electrons in copper numerical

The electrons that carry current in copper have kinetic energies of about 7 eV. Calculate their wavelength and compare it with the spacing between copper atoms, to decide whether their wave nature matters as they move through the metal. (Density $8.9\times10^3\ \mathrm{kg\,m^{-3}}$ ; take the mass of a copper atom as 60 amu.)

💡 Phillips' own hint

Show that the de Broglie wavelength of a 7 eV electron is 0.46 nm, and that this is comparable with the distance between atoms in copper.

✓ Worked solution

Wavelength. $\lambda = \sqrt{1.5/7} = 0.46\ \mathrm{nm}$ .

Atomic spacing. Number density $n = \rho/m_{\text{atom}} = 8.9\times10^3 / (60 \times 1.661\times10^{-27}) = 8.9\times10^{28}\ \mathrm{m^{-3}}$ , so the typical separation is

d = n^{-1/3} = 0.22\ \mathrm{nm}

The wavelength is about twice the atomic spacing — the same order, not remotely negligible. A conduction electron is therefore strongly diffracted by the lattice, and treating it as a little ball bouncing between atoms is hopeless.

What this really says — a classical electron gas gets copper’s resistance wrong in size and in sign

This is the most consequential answer in the set. If conduction electrons were classical particles ricocheting off ions, copper’s resistance would be enormous and would fall with temperature. Both predictions are wrong.

What actually happens is that a wave in a periodic lattice propagates without scattering at all — a perfect crystal is transparent to it, exactly as a periodic filter passes certain frequencies unattenuated. Resistance comes only from departures from periodicity: thermal vibration and impurities. That is why resistance rises with temperature, and why a purer, colder wire conducts better.

The entire band theory of solids — conductors, insulators, semiconductors, every transistor you have used — starts from the observation in this problem.

Reproduce it in NumPy

import numpy as np
h, me, eV, u = 6.626e-34, 9.109e-31, 1.602e-19, 1.661e-27

lam = h / np.sqrt(2*me*7*eV)
print(f"7 eV electron:      lambda = {lam*1e9:.4f} nm")
n = 8.9e3 / (60*u)                 # the book's 60 amu
d = n**(-1/3)
print(f"copper: n = {n:.3e} atoms/m^3  ->  spacing d = {d*1e9:.3f} nm")
print(f"lambda / d = {lam/d:.2f}   -- same order, so the lattice diffracts it strongly")

prints7 eV electron: lambda = 0.4636 nm copper: n = 8.930e+28 atoms/m^3 -> spacing d = 0.224 nm lambda / d = 2.07 -- same order, so the lattice diffracts it strongly

11 Thermal neutrons from a reactor numerical

Neutrons from a reactor are brought to thermal equilibrium by repeated collisions in heavy water at $T = 300\ \mathrm{K}$ . What is their average energy in eV and their typical wavelength? Explain why they are diffracted by crystalline solids.

💡 Phillips' own hint

Show that the de Broglie wavelength of a neutron with thermal energy $\tfrac{3}{2}kT$ is comparable with the distance between atoms in a solid when $T \approx 300\ \mathrm{K}$ .

✓ Worked solution

E = \tfrac{3}{2}kT = \tfrac{3}{2}(1.381\times10^{-23})(300) = 6.2\times10^{-21}\ \mathrm{J} = 0.039\ \mathrm{eV}

\lambda = \frac{h}{\sqrt{2m_nE}} = 0.145\ \mathrm{nm}

Atomic spacings in solids are 0.1–0.3 nm, so a room-temperature neutron’s wavelength lands squarely in the range where a crystal lattice acts as a diffraction grating.

Reproduce it in NumPy

import numpy as np
h, mn, eV, k = 6.626e-34, 1.675e-27, 1.602e-19, 1.381e-23

E = 1.5 * k * 300
lam = h / np.sqrt(2*mn*E)
print(f"neutron at 300 K: E = 3/2 kT = {E/eV:.4f} eV")
print(f"                  lambda      = {lam*1e9:.4f} nm")
print("atomic spacings are 0.1-0.3 nm -> strongly diffracted")
print("(and note: no accelerator needed -- moderation alone tunes the wavelength)")

printsneutron at 300 K: E = 3/2 kT = 0.0388 eV lambda = 0.1452 nm atomic spacings are 0.1-0.3 nm -> strongly diffracted (and note: no accelerator needed -- moderation alone tunes the wavelength)

12 Oxygen molecules in the air numerical

Estimate the de Broglie wavelength of an oxygen molecule in air at NTP. Compare it with the average separation between molecules, and explain why the motion of oxygen molecules in air is unaffected by their wave nature.

💡 Phillips' own hint

Estimate the thermal energy of an oxygen molecule at $T = 273\ \mathrm{K}$ and show its de Broglie wavelength is much smaller than the typical distance between molecules in air.

✓ Worked solution

Wavelength. $E = \tfrac{3}{2}kT = 0.035\ \mathrm{eV}$ at 273 K, and with $m(\mathrm{O_2}) = 32\ \mathrm{u}$ :

\lambda = \frac{h}{\sqrt{2mE}} = 0.027\ \mathrm{nm}

Separation. At NTP, $n = P/kT = 2.7\times10^{25}\ \mathrm{m^{-3}}$ , so

d = n^{-1/3} = 3.3\ \mathrm{nm}

The separation is about 120 times the wavelength. Each molecule’s wave is tiny compared with the space it moves through and with everything it might diffract from, so it travels in a straight line between collisions like a classical billiard ball — and the kinetic theory of gases works.

Line up the last five problems

Problems 8 through 12 are one question asked five times. In each, compute $\lambda$ and compare it with the relevant size:

Case	$\lambda$	compared with	ratio	quantum?
200 eV electron, 2 cm gap (P8)	0.087 nm	20 mm	$4\times10^{-9}$	no
54 eV electron, Ni surface (P9)	0.167 nm	0.215 nm	0.8	yes
7 eV electron in copper (P10)	0.46 nm	0.22 nm	2.1	yes
300 K neutron in a crystal (P11)	0.145 nm	~0.2 nm	0.7	yes
O₂ molecule in air (P12)	0.027 nm	3.3 nm	0.008	no

Nothing in the physics changes down this column — every row uses the same $\lambda = h/\sqrt{2mE}$ . What changes is only whether the wavelength is comparable to something. That ratio is the whole of chapter 1 in one number.

Reproduce it in NumPy

import numpy as np
h, u, eV, k = 6.626e-34, 1.661e-27, 1.602e-19, 1.381e-23
mO2, T, P = 32*u, 273.0, 101325.0

E = 1.5 * k * T
lam = h / np.sqrt(2*mO2*E)
print(f"O2 at NTP ({T:.0f} K): E = 3/2 kT = {E/eV:.4f} eV")
print(f"                   lambda      = {lam*1e9:.5f} nm")
n = P / (k*T)
d = n**(-1/3)
print(f"air: n = {n:.3e} /m^3  ->  molecule spacing = {d*1e9:.2f} nm")
print(f"spacing / lambda = {d/lam:.0f}  -> wave effects utterly negligible")

printsO2 at NTP (273 K): E = 3/2 kT = 0.0353 eV lambda = 0.02702 nm air: n = 2.688e+25 /m^3 -> molecule spacing = 3.34 nm spacing / lambda = 124 -> wave effects utterly negligible

Check yourself

0 / 6 answered

1.Problems 8–12 all ask the same underlying question. What is it?
2.Problem 4 builds a length from $m_{e}$ , $e^{2} /4 π ϵ_{0}$ and $c$ and gets $2.8 \times 1 0^{- 15}$ m. What does that establish?
3.Problem 3 gets $I = f k^{4} T^{4} / h^{3} c^{2}$ with $f = 40.8$ from experiment. The exact answer is $2 π^{5} /15 = 40.80$ . What is the lesson?
4.Problem 10 finds conduction electrons in copper have $λ = 0.46$ nm against a 0.22 nm atomic spacing. Why does that matter beyond this problem?
5.In problem 7, $ℏ c = 197 MeV fm$ does most of the work. What is that number saying?
6.Problem 6 gives the electron in an atom a kinetic energy of about 13.6 eV — exactly the Rydberg energy. How much should you read into that?