Physics · Book 2 · Grades 10–12

High School Physics

High School Physics · Grades 10–12

34The Quantum World: Photons and Energy Levels

An infrared heat lamp can bathe your skin all evening and never mark it; ten minutes of thin mountain sunlight leaves a burn. The lamp delivers far more energy — but energy is not what marks skin. Light, this chapter shows, arrives in grains, and only a grain that hits hard enough does chemistry: one ultraviolet grain snaps a bond that a billion infrared grains only warm. Follow the grains and a century opens: metals leaking electrons, atoms singing in spectral lines, electrons interfering like waves.

34.1 Light arrives in grains: the photon

Definition 34.1 (The photon)

Light of frequency ν\nu exchanges energy with matter only in whole grains called photons, each of energy

E=hν=hcλ,h=6.63×10−34 J s,E = h\nu = \frac{hc}{\lambda}, \qquad h = 6.63 \times 10^{-34}\,\mathrm{J}\,\mathrm{s},

where hh is Planck’s constant (Planck, 1900; Einstein, 1905) and λ\lambda the wavelength in vacuum. A beam’s intensity measures the number of photons per second; each grain’s energy is fixed by the frequency alone.

Notation 34.2 (The electronvolt)

Atomic physics uses the electronvolt, 1 eV=1.60×10−19 J1\,\mathrm{eV} = 1.60 \times 10^{-19}\,\mathrm{J}: the energy one elementary charge gains crossing 1 V1\,\mathrm{V} (Chapter 14). Handy: E (eV)≈1240/λ (nm)E \,(\mathrm{eV}) \approx 1240/\lambda\,(\mathrm{nm}).

One scale for the electromagnetic family (): wavelength (top), energy per photon hc/ (blue), from radio grains of a microelectronvolt to the MeV gamma grains of — yet a 1.0\, mW pointer pours 3 × 1015 grains per second: everyday light flows as smoothly as poured sand.
One scale for the electromagnetic family (Chapter 22): wavelength (top), energy per photon hc/λhc/\lambda (blue), from radio grains of a microelectronvolt to the MeV\mathrm{MeV} gamma grains of Chapter 19 — yet a 1.0 mW1.0\,\mathrm{mW} pointer pours 3×10153 \times 10^{15} grains per second: everyday light flows as smoothly as poured sand.

34.2 The photoelectric effect

Shine light on a clean metal plate in vacuum and electrons pop out — the photoelectric effect. Its rules broke classical physics.

Proposition 34.3 (What the experiment shows)

For each metal there is a threshold frequency ν0\nu_0:

  1. below ν0\nu_0, no electron leaves, however intense the light;
  2. above ν0\nu_0, emission starts without delay, however faint;
  3. intensity raises the number of electrons per second, not their maximum kinetic energy; frequency raises Ek,max⁡E_{k,\max}.

None of it fits a continuous wave, which any color should let charge up — over months in faint light (Exercise 34.14) — and whose intensity should set the electrons’ energy: all three predictions fail.

Proof. Admitted at this level. ∎

Theorem 34.4 (Einstein’s photoelectric relation)

An electron needs at least the work function W0W_0 of the metal to escape its surface. One photon is absorbed whole by one electron, so

hν=W0+Ek,max⁡,hence ν0=W0h.h\nu = W_0 + E_{k,\max}, \qquad \text{hence } \nu_0 = \frac{W_0}{h}.

Proof. Energy bookkeeping: hνh\nu in, W0W_0 spent on escape, the rest kinetic — at most hν−W0h\nu - W_0, less for electrons starting deeper. Below ν0\nu_0 one grain cannot pay the exit toll, whatever their number. ∎

Definition 34.5 (Stopping potential)

To measure Ek,max⁡E_{k,\max}, a reverse voltage brakes the electrons (Chapter 14); the stopping potential UsU_s is the smallest voltage canceling the current: the fastest electrons just fail to cross, so Ek,max⁡=e UsE_{k,\max} = e\,U_s.

Measured E_k, = eU_s against frequency (sodium): a straight line of slope h, cutting the axis at _0; extrapolated (dashes), it reveals -W_0 at = 0.
Measured Ek,max⁡=eUsE_{k,\max} = eU_s against frequency (sodium): a straight line of slope hh, cutting the axis at ν0\nu_0; extrapolated (dashes), it reveals −W0-W_0 at ν=0\nu = 0.

Method 34.6 (Measuring Planck’s constant)

  1. Illuminate a photocell through filters of known wavelengths; compute each ν=c/λ\nu = c/\lambda; record each stopping potential.
  2. Plot Ek,max⁡=eUsE_{k,\max} = eU_s against ν\nu: Einstein promises the straight line hν−W0h\nu - W_0.
  3. Read the slope: it is hh; the intercepts give ν0\nu_0 and −W0-W_0.

34.3 Energy levels: why atoms emit lines

Definition 34.7 (Energy levels)

The energy of an electron bound in an atom is quantized: restricted to a discrete ladder of energy levels E1<E2<⋯<0E_1 < E_2 < \dots < 0, counted negative from the free electron. The lowest rung is the ground state, the others excited states; reaching 00 is ionization.

Proposition 34.8 (The hydrogen ladder)

The energy levels of the hydrogen atom are

En=−13.6 eVn2,n=1,2,3,…E_n = -\frac{13.6\,\mathrm{eV}}{n^2}, \qquad n = 1, 2, 3, \dots

so E1=−13.6 eVE_1 = -13.6\,\mathrm{eV}, E2=−3.40 eVE_2 = -3.40\,\mathrm{eV}, E3=−1.51 eVE_3 = -1.51\,\mathrm{eV}, E4=−0.85 eVE_4 = -0.85\,\mathrm{eV}, crowding toward 00.

Proof. Admitted at this level. ∎

Remark 34.9 (Where the ladder comes from)

Why these numbers is quantum mechanics, kept for the university volumes — as is de Broglie’s formula: the electron’s matter wave (Definition 34.15 below) must close on itself around the nucleus, and only certain wavelengths fit — like a guitar string, an atom has a discrete set of notes.

Theorem 34.10 (Photon of a transition)

An atom changes level only by emitting or absorbing one photon carrying exactly the difference: dropping from level pp to a lower level nn emits hν=Ep−Enh\nu = E_p - E_n; climbing from nn to pp absorbs that same energy.

Proof. Energy conservation, one grain at a time: the photon is the only currency light accepts, and the atom’s books must balance. ∎

Remark 34.11 (Line spectra explained)

Here is the answer promised by Chapter 2: a hot gas emits only the frequencies (Ep−En)/h(E_p - E_n)/h of its own ladder — a barcode of bright lines — and a cool gas steals exactly those frequencies from white light, printing matching dark absorption lines. Each element has its own ladder, hence its own barcode — how we read starlight.

Example 34.12 (The visible lines of hydrogen)

Transitions ending on n=2n = 2 fall in the visible. With E (eV)=1240/λ (nm)E \,(\mathrm{eV}) = 1240/\lambda\,(\mathrm{nm}): 3→23 \to 2: 1.89 eV1.89\,\mathrm{eV}, 656 nm656\,\mathrm{nm} (red); 4→24 \to 2: 2.55 eV2.55\,\mathrm{eV}, 486 nm486\,\mathrm{nm} (blue-green); 5→25 \to 2: 2.86 eV2.86\,\mathrm{eV}, 434 nm434\,\mathrm{nm} (violet). Transitions to n=1n = 1 exceed 10 eV10\,\mathrm{eV}: all ultraviolet.

The hydrogen ladder, to scale: the visible lines all end on n = 2; the drop to the ground state is deep ultraviolet.
The hydrogen ladder, to scale: the visible lines all end on n=2n = 2; the drop to the ground state is deep ultraviolet.

Remark 34.13 (Lasers, in one breath)

A photon of exactly Ep−EnE_p - E_n passing an excited atom can trigger it to emit an identical photon — same energy, same direction, in step. Keep more atoms up than down and one photon becomes an avalanche of clones: stimulated emission — and a laser is little more.

34.4 Matter waves

Proposition 34.14 (Electrons interfere)

Send electrons one at a time through a double slit (Chapter 22): each lands as a single sharp dot — a particle. But as thousands accumulate, the dots draw interference fringes: each electron, alone, passes as a wave through both slits. Crystals diffract electron beams too (Davisson and Germer, 1927).

Proof. Admitted at this level. ∎

One electron, one dot; thousands, fringes. Where the next dot lands is random — like one nucleus’s decay () — but the wave fixes the probability: bright fringe, likely; dark, nearly never. Individual chance ruled by an exact wave: the heart of quantum physics.
One electron, one dot; thousands, fringes. Where the next dot lands is random — like one nucleus’s decay (Chapter 19) — but the wave fixes the probability: bright fringe, likely; dark, nearly never. Individual chance ruled by an exact wave: the heart of quantum physics.

Definition 34.15 (De Broglie wavelength)

Every particle of momentum p=mvp = mv travels as a matter wave of de Broglie wavelength

λ=hp=hmv.\lambda = \frac{h}{p} = \frac{h}{mv}.

Proof. Admitted at this level. ∎

Example 34.16 (Why you never diffract)

An electron accelerated through 100 V100\,\mathrm{V} has Ek=1.60×10−17 JE_k = 1.60 \times 10^{-17}\,\mathrm{J}, p=2meEk=5.4×10−24 kg m/sp = \sqrt{2m_e E_k} = 5.4 \times 10^{-24}\,\mathrm{kg}\,\mathrm{m}/\mathrm{s}, so λ≈0.12 nm\lambda \approx 0.12\,\mathrm{nm} — the spacing of atoms in a crystal, which therefore acts as its double slit. A 58 g58\,\mathrm{g} tennis ball at 25 m/s25\,\mathrm{m}/\mathrm{s}: λ=6.63×10−34/1.45≈4.6×10−34 m\lambda = 6.63\times10^{-34}/1.45 \approx 4.6 \times 10^{-34}\,\mathrm{m}, some 101910^{19} times smaller than a nucleus — no slit will ever show it fringing. hh is so small that quantum graininess surfaces only at the atom’s scale.

Example 34.17 (The electron microscope)

Diffraction forbids resolving detail much smaller than the wavelength used (Chapter 22): visible light stops near 0.5 µm0.5\,\text{µ}\mathrm{m}. Electrons at 10 kV10\,\mathrm{kV} have λ≈12 pm\lambda \approx 12\,\mathrm{pm} — tens of thousands of times shorter: electron microscopes image viruses, cell machinery, even single atoms. The matter wave handed physics a new eye.

34.5 Exercises

Exercise 34.1 ★

A green laser pointer emits 1.0 mW1.0\,\mathrm{mW} at 530 nm530\,\mathrm{nm}. Compute one photon’s energy in joules and electronvolts, then the photons per second.

Solution

Solution of Exercise 34.1.

E=hc/λ=1.99×10−25/5.30×10−7=3.75×10−19 J=2.35 eVE = hc/\lambda = 1.99 \times 10^{-25}/5.30 \times 10^{-7} = 3.75 \times 10^{-19}\,\mathrm{J} = 2.35\,\mathrm{eV}. N=1.0×10−3/3.75×10−19≈2.7×1015N = 1.0 \times 10^{-3}/3.75 \times 10^{-19} \approx 2.7 \times 10^{15} photons per second.

Exercise 34.2 ★

An FM station broadcasts at 100 MHz100\,\mathrm{MHz}. Compute its photon energy in eV\mathrm{eV} and compare it to a visible photon: why is radio graininess undetectable?

Solution

Solution of Exercise 34.2.

E=hν=6.63×10−34×108=6.63×10−26 J≈4.1×10−7 eVE = h\nu = 6.63\times10^{-34} \times 10^{8} = 6.63 \times 10^{-26}\,\mathrm{J} \approx 4.1 \times 10^{-7}\,\mathrm{eV} — about five million times less than a visible photon. Any detectable radio signal involves so many photons per second that the flow is perfectly smooth.

Exercise 34.3 ★

Cesium has W0=1.9 eVW_0 = 1.9\,\mathrm{eV}. Compute its threshold frequency and wavelength. Does a red laser at 633 nm633\,\mathrm{nm} eject electrons from cesium — and if so, with what maximum kinetic energy?

Solution

Solution of Exercise 34.3.

ν0=W0/h=3.04×10−19/6.63×10−34≈4.6×1014 Hz\nu_0 = W_0/h = 3.04 \times 10^{-19}/6.63 \times 10^{-34} \approx 4.6 \times 10^{14}\,\mathrm{Hz}; λ0=c/ν0≈654 nm\lambda_0 = c/\nu_0 \approx 654\,\mathrm{nm}. At 633 nm633\,\mathrm{nm}, E=1240/633=1.96 eV>1.9 eVE = 1240/633 = 1.96\,\mathrm{eV} > 1.9\,\mathrm{eV}: yes, barely, with Ek,max⁡≈0.06 eV≈1×10−20 JE_{k,\max} \approx 0.06\,\mathrm{eV} \approx 1 \times 10^{-20}\,\mathrm{J}.

Exercise 34.4 ★

Using En=−13.6/n2E_n = -13.6/n^2 eV\mathrm{eV}, compute the energy and wavelength of the photon emitted in the hydrogen transition 3→23 \to 2. Its color?

Solution

Solution of Exercise 34.4.

E3−E2=−1.51+3.40=1.89 eVE_3 - E_2 = -1.51 + 3.40 = 1.89\,\mathrm{eV}; λ=1240/1.89≈656 nm\lambda = 1240/1.89 \approx 656\,\mathrm{nm}: the red line.

Exercise 34.5 ★

From the Ek,max⁡E_{k,\max}-versus-ν\nu graph of the course, read off sodium’s threshold frequency and work function (in eV\mathrm{eV}); what constant is the slope?

Solution

Solution of Exercise 34.5.

ν0≈5.6×1014 Hz\nu_0 \approx 5.6 \times 10^{14}\,\mathrm{Hz}; W0=hν0≈3.7×10−19 J≈2.3 eVW_0 = h\nu_0 \approx 3.7 \times 10^{-19}\,\mathrm{J} \approx 2.3\,\mathrm{eV} (the −W0-W_0 intercept). The slope is Planck’s constant hh.

Exercise 34.6 ★★

Light of wavelength 400 nm400\,\mathrm{nm} falls on potassium (W0=2.3 eVW_0 = 2.3\,\mathrm{eV}). Compute the photon energy, the maximum kinetic energy of the electrons (in eV\mathrm{eV} and joules), and the stopping potential.

Solution

Solution of Exercise 34.6.

E=1240/400=3.10 eVE = 1240/400 = 3.10\,\mathrm{eV}; Ek,max⁡=3.10−2.3=0.80 eV=1.3×10−19 JE_{k,\max} = 3.10 - 2.3 = 0.80\,\mathrm{eV} = 1.3 \times 10^{-19}\,\mathrm{J}; Us=Ek,max⁡/e=0.80 VU_s = E_{k,\max}/e = 0.80\,\mathrm{V}.

Exercise 34.7 ★★

A photocell is lit above threshold. What happens to (a) the current, (b) the stopping potential, when the intensity is doubled at fixed frequency? When the frequency is raised? Which fact contradicts the wave picture, and why?

Solution

Solution of Exercise 34.7.

Intensity doubled: current doubles (twice the photons), UsU_s unchanged (same energy per photon). Frequency raised: UsU_s rises, Ek,max⁡=hν−W0E_{k,\max} = h\nu - W_0. The wave picture ties the electrons’ energy to the intensity, not the frequency — exactly what is not observed.

Exercise 34.8 ★★

What minimum photon energy ionizes a hydrogen atom from its ground state? From n=2n = 2? Compute both threshold wavelengths and name their spectral domains.

Solution

Solution of Exercise 34.8.

From n=1n = 1: 13.6 eV13.6\,\mathrm{eV}, λ=1240/13.6≈91 nm\lambda = 1240/13.6 \approx 91\,\mathrm{nm}, far ultraviolet. From n=2n = 2: 3.40 eV3.40\,\mathrm{eV}, λ≈365 nm\lambda \approx 365\,\mathrm{nm}, near ultraviolet.

Exercise 34.9 ★★

White light crosses a jar of cool hydrogen (all atoms in the ground state). Explain why the transmitted spectrum shows no dark line in the visible, yet Balmer absorption lines appear in the spectra of hot stars.

Solution

Solution of Exercise 34.9.

From the ground state the smallest jump costs E2−E1=10.2 eVE_2 - E_1 = 10.2\,\mathrm{eV} (122 nm122\,\mathrm{nm}): every absorption line is ultraviolet, none visible. In a hot stellar atmosphere collisions keep some atoms in n=2n = 2, from which visible Balmer photons (1.89 to 3.40 eV1.89\text{ to }3.40\,\mathrm{eV}) can be absorbed.

Exercise 34.10 ★★

Compute the de Broglie wavelength of (a) an electron accelerated through 100 V100\,\mathrm{V}, (b) a 0.10 g0.10\,\mathrm{g} fly at 1.0 m/s1.0\,\mathrm{m}/\mathrm{s}. Compare each to the size of an atom (∼10−10 m\sim 10^{-10}\,\mathrm{m}); conclude.

Solution

Solution of Exercise 34.10.

(a) Ek=1.6×10−17 JE_k = 1.6 \times 10^{-17}\,\mathrm{J}, p=2meEk=5.4×10−24 kg m/sp = \sqrt{2m_e E_k} = 5.4 \times 10^{-24}\,\mathrm{kg}\,\mathrm{m}/\mathrm{s}, λ=h/p≈1.2×10−10 m\lambda = h/p \approx 1.2 \times 10^{-10}\,\mathrm{m} — atom-sized: crystals diffract it. (b) p=1.0×10−4×1.0=1.0×10−4 kg m/sp = 1.0 \times 10^{-4} \times 1.0 = 1.0 \times 10^{-4}\,\mathrm{kg}\,\mathrm{m}/\mathrm{s}, λ≈6.6×10−30 m\lambda \approx 6.6 \times 10^{-30}\,\mathrm{m} — 102010^{20} times smaller than an atom: the fly’s wave is forever invisible.

Exercise 34.11 ★★

A dark-adapted eye responds to about 9090 photons of 505 nm505\,\mathrm{nm} light. Compute that energy, compare it to the kinetic energy of a 2.5 mg2.5\,\mathrm{mg} mosquito flying at 0.50 m/s0.50\,\mathrm{m}/\mathrm{s}, and interpret the ratio.

Solution

Solution of Exercise 34.11.

One photon: 1.99×10−25/5.05×10−7=3.94×10−19 J1.99 \times 10^{-25}/5.05 \times 10^{-7} = 3.94 \times 10^{-19}\,\mathrm{J}; ninety: ≈3.5×10−17 J\approx 3.5 \times 10^{-17}\,\mathrm{J}. Mosquito: 12×2.5×10−6×0.502=3.1×10−7 J\tfrac12 \times 2.5 \times 10^{-6} \times 0.50^2 = 3.1 \times 10^{-7}\,\mathrm{J} — 101010^{10} times more. The eye works within a hundred grains of the absolute limit: almost a photon counter.

Exercise 34.12 ★★★

A photocell gives Us=1.19 VU_s = 1.19\,\mathrm{V} at 405 nm405\,\mathrm{nm} and Us=0.40 VU_s = 0.40\,\mathrm{V} at 546 nm546\,\mathrm{nm}. From these two points measure hh, then the work function in eV\mathrm{eV}. Which metal of this chapter is the cathode?

Solution

Solution of Exercise 34.12.

ν1=7.41×1014 Hz\nu_1 = 7.41 \times 10^{14}\,\mathrm{Hz}, ν2=5.49×1014 Hz\nu_2 = 5.49 \times 10^{14}\,\mathrm{Hz}; h=e ΔUs/Δν=1.60×10−19×0.79/1.91×1014≈6.6×10−34 J sh = e\,\Delta U_s/\Delta\nu = 1.60 \times 10^{-19} \times 0.79 / 1.91 \times 10^{14} \approx 6.6 \times 10^{-34}\,\mathrm{J}\,\mathrm{s}. W0=hν2−eUs,2=3.63×10−19−0.64×10−19=3.0×10−19 J≈1.9 eVW_0 = h\nu_2 - eU_{s,2} = 3.63 \times 10^{-19} - 0.64 \times 10^{-19} = 3.0 \times 10^{-19}\,\mathrm{J} \approx 1.9\,\mathrm{eV}: cesium.

Exercise 34.13 ★★★

A hydrogen lamp shows lines at 486 nm486\,\mathrm{nm} and 434 nm434\,\mathrm{nm}. Convert each to a photon energy and identify the two transitions by matching differences of the levels En=−13.6/n2E_n = -13.6/n^2 eV\mathrm{eV}.

Solution

Solution of Exercise 34.13.

1240/486=2.55 eV=E4−E2=3.40−0.851240/486 = 2.55\,\mathrm{eV} = E_4 - E_2 = 3.40 - 0.85: transition 4→24 \to 2. 1240/434=2.86 eV=E5−E2=3.40−0.541240/434 = 2.86\,\mathrm{eV} = E_5 - E_2 = 3.40 - 0.54: transition 5→25 \to 2.

Exercise 34.14 ★★★

Classical delay estimate: faint light of intensity 1.0×10−6 W/m21.0 \times 10^{-6}\,\mathrm{W}/\mathrm{m}^{2} falls on potassium (W0=2.3 eVW_0 = 2.3\,\mathrm{eV}). If an atom collected only the energy crossing its cross-section (radius 1.0×10−10 m1.0 \times 10^{-10}\,\mathrm{m}), how long would it need to gather W0W_0? Compare with the observed delay (under 10−9 s10^{-9}\,\mathrm{s}).

Solution

Solution of Exercise 34.14.

Cross-section πr2≈3.1×10−20 m2\pi r^2 \approx 3.1 \times 10^{-20}\,\mathrm{m}^{2}, collected power 3.1×10−26 W3.1 \times 10^{-26}\,\mathrm{W}; t=3.7×10−19/3.1×10−26≈1.2×107 st = 3.7 \times 10^{-19}/3.1 \times 10^{-26} \approx 1.2 \times 10^{7}\,\mathrm{s} — four months, against under a nanosecond observed. The energy cannot be spread over the wavefront: it travels concentrated, in grains.

Exercise 34.15 ★★★

In an electron microscope the electrons are accelerated through 5.0 kV5.0\,\mathrm{kV}. Compute their de Broglie wavelength, compare it to green light (550 nm550\,\mathrm{nm}), and explain via diffraction (Chapter 22) why this buys resolution.

Solution

Solution of Exercise 34.15.

Ek=8.0×10−16 JE_k = 8.0 \times 10^{-16}\,\mathrm{J}, p=2meEk=3.8×10−23 kg m/sp = \sqrt{2m_e E_k} = 3.8 \times 10^{-23}\,\mathrm{kg}\,\mathrm{m}/\mathrm{s}, λ=h/p≈1.7×10−11 m=17 pm\lambda = h/p \approx 1.7 \times 10^{-11}\,\mathrm{m} = 17\,\mathrm{pm} — about 3×1043 \times 10^{4} times shorter than 550 nm550\,\mathrm{nm}. Diffraction blurs detail smaller than ∼λ\sim\lambda, so shrinking λ\lambda by 10410^4 buys 10410^4 in resolution.

34.6 Problem: The Solar-Panel Physics Lab

Problem 34.1

Weekend problem — the solar-panel physics lab: the 102110^{21} grains a rooftop drinks each second, a photocell that measures Planck’s constant, a hydrogen lamp read line by line — one weekend, one hh

Your class instruments the school’s rooftop solar panel for a weekend. Data: solar irradiance at noon 1.0×103 W/m21.0 \times 10^{3}\,\mathrm{W}/\mathrm{m}^{2}; mean solar wavelength 550 nm550\,\mathrm{nm}; panel area 1.0 m21.0\,\mathrm{m}^{2}, efficiency 20%20\%; the day delivers the equivalent of 5.0 h5.0\,\mathrm{h} of full sun.

Part I — Photon bookkeeping.

  1. Compute the energy of one mean solar photon, in joules and electronvolts.
  2. Deduce the number of photons striking the panel per second at noon.
  3. Compute the panel’s electrical power at noon, then the energy produced over the day, in kW h\mathrm{kW}\,\mathrm{h}.
  4. The panel’s current looks perfectly smooth. Using question 2, explain why the grain-by-grain arrival is invisible.
  5. Sunlight carries more infrared energy than ultraviolet, yet only ultraviolet burns skin (bonds ∼3 to 4 eV\sim 3\text{ to }4\,\mathrm{eV}). Explain in grains.

Part II — Measuring hh with the kit’s photocell. The kit’s photocell has an unknown cathode. Filters give:

λ\lambda (nm\mathrm{nm})365405436546
UsU_s (V\mathrm{V})1.501.160.940.37
  1. What does the stopping potential measure? Relate UsU_s, ν\nu, W0W_0, hh.
  2. Convert the four wavelengths to frequencies (in 1014 Hz10^{14}\,\mathrm{Hz}).
  3. Why should plotting eUseU_s against ν\nu give a straight line? Check the alignment of the points.
  4. From the extreme points, compute the slope: your measured hh.
  5. Deduce the cathode’s work function (eV\mathrm{eV}), threshold frequency and threshold wavelength.
  6. Which part of the sunlight could eject electrons from this cathode — and which half of the spectrum is useless for it?
  7. Compare your hh to 6.63×10−34 J s6.63 \times 10^{-34}\,\mathrm{J}\,\mathrm{s}: relative error in percent?

Part III — The hydrogen calibration lamp. The kit’s wavelength scale is checked against a hydrogen lamp, En=−13.6/n2E_n = -13.6/n^2 eV\mathrm{eV}.

  1. Compute E2E_2, E3E_3, E4E_4 and E5E_5.
  2. Compute the photon energies of the transitions 3→23 \to 2, 4→24 \to 2, 5→25 \to 2.
  3. Deduce the three wavelengths and give each line’s color.
  4. Why does the lamp emit lines rather than a continuous rainbow? One sentence, naming the key idea.
  5. Could the lamp’s hydrogen absorb light at 500 nm500\,\mathrm{nm}? Justify with the ladder.

Part IV — Inspecting the cells: the electron wave. Micro-cracks in a cell lie far below the 0.5 µm0.5\,\text{µ}\mathrm{m} reach of light microscopes; the lab’s scanning electron microscope runs at 10 kV10\,\mathrm{kV}.

  1. Compute the electrons’ speed; check it stays below 0.25 c0.25\,c (classical formula).
  2. Compute their momentum and de Broglie wavelength.
  3. Resolution scales with wavelength: compute the gain over green light (550 nm550\,\mathrm{nm}); conclude with the weekend’s two numbers — your hh and this gain.
Solution

Solution of Problem 34.1.

1. E=hc/λ=1.99×10−25/5.50×10−7=3.62×10−19 J=2.26 eVE = hc/\lambda = 1.99 \times 10^{-25}/5.50 \times 10^{-7} = 3.62 \times 10^{-19}\,\mathrm{J} = 2.26\,\mathrm{eV}.

2. N=1.0×103/3.62×10−19≈2.8×1021N = 1.0 \times 10^{3}/3.62 \times 10^{-19} \approx 2.8 \times 10^{21} photons per second.

3. P=0.20×1.0×103 W=200 WP = 0.20 \times 1.0 \times 10^{3}\,\mathrm{W} = 200\,\mathrm{W}; E=200×5.0=1.0 kW hE = 200 \times 5.0 = 1.0\,\mathrm{kW}\,\mathrm{h} (3.6×106 J3.6 \times 10^{6}\,\mathrm{J}).

4. With 102110^{21} grains per second, the statistical flicker is fantastically small: graininess averages into a perfectly smooth current, as sand poured fast pours like water.

5. Damage needs one grain of 3 to 4 eV3\text{ to }4\,\mathrm{eV}: ultraviolet grains qualify, infrared grains (∼0.1 eV\sim 0.1\,\mathrm{eV}) never do — numbers cannot substitute for punch per grain.

6. UsU_s is the reverse voltage that just stops the fastest electrons: eUs=Ek,max⁡=hν−W0eU_s = E_{k,\max} = h\nu - W_0.

7. ν=c/λ\nu = c/\lambda: 8.228.22, 7.417.41, 6.886.88 and 5.495.49 × 1014 Hz\times\,10^{14}\,\mathrm{Hz}.

8. eUs=hν−W0eU_s = h\nu - W_0 is affine in ν\nu with slope hh. Successive slopes: (0.94−0.37)/1.39=0.41(0.94 - 0.37)/1.39 = 0.41 and (1.50−0.94)/1.34=0.42(1.50 - 0.94)/1.34 = 0.42 eV\mathrm{eV} per 1014 Hz10^{14}\,\mathrm{Hz} — the three points align.

9. h=e(Us,1−Us,4)/(ν1−ν4)=1.60×10−19×1.13/2.72×1014=6.64×10−34 J sh = e(U_{s,1}-U_{s,4})/(\nu_1-\nu_4) = 1.60 \times 10^{-19} \times 1.13/2.72 \times 10^{14} = 6.64 \times 10^{-34}\,\mathrm{J}\,\mathrm{s}.

10. At 546 nm546\,\mathrm{nm}: W0=hν−eUs=3.64×10−19−0.59×10−19=3.05×10−19 J=1.90 eVW_0 = h\nu - eU_s = 3.64 \times 10^{-19} - 0.59 \times 10^{-19} = 3.05 \times 10^{-19}\,\mathrm{J} = 1.90\,\mathrm{eV}; ν0=W0/h=4.60×1014 Hz\nu_0 = W_0/h = 4.60 \times 10^{14}\,\mathrm{Hz}; λ0≈653 nm\lambda_0 \approx 653\,\mathrm{nm}.

11. Only λ<653 nm\lambda < 653\,\mathrm{nm} ejects: the violet-to-red visible and the ultraviolet. The whole infrared side — roughly half the Sun’s energy — is useless to this cathode.

12. (6.64−6.63)/6.63≈0.2%(6.64 - 6.63)/6.63 \approx 0.2\% — a two-permille physics-lab measurement of a universal constant.

13. E2=−3.40 eVE_2 = -3.40\,\mathrm{eV}, E3=−1.51 eVE_3 = -1.51\,\mathrm{eV}, E4=−0.85 eVE_4 = -0.85\,\mathrm{eV}, E5=−0.54 eVE_5 = -0.54\,\mathrm{eV}.

14. 3→23 \to 2: 1.89 eV1.89\,\mathrm{eV}; 4→24 \to 2: 2.55 eV2.55\,\mathrm{eV}; 5→25 \to 2: 2.86 eV2.86\,\mathrm{eV}.

15. λ=1240/E\lambda = 1240/E: 656 nm656\,\mathrm{nm} (red), 486 nm486\,\mathrm{nm} (blue-green), 434 nm434\,\mathrm{nm} (violet).

16. The atom’s energies are quantized, so it can only emit the discrete differences Ep−EnE_p - E_n: a ladder makes lines, not a rainbow.

17. No: 500 nm500\,\mathrm{nm} means 2.48 eV2.48\,\mathrm{eV}, which matches no level difference (E2E_2 upward gives 1.891.89, 2.552.55, 2.862.86, 3.40 eV3.40\,\mathrm{eV}; from E1E_1 everything costs at least 10.2 eV10.2\,\mathrm{eV}) — the gas is transparent there.

18. eU=12mev2eU = \tfrac12 m_e v^2: v=2×1.60×10−19×1.0×104/9.11×10−31≈5.9×107 m/s≈0.20 cv = \sqrt{2 \times 1.60 \times 10^{-19} \times 1.0 \times 10^{4}/9.11 \times 10^{-31}} \approx 5.9 \times 10^{7}\,\mathrm{m}/\mathrm{s} \approx 0.20\,c — classical treatment acceptable.

19. p=mev=5.4×10−23 kg m/sp = m_e v = 5.4 \times 10^{-23}\,\mathrm{kg}\,\mathrm{m}/\mathrm{s}; λ=h/p=1.2×10−11 m=12 pm\lambda = h/p = 1.2 \times 10^{-11}\,\mathrm{m} = 12\,\mathrm{pm}.

20. Gain =5.50×10−7/1.23×10−11≈4.5×104= 5.50 \times 10^{-7}/1.23 \times 10^{-11} \approx 4.5 \times 10^{4}. The weekend’s two numbers: h=6.64×10−34 J sh = 6.64 \times 10^{-34}\,\mathrm{J}\,\mathrm{s}, measured to 0.2%0.2\% with filters and a voltmeter, and a microscope forty-five-thousand times sharper because electrons wave.