E.2 Quantum physics HL only

1. Photoelectrons are emitted without any measurable delay, even in very dim light. — In the wave theory, energy spreads over the wavefront, so in dim light an electron would need time to absorb enough energy to escape. Experiment shows emission begins at once, because a single photon delivers all the energy one electron needs.
2. More photoelectrons are emitted each second when the light is made much more intense. — A student who counts every photon-model result as evidence against waves picks this. The wave theory also predicts it: more intense light delivers more energy per second and so releases more electrons per second. It does not distinguish the two models.
3. More photoelectrons are emitted each second when the frequency of the light is raised. — A student who thinks frequency controls the number of electrons picks this. It is not one of the features of the effect: the number of photoelectrons per second is proportional to the number of photons arriving per second, which is set by the intensity, and at a fixed intensity a higher frequency means fewer photons per second, not more. Frequency sets the maximum kinetic energy, E_max = hf − Φ.
4. Every photoelectron leaves the metal with kinetic energy exactly hf − Φ. — A student who reads Einstein's equation as applying to every electron picks this. It is not observed: photoelectrons have a range of kinetic energies up to a maximum, E_max = hf − Φ, because electrons from below the surface lose extra energy on the way out.

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1. The maximum frequency of light that can release photoelectrons from the metal — A student who thinks lower-frequency, longer-wavelength light carries more energy picks this. Photon energy E = hf rises with frequency, so there is a MINIMUM frequency for emission. The threshold WAVELENGTH is the maximum wavelength.
2. The frequency of light that releases the most photoelectrons per second from the metal — A student who thinks frequency controls the number of photoelectrons picks this. The number emitted per second is set by the intensity (photons per second). The threshold frequency is simply the lowest frequency at which any emission happens.
3. The frequency below which photoelectrons leave the metal only after a long delay — A student who thinks electrons can build up energy from the light over time picks this. Below the threshold frequency no photoelectrons are emitted at all, however long the light shines, because one photon cannot supply the work function and energy does not accumulate.
4. The lowest frequency of light that can release photoelectrons from the metal — Below the threshold frequency f₀, each photon has less energy than the work function (hf < Φ), so no electron can be released, whatever the intensity. At f₀, hf₀ = Φ.

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1. 6.22 V — A student who gives the whole photon energy to the electron uses E_max = hf = 6.22 eV. At least the work function, 4.30 eV, is needed to free the electron, so E_max = 6.22 − 4.30 = 1.92 eV and V_s = 1.92 V.
2. 4.30 V — A student who thinks the stopping potential measures the work function writes eV_s = Φ. The stopping potential stops electrons that have already escaped; it measures their maximum kinetic energy, eV_s = hf − Φ = 1.92 eV.
3. 1.92 V — hf = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸)/(200 × 10⁻⁹) = 9.95 × 10⁻¹⁹ J = 6.22 eV. E_max = hf − Φ = 6.22 − 4.30 = 1.92 eV, and eV_s = E_max, so V_s = 1.92 V.
4. 10.5 V — A student who adds the work function uses E_max = hf + Φ = 6.22 + 4.30 = 10.5 eV. The electron must use energy Φ to escape, so it is subtracted: E_max = 1.92 eV.

Working hf = hc/λ = (6.63 × 10⁻³⁴ J s)(3.00 × 10⁸ m s⁻¹)/(200 × 10⁻⁹ m) = 9.95 × 10⁻¹⁹ J. In eV: 9.95 × 10⁻¹⁹ J / 1.60 × 10⁻¹⁹ J eV⁻¹ = 6.22 eV. E_max = hf − Φ = 6.22 eV − 4.30 eV = 1.92 eV. eV_s = E_max, so V_s = 1.92 V.

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1. The rings form where electrons bounce off the carbon atoms like balls, at particular angles. — A student who treats electrons purely as particles picks this. Particle collisions with atoms would scatter electrons over a spread of angles, not into sharp rings whose size depends on the electrons' momentum as λ = h/p predicts.
2. The rings are a diffraction pattern, at angles that fit a wavelength λ = h/p for the electrons. — Diffraction is a wave property. The rings are at angles where waves scattered by the regularly spaced carbon atoms reinforce, and those angles match the de Broglie wavelength h/p. Changing the electrons' momentum changes the ring sizes as this wavelength predicts.
3. Electrons in the beam interfere with one another beyond the film, and the rings mark where they reinforce. — A student who thinks electrons must meet to interfere picks this. The same pattern builds up even when electrons pass through one at a time; each electron's own wave is diffracted by the atoms.
4. The rings form because each electron follows a wavy path between the rows of atoms. — A student who pictures an electron moving along a wave-shaped path picks this. The electron's wave is not a trajectory; it describes where the electron is likely to be detected, and diffraction of that wave produces the rings.

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1. Particles such as electrons show wave behaviour, such as diffraction, in some experiments and particle behaviour in others. — Electrons diffract and interfere like waves, yet each is detected at a single point with its whole charge and mass, like a particle. Neither classical model alone describes them; this is also true of light.
2. Particles such as electrons remain particles, but they travel through space along wave-shaped paths. — A student who pictures a particle following a sinusoidal line picks this. The electron's wave is not a path; it governs where the electron is likely to be detected, and a wavy trajectory could not produce diffraction patterns.
3. Only the smallest particles, such as electrons, have a wave nature; objects such as dust grains are purely particles. — A student who thinks wave properties belong only to subatomic particles picks this. All matter has a de Broglie wavelength λ = h/p; for large objects it is far too short for diffraction to be observed, but it is not zero.
4. Particles such as electrons are really spread-out waves, and the particle description no longer applies to them. — A student who replaces the particle model with a classical wave model picks this. Each electron is still detected at a single point, as a particle; duality means that both descriptions are needed.

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1. 6.2 × 10⁻¹⁰ m — A student who uses the photon relation λ = hc/E gets (6.63 × 10⁻³⁴ × 3.00 × 10⁸)/(3.20 × 10⁻¹⁶) = 6.2 × 10⁻¹⁰ m. That holds only for photons; for an electron λ = h/p with p = √(2m_eE_k).
2. 3.9 × 10⁻¹¹ m — A student who uses p = √(m_eE_k), leaving out the factor 2, gets 3.9 × 10⁻¹¹ m. From E_k = p²/2m, p = √(2m_eE_k), giving λ = 2.7 × 10⁻¹¹ m.
3. 1.1 × 10⁻²⁰ m — A student who uses E_k = 2000 (the energy in eV) as if it were in joules gets p = √(2 × 9.11 × 10⁻³¹ × 2000) and λ = 1.1 × 10⁻²⁰ m. The kinetic energy is 2000 eV = 3.20 × 10⁻¹⁶ J.
4. 2.7 × 10⁻¹¹ m — E_k = eV = 1.60 × 10⁻¹⁹ × 2000 = 3.20 × 10⁻¹⁶ J. p = √(2m_eE_k) = √(2 × 9.11 × 10⁻³¹ × 3.20 × 10⁻¹⁶) = 2.41 × 10⁻²³ kg m s⁻¹. λ = h/p = 2.7 × 10⁻¹¹ m.

Working E_k = eV = (1.60 × 10⁻¹⁹ C)(2.0 × 10³ V) = 3.20 × 10⁻¹⁶ J. p = √(2m_eE_k) = √(2 × 9.11 × 10⁻³¹ kg × 3.20 × 10⁻¹⁶ J) = 2.41 × 10⁻²³ kg m s⁻¹. λ = h/p = (6.63 × 10⁻³⁴ J s)/(2.41 × 10⁻²³ kg m s⁻¹) = 2.7 × 10⁻¹¹ m.

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1. The scattered X-rays included a longer wavelength, shifted by an amount that increased with angle. — Classical wave theory predicts scattered radiation of unchanged wavelength. The shifted wavelength, increasing with angle as Δλ = (h/m_ec)(1 − cos θ), is exactly what a collision between a photon of momentum h/λ and an electron predicts.
2. The scattered X-rays all had the same wavelength as the incident X-rays, at every angle. — A student who pictures a photon bouncing off an electron like a ball off a wall expects no change. This was not observed, and an unchanged wavelength is what the classical WAVE model predicts, so it could not be evidence for photons.
3. X-rays were scattered in all directions, as balls would be when they hit a target. — A student who takes any particle-like description as evidence picks this. Waves are also scattered in all directions by electrons, so scattering alone does not distinguish the models. The evidence is the angle-dependent increase in wavelength.
4. The scattered X-rays included a longer wavelength, shifted in proportion to the incident wavelength. — A student who expects the shift to be a fixed fraction of the wavelength picks this. The Compton shift depends only on the scattering angle, Δλ = (h/m_ec)(1 − cos θ), not on the incident wavelength.

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1. The photon loses some of its speed in the collision, which stretches its wavelength. — A student who treats a photon like a material particle whose energy depends on speed picks this. A photon always travels at c in a vacuum; its energy falls by a decrease in frequency, which means an increase in wavelength.
2. The photon transfers some of its energy to the recoiling electron, so its energy hc/λ decreases. — Energy and momentum are conserved in the collision. The electron recoils with kinetic energy taken from the photon, so the scattered photon has less energy E = hc/λ, and therefore a longer wavelength.
3. Some photons are absorbed, so the scattered beam is weaker and has a longer wavelength. — A student who confuses the intensity of a beam with the energy of each photon picks this. Each scattered photon's wavelength is longer because that photon lost energy to an electron; the number of photons has no effect on the wavelength.
4. The photon gains energy from the electron, and a photon with more energy has a longer wavelength. — A student who thinks longer wavelength means more energy picks this. E = hc/λ: a longer wavelength means LESS energy. The photon loses energy to the electron, which is initially at rest.

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1. θ = 90°, where the photon turns at right angles — A student who thinks a right-angle deflection is the largest change picks this. At 90°, 1 − cos θ = 1, so Δλ = h/m_ec; at 180°, 1 − cos θ = 2, twice as large.
2. θ = 0°, where the photon carries straight on — A student who uses cos θ in place of (1 − cos θ) picks this, since cos θ is greatest at 0°. A photon that is not deflected is not changed: 1 − cos 0° = 0, so the shift is zero.
3. θ = 180°, where the photon is scattered straight back — Δλ = (h/m_ec)(1 − cos θ) is greatest when cos θ = −1, at θ = 180°, where Δλ = 2h/m_ec = 4.85 × 10⁻¹² m. A photon that rebounds straight back has the largest change of momentum and gives the most energy to the electron.
4. No single angle: the shift is h/m_ec at every angle — A student who takes the constant h/m_ec to be the shift itself picks this. The shift is (h/m_ec)(1 − cos θ), which rises from zero at 0° to 2h/m_ec at 180°.

Working Δλ = (h/m_ec)(1 − cos θ). 1 − cos θ ranges from 0 (θ = 0°) through 1 (θ = 90°) to 2 (θ = 180°), so the shift is greatest at θ = 180°: Δλ_max = 2h/m_ec = 2 × 2.43 × 10⁻¹² m = 4.85 × 10⁻¹² m.

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1. It is valid: a wave model cannot explain why a more intense beam releases more photoelectrons. — A student who treats every photon-model result as evidence against waves picks this. A wave model does explain it: more intense light carries more energy per second. The argument uses a result that does not distinguish the models.
2. It is flawed: the photocurrent depends on the frequency of the light, not on its intensity. — A student who swaps the roles of frequency and intensity picks this. Above the threshold, the photocurrent is set by the number of photons per second, which is the intensity; the frequency sets the maximum kinetic energy of the photoelectrons.
3. It is flawed: a wave model also predicts that more intense light releases more photoelectrons each second. — A result is evidence for photons only if the wave model cannot explain it. More intense waves deliver more energy per second and would also release more electrons per second, so this result fits both models. The threshold frequency, E_max independent of intensity and the lack of delay are the results that rule out the wave model.
4. It is flawed: the photocurrent is fixed by the potential difference across the photocell, not by the light. — A student who applies Ohm's law to a photocell picks this. Once the collector is positive enough to collect every photoelectron, the current is the saturation current, set by the number of photoelectrons emitted per second, which doubles when the intensity doubles.

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1. Only the red light, from the 650 nm laser — A student who thinks longer-wavelength light carries more energy per photon picks this. E = hc/λ: the 650 nm photons have only 1.91 eV, below the 2.28 eV work function, while the 400 nm photons have 3.11 eV. Light releases electrons only if its wavelength is SHORTER than the threshold wavelength, here 545 nm.
2. Only the violet light, despite its lower power — Violet photon: E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸)/(400 × 10⁻⁹) = 4.97 × 10⁻¹⁹ J = 3.11 eV, more than 2.28 eV. Red photon: 3.06 × 10⁻¹⁹ J = 1.91 eV, less than 2.28 eV. The power of the laser only sets how many red photons arrive; none of them can free an electron.
3. Both the red light and the violet light — A student who thinks intense light can release electrons at any frequency picks this. The violet light does release electrons, but the red photons each carry 1.91 eV, below the 2.28 eV work function, and energy from many photons does not accumulate in an electron, however powerful the laser.
4. Neither the red light nor the violet light — A student who compares photon energies in joules (3.06 × 10⁻¹⁹ J and 4.97 × 10⁻¹⁹ J) directly with 2.28 concludes that both are far too small. Converting first: the violet photons have 4.97 × 10⁻¹⁹ ÷ 1.60 × 10⁻¹⁹ = 3.11 eV, above the work function, so violet light does release electrons.

Working Red: E = hc/λ = (6.63 × 10⁻³⁴ J s)(3.00 × 10⁸ m s⁻¹)/(650 × 10⁻⁹ m) = 3.06 × 10⁻¹⁹ J = (3.06 × 10⁻¹⁹)/(1.60 × 10⁻¹⁹) eV = 1.91 eV < 2.28 eV, so no emission at any power. Violet: E = (6.63 × 10⁻³⁴)(3.00 × 10⁸)/(400 × 10⁻⁹) = 4.97 × 10⁻¹⁹ J = 3.11 eV > 2.28 eV, so photoelectrons are emitted (E_max = 3.11 − 2.28 = 0.83 eV). Equivalently, λ₀ = hc/Φ = (6.63 × 10⁻³⁴)(3.00 × 10⁸)/(2.28 × 1.60 × 10⁻¹⁹) = 5.45 × 10⁻⁷ m = 545 nm; only wavelengths shorter than this release electrons.

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1. Q's line is steeper, because its gradient is set by the larger work function. — A student who reads the work function from the gradient picks this. The gradient of E_max against f is h for every metal. The work function appears in the intercepts: the line meets the frequency axis at f₀ = Φ/h and, extended, meets the E_max axis at −Φ.
2. Each line shows the kinetic energy of every photoelectron at that frequency. — A student who thinks every photoelectron leaves with the same energy picks this. Each point on a line is the MAXIMUM kinetic energy, hf − Φ. Φ is the minimum energy needed to remove an electron; electrons from below the surface lose extra energy on the way out and leave with less, down to zero.
3. Each line would be shifted upwards if light of greater intensity were used. — A student who thinks brighter light gives each photoelectron more energy picks this. E_max = hf − Φ contains no intensity: more intense light of the same frequency releases more electrons per second, not faster ones, so each line stays where it is.
4. They have equal gradients, each equal to the Planck constant h. — E_max = hf − Φ has the form y = mx + c with gradient h for every metal, so the lines are parallel. They are displaced because Q has the larger work function: Φ = hf₀, so Φ_Q − Φ_P = h × 2.0 × 10¹⁴ Hz.

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1. The current is zero at −1.2 V and below, as before, and levels off at 6.0 μA. — Doubling the intensity at the same frequency doubles the number of photons, and so of photoelectrons, per second: the saturation current doubles to 6.0 μA. Each photon still has energy hf, so E_max and the stopping potential are unchanged.
2. It becomes zero only at a more negative potential, and levels off at 6.0 μA. — A student who thinks brighter light gives electrons more energy picks this. The saturation current does double, but the energy of each photon, hf, is unchanged, so E_max = hf − Φ and the stopping potential of 1.2 V stay the same.
3. It becomes zero only at a more negative potential, and still levels off at 3.0 μA. — A student who swaps the roles of intensity and frequency picks this, making intensity control the energy and not the number of electrons. It is the reverse: intensity sets the number per second, doubling the saturation current, and frequency sets E_max.
4. The current is zero at −1.2 V and below, and still levels off at 3.0 μA, set by the potential. — A student who applies Ohm's law to the photocell thinks the potential alone sets the current. The saturation current is the charge of all the photoelectrons emitted per second, which doubles when the number of photons per second doubles.

Working At a fixed frequency each photon has the same energy hf, so E_max = hf − Φ is unchanged and the stopping potential is still 1.2 V (the current is zero at −1.2 V). Doubling the intensity doubles the number of photons, and so of photoelectrons, per second: saturation current = 2 × 3.0 μA = 6.0 μA.

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1. 12 mm — A student who uses the photon relation λ = hc/E takes λ ∝ 1/E_k and halves the radius. For a particle with mass, p = √(2mE_k), so λ ∝ 1/√E_k and the radius falls by √2, to 17 mm.
2. 34 mm — A student who thinks faster particles have longer wavelengths multiplies by √2. λ = h/p: the electrons gain momentum, so their wavelength and the ring radius DECREASE, to 24/√2 = 17 mm.
3. 24 mm — A student who thinks the ring pattern is set only by the arrangement of the atoms, as for particles deflected by them, expects no change. The ring angles depend on the electrons' de Broglie wavelength, which decreases as the potential difference increases.
4. 17 mm — E_k = eV and p = √(2m_e eV), so λ = h/p ∝ 1/√V. Doubling V divides λ, and so the radius, by √2: 24 mm/√2 = 17 mm.

Working E_k = eV, so p = √(2m_e eV) and λ = h/√(2m_e eV) ∝ 1/√V. With r ∝ λ: r_new = r × √(V₁/V₂) = 24 mm × √(3.0 kV/6.0 kV) = 24 mm/√2 = 17 mm.

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1. 4.7 × 10⁻¹⁵ m — A student who uses cos 51° = 0.629 in place of sin 51° gets 4.7 × 10⁻¹⁵ m. The first minimum is given by sin θ ≈ λ/D, with θ measured from the direction of the beam.
2. 4.4 × 10⁻¹⁵ m — A student whose calculator is in radian mode gets sin 51 = 0.670 and D = 4.4 × 10⁻¹⁵ m. The angle is in degrees: sin 51° = 0.777.
3. 3.8 × 10⁻¹⁵ m — λ = h/p = 6.63 × 10⁻³⁴/2.24 × 10⁻¹⁹ = 2.96 × 10⁻¹⁵ m. D ≈ λ/sin θ = 2.96 × 10⁻¹⁵/sin 51° = 2.96 × 10⁻¹⁵/0.777 = 3.8 × 10⁻¹⁵ m.
4. 3.0 × 10⁻¹⁵ m — A student who thinks diffraction happens when the wavelength equals the size of the object sets D = λ = 3.0 × 10⁻¹⁵ m. The size is found from the angle of the first minimum: D ≈ λ/sin θ = 3.8 × 10⁻¹⁵ m.

Working λ = h/p = (6.63 × 10⁻³⁴ J s)/(2.24 × 10⁻¹⁹ kg m s⁻¹) = 2.96 × 10⁻¹⁵ m. sin θ ≈ λ/D, so D ≈ λ/sin θ = (2.96 × 10⁻¹⁵ m)/sin 51° = (2.96 × 10⁻¹⁵ m)/0.777 = 3.8 × 10⁻¹⁵ m.

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1. It is deflected by the electrons that went through before it, and this builds up the fringes over time. — A student who thinks the pattern needs electrons to interact with one another picks this. Only one electron is in the apparatus at a time, and the pattern does not depend on the rate; each electron interferes only with its own wave.
2. It is detected at one point, like a particle, but wave interference sets where it is likely to land. — Each detection is a single dot, which is particle behaviour. Yet the dots build up an interference pattern even though the electrons pass one at a time, so each electron's own wave, passing both slits, sets the probability of where it lands. This is wave–particle duality.
3. It follows a wavy path from the slits, which carries it to one of the bright fringes. — A student who pictures the wave as a wavy trajectory picks this. There is no such path; the wave determines only the probability of detection at each point, and the dots land at random positions within that distribution.
4. It spreads out over the whole screen as a wave, and each dot marks the place where most of it arrived. — A student who treats the electron as a spread-out classical wave picks this. Each dot is a whole electron, with all its charge and mass, detected at one point; nothing of it arrives anywhere else.

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1. The proton's wavelength is about 43 times as long as the electron's. — A student who thinks a heavier particle has a longer wavelength inverts the ratio. At equal kinetic energy the proton has the larger momentum, √(2m_pE_k), so it has the SHORTER wavelength.
2. The electron's wavelength is about 1830 times the proton's. — A student who assumes equal kinetic energy means equal speed uses λ = h/mv with the same v and gets the mass ratio. The lighter electron moves faster, so its momentum is only √1830 ≈ 43 times smaller.
3. The electron's wavelength is about 43 times the proton's. — λ = h/p = h/√(2mE_k). At equal E_k, λ ∝ 1/√m, so λ_e/λ_p = √(m_p/m_e) = √1830 ≈ 43.
4. They have equal wavelengths, as the particles have equal energies. — A student who uses the photon relation λ = hc/E for particles thinks equal energies give equal wavelengths. For particles with mass, λ = h/√(2mE_k), which depends on mass as well as energy.

Working λ = h/p and p = √(2mE_k). With equal E_k: λ_e/λ_p = √(2m_pE_k)/√(2m_eE_k) = √(m_p/m_e) = √1830 = 42.8 ≈ 43.

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1. It has no de Broglie wavelength, since it is not a subatomic particle. — A student who thinks only tiny particles have wave properties picks this. λ = h/p applies to any object with momentum; the ball's wavelength is 1.4 × 10⁻³⁴ m, not zero, but far too short to show diffraction.
2. Its de Broglie wavelength is far too short compared with the width of the gap. — λ = h/p = 6.63 × 10⁻³⁴/(0.16 × 30) = 1.4 × 10⁻³⁴ m. Diffraction is noticeable only when the wavelength is comparable with the gap; here it is about 10³⁴ times smaller, so the spreading is immeasurably small.
3. It is uncharged, and only charged particles such as electrons diffract. — A student who thinks diffraction of particles is an electrical deflection by atoms picks this. Diffraction is a wave effect: uncharged neutrons, atoms and molecules diffract. The ball does not because its wavelength is so short.
4. Its de Broglie wavelength does not equal the gap width, and diffraction needs them equal. — A student who thinks diffraction happens only when the wavelength equals the gap picks this. Diffraction is noticeable whenever the wavelength is comparable with the gap, not only when the two are equal. The ball's λ = h/p = 6.63 × 10⁻³⁴/(0.16 × 30) = 1.4 × 10⁻³⁴ m, about 10³⁴ times smaller than the gap, which is why no diffraction is seen.

Working p = mv = 0.16 kg × 30 m s⁻¹ = 4.8 kg m s⁻¹. λ = h/p = (6.63 × 10⁻³⁴ J s)/(4.8 kg m s⁻¹) = 1.4 × 10⁻³⁴ m, about 10³⁴ times smaller than a gap of a few metres.

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1. 6.74 × 10⁻¹¹ m — A student who subtracts the shift gets 7.10 × 10⁻¹¹ − 0.364 × 10⁻¹¹ = 6.74 × 10⁻¹¹ m. Δλ = λ_f − λ_i is positive: the scattered photon has less energy and a LONGER wavelength.
2. 6.98 × 10⁻¹¹ m — A student who uses cos θ in place of (1 − cos θ) gets Δλ = 2.43 × 10⁻¹² × (−0.50) = −1.21 × 10⁻¹² m and λ_f = 6.98 × 10⁻¹¹ m. The factor is 1 − cos 120° = 1.50.
3. 7.15 × 10⁻¹¹ m — A student whose calculator is in radian mode gets cos 120 = 0.814, so 1 − cos θ = 0.186 and Δλ = 4.5 × 10⁻¹³ m. In degrees, cos 120° = −0.50 and 1 − cos θ = 1.50.
4. 7.46 × 10⁻¹¹ m — h/m_ec = 2.43 × 10⁻¹² m. Δλ = 2.43 × 10⁻¹² × (1 − cos 120°) = 2.43 × 10⁻¹² × 1.50 = 3.64 × 10⁻¹² m. λ_f = 7.10 × 10⁻¹¹ + 0.364 × 10⁻¹¹ = 7.46 × 10⁻¹¹ m.

Working h/(m_ec) = (6.63 × 10⁻³⁴ J s)/(9.11 × 10⁻³¹ kg × 3.00 × 10⁸ m s⁻¹) = 2.43 × 10⁻¹² m. Δλ = (h/m_ec)(1 − cos 120°) = 2.43 × 10⁻¹² m × (1 − (−0.50)) = 3.64 × 10⁻¹² m. λ_f = λ_i + Δλ = 7.10 × 10⁻¹¹ m + 0.364 × 10⁻¹¹ m = 7.46 × 10⁻¹¹ m.

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1. The shift is the same for both, but it is a tiny fraction of the visible wavelength. — Δλ = (h/m_ec)(1 − cos 90°) = 2.4 × 10⁻¹² m whatever the incident wavelength. It is 2.4 × 10⁻¹²/7.1 × 10⁻¹¹ ≈ 3% of the X-ray wavelength but only 2.4 × 10⁻¹²/5.0 × 10⁻⁷ ≈ 5 × 10⁻⁶ of the visible wavelength, too small to measure.
2. The shift is far smaller for visible light, as visible photons carry much less energy. — A student who expects the shift to depend on the photon's energy picks this. The formula contains no incident wavelength: at 90° the shift is 2.4 × 10⁻¹² m for both. Only its size relative to the wavelength differs.
3. Visible light is scattered only as a wave, since just high-frequency radiation acts as photons. — A student who thinks only X-rays and gamma rays are photons picks this. All electromagnetic radiation consists of photons; visible light even shows the photoelectric effect. Visible photons undergo the same shift, but it is negligible compared with their wavelength.
4. Visible photons carry more energy than X-rays, so a small energy loss alters their wavelength less. — A student who thinks longer-wavelength photons carry more energy picks this. E = hc/λ: a 500 nm photon has about 7000 times LESS energy than a 0.071 nm X-ray photon. In fact the shift at 90°, 2.4 × 10⁻¹² m, is the same for both; it is simply a tiny fraction of the visible wavelength.

Working Δλ = (h/m_ec)(1 − cos 90°) = 2.4 × 10⁻¹² m × 1 = 2.4 × 10⁻¹² m for both. Fraction of X-ray wavelength: 2.4 × 10⁻¹²/7.1 × 10⁻¹¹ = 0.034 (3%). Fraction of visible wavelength: 2.4 × 10⁻¹²/5.0 × 10⁻⁷ = 4.8 × 10⁻⁶ (about 0.0005%).

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1. It shows that a photon also carries momentum h/λ, exchanged with one electron in a collision that conserves momentum and energy. — The photoelectric effect tests only the photon's energy. Compton's results, including the dependence of the shift on angle, are reproduced by treating the event as a two-body collision in which a photon of momentum h/λ and energy hc/λ collides with one electron and both quantities are conserved. This is additional, independent evidence that light behaves as particles.
2. X-rays are scattered in every direction by the electrons in the target, which a wave could not do, so the X-rays must be particles. — A student who takes any particle-sounding description as evidence picks this. Waves are also scattered in all directions by electrons, so the direction of scattering distinguishes nothing. The evidence lies in the change of wavelength and its dependence on angle.
3. It shows that X-rays consist of photons, whereas the photoelectric effect shows this only for visible and ultraviolet light. — A student who thinks the photon model applies only to certain parts of the spectrum picks this. The photoelectric effect occurs with X-rays too, and the photon description applies to radiation of every frequency. The new evidence is about momentum, not about which radiation counts as photons.
4. Photons must have mass, because only a particle with mass can carry momentum into a collision with an electron. — A student who reasons from p = mv that momentum needs mass picks this. A photon has zero rest mass but carries momentum h/λ = E/c; Compton scattering demonstrates that this momentum is exchanged with the electron, not that photons are massive.

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1. 1.39 × 10⁻¹⁵ J — The photon loses energy to the electron. E_i = hc/λ_i = 9.04 × 10⁻¹⁵ J and E_f = hc/λ_f = 7.65 × 10⁻¹⁵ J, so the electron gains E_i − E_f = 1.39 × 10⁻¹⁵ J as kinetic energy.
2. 9.04 × 10⁻¹⁵ J — A student who treats the event like the photoelectric effect, with the whole photon energy hc/λ_i absorbed by the electron, picks this. In Compton scattering the photon survives with energy hc/λ_f = 7.65 × 10⁻¹⁵ J, and the electron receives only the difference.
3. 4.97 × 10⁻¹⁴ J — A student who substitutes the change in wavelength into E = hc/λ, giving hc/Δλ with Δλ = 4.0 × 10⁻¹² m, picks this. That is five times the incident photon's whole energy, which is impossible. E is inversely proportional to λ, so subtract the two photon energies.
4. 4.64 × 10⁻²⁴ J — A student who uses h/λ as the photon energy picks this. h/λ is the photon's momentum, so h/λ_i − h/λ_f = 4.64 × 10⁻²⁴ kg m s⁻¹ is a momentum, not an energy. The photon energy is hc/λ, a factor c larger.

Working Incident photon energy E_i = hc/λ_i = (6.63 × 10⁻³⁴ J s)(3.00 × 10⁸ m s⁻¹)/(2.20 × 10⁻¹¹ m) = 9.04 × 10⁻¹⁵ J. Scattered photon energy E_f = hc/λ_f = (1.989 × 10⁻²⁵ J m)/(2.60 × 10⁻¹¹ m) = 7.65 × 10⁻¹⁵ J. Energy is conserved and the electron starts at rest, so E_k = E_i − E_f = 9.04 × 10⁻¹⁵ − 7.65 × 10⁻¹⁵ = 1.39 × 10⁻¹⁵ J (about 8.7 keV). The shift of 4.0 × 10⁻¹² m is less than the maximum Compton shift 2h/m_e c = 4.85 × 10⁻¹² m, so the data are physically possible.

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