15.1 Quantum Theory and Wave-Particle Duality

1. AHow the intensity of the radiation emitted by a hot, glowing object varies with wavelength Correct
   Classical physics could not account for the way the intensity of blackbody radiation is distributed over wavelengths; explaining it was one of the problems that led to quantum theory, along with atomic spectra and the photoelectric effect.
2. BThe pattern of bright and dark bands formed by light passing through two narrow slits
   A student who links the double-slit experiment with quantum physics picks this. The interference pattern of light is explained by the classical wave model; it was the evidence, in Young's experiment, that light behaves as a wave.
3. CThe curved path followed by a beam of electrons as it moves through a magnetic field
   A student who thinks anything involving electrons needs quantum theory picks this. The curved path of an electron beam many centimeters across is explained by the classical magnetic force on a moving charge.
4. DThe random, jittery motion of pollen grains suspended in a drop of water under a microscope
   A student who thinks anything microscopic is quantum picks this. The jittery motion of pollen grains comes from collisions with water molecules and is explained by classical physics.

CED 15.1.A.1 · Read this in Fix

1. AThe photon model replaced the wave model, which experiments showed to be wrong.
   A student who thinks newer models overturn older ones picks this. The wave model still correctly explains interference and diffraction; the photon model adds what the wave model cannot explain.
2. BThe two combine: photons are particles traveling along wave-shaped paths.
   A student who pictures the wave as a wiggling path picks this. A photon in empty space travels in a straight line; light's wave behavior shows up in interference patterns, not in the shape of a path.
3. CEach model explains some observations of light, so both models are needed. Correct
   Light can be modeled both as a wave and as discrete photons. The wave model explains interference and diffraction; the photon model explains observations in which light delivers energy in discrete amounts. Both are needed.
4. DThe wave model explains energy transfer; photons explain double-slit bands.
   A student who thinks the double-slit pattern is evidence for photons picks this. The bright and dark bands of a double slit are explained by the wave model; photons are needed to explain energy delivered in discrete amounts.

CED 15.1.A.2 · Read this in Fix

1. Av_X = 2v_Y
   A student who reads v = fλ as 'speed increases with frequency' picks this: X has twice the frequency. Its wavelength is also half as long, so fλ, the speed, is c for both.
2. Bv_X = √2 v_Y
   A student who treats a photon's energy as ½mv² picks this: X has twice the energy, so its speed would be √2 times as great. Photons are massless, and all move at c in free space.
3. Cv_X = v_Y/2
   A student who reads v = fλ as 'speed increases with wavelength' picks this: X has half the wavelength. Its frequency is twice as great, so fλ, the speed, is c for both.
4. Dv_X = v_Y Correct
   All photons travel at c in free space. Photon X has half the wavelength and so twice the frequency of Y, and fλ = c for both.

Working In vacuum v = c for every photon. λ_X = λ_Y/2 means f_X = 2f_Y, and f_Xλ_X = f_Yλ_Y = c, so v_X/v_Y = 1.

CED 15.1.A.3.i · Read this in Fix

1. A7.0 × 10⁻¹⁰ m
   A student who uses the photon relationship λ = hc/E with the electron's kinetic energy picks this: K = ½mv² = 2.8 × 10⁻¹⁶ J and hc/K = 7.0 × 10⁻¹⁰ m. That relationship holds only for photons; for the electron, λ = h/(mv).
2. B2.4 × 10⁻¹² m
   A student who uses the speed of light in the momentum picks this: h/(mc) = 2.4 × 10⁻¹² m. The electron moves at 2.5 × 10⁷ m/s, not c, so p = mv.
3. C2.3 × 10⁻¹⁸ m
   A student who puts the kinetic energy in place of the momentum picks this: h/K = h/(½mv²) = 2.3 × 10⁻¹⁸ m. The de Broglie wavelength uses the momentum, p = mv.
4. D2.9 × 10⁻¹¹ m Correct
   λ = h/p = h/(mv) = (6.63 × 10⁻³⁴ J·s)/[(9.11 × 10⁻³¹ kg)(2.5 × 10⁷ m/s)] = 2.9 × 10⁻¹¹ m.

Working p = mv = (9.11 × 10⁻³¹ kg)(2.5 × 10⁷ m/s) = 2.28 × 10⁻²³ kg·m/s. λ = h/p = (6.63 × 10⁻³⁴ J·s)/(2.28 × 10⁻²³ kg·m/s) = 2.9 × 10⁻¹¹ m.

CED 15.1.A.4.i · Read this in Fix

1. AAn electron moving freely in a vacuum tube
   A student who thinks every electron has quantized energy picks this. A free electron is not bound, and its kinetic energy can have any value in a continuous range.
2. BA photon of light from a glowing filament
   A student who thinks photon energies are restricted to fixed values picks this. Each photon has energy hf, but a glowing filament emits photons with a continuous range of frequencies and so of energies.
3. CA bacterium swimming in a drop of water
   A student who thinks anything microscopic is quantized picks this. A bacterium is made of billions of atoms, and its kinetic energy can have any value.
4. DAn electron bound to the nucleus of an atom Correct
   Energy has discrete, quantized values for bound systems described by quantum theory, such as an electron bound to a nucleus.

CED 15.1.A.5 · Read this in Fix

1. AThe motion of a spacecraft that moves at 90% of the speed of light
   A student who thinks quantum theory is the physics of very fast objects picks this. A spacecraft near the speed of light needs special relativity, a different theory; it is far too large for quantum effects to matter.
2. BThe energy of an electron bound to the proton in a hydrogen atom Correct
   Quantum theory is necessary to describe matter at atomic and subatomic scales. An electron bound in an atom is such a system: its energy can take only certain values, which classical physics cannot explain.
3. CThe motion of a bacterium that swims through a drop of pond water
   A student who thinks anything microscopic needs quantum theory picks this. A bacterium is made of billions of atoms, and its motion is described well by classical physics.
4. DThe path of an electron beam crossing a tube 30 cm long
   A student who thinks anything involving electrons needs quantum theory picks this. The beam's path across a tube tens of centimeters long is described well by classical physics; quantum theory is needed at atomic scales.

CED 15.1.A.1.i · Read this in Fix

1. AThey are tiny particles, and only light, not matter, can form an interference pattern.
   A student who thinks only light can behave as a wave picks this. Electron beams produce interference patterns when they pass through two slits or a crystal, which shows that electrons have wave properties.
2. BThey are particles moving along wavy paths, and this wiggling is what makes them wave-like.
   A student who pictures the wave as a wiggling path picks this. Wave behavior shows up as interference and diffraction patterns, not as a sideways wiggle in the electron's path.
3. CThey are more wave-like the faster they move, as their wavelength grows with speed.
   A student who reads λ = h/p as a direct relationship picks this. The de Broglie wavelength decreases as the momentum increases, so faster electrons have shorter wavelengths.
4. DEach is detected at a single point, yet an electron beam can form an interference pattern. Correct
   In quantum theory fundamental particles such as electrons show both particle-like behavior (each electron is detected at one point) and wave-like behavior (a beam of electrons produces interference and diffraction patterns).

CED 15.1.A.1.ii · Read this in Fix

1. AGraph A
   A student who thinks longer-wavelength photons have more energy picks this: energy falling as frequency rises. In fact E = hf rises in proportion to f.
2. BGraph B Correct
   A photon's energy is proportional to its frequency, E = hf, so the graph is a straight line through the origin with slope h: doubling f doubles E.
3. CGraph C
   A student who thinks a photon's energy depends on the brightness of the light rather than its frequency picks this horizontal line. E = hf, so the energy rises with frequency.
4. DGraph D
   A student who takes any rising straight line to show proportionality picks this. Graph D has a positive intercept, so doubling f would not double E; E = hf is a straight line through the origin.

Working E = hf: E is proportional to f, so the graph is a straight line through the origin (slope h). Graph B.

CED 15.1.A.2.i · Read this in Fix

1. A3.98 × 10⁻¹⁹ J Correct
   A photon's energy is E = hf, and the frequency does not change when light enters water. f = c/λ = (3.00 × 10⁸ m/s)/(500 × 10⁻⁹ m) = 6.00 × 10¹⁴ Hz, so E = (6.63 × 10⁻³⁴ J·s)(6.00 × 10¹⁴ Hz) = 3.98 × 10⁻¹⁹ J, in air and in water.
2. B5.29 × 10⁻¹⁹ J
   A student who uses E = hc/λ with the shorter wavelength in water, 500 nm/1.33 = 376 nm, picks this. E = hc/λ holds with the wavelength in vacuum (or air); in water the frequency, and so the energy, is unchanged.
3. C2.25 × 10⁻¹⁹ J
   A student who treats a photon's energy as kinetic energy picks this: the light slows by a factor of 1.33, so the energy is divided by 1.33². A photon's energy is hf, and f does not change in water.
4. D1.33 × 10⁻²⁷ J
   A student who takes h/λ as the photon's energy picks this. h/λ has units of kg·m/s, not joules; the photon's energy is hc/λ = 3.98 × 10⁻¹⁹ J.

Working f = c/λ = (3.00 × 10⁸ m/s)/(500 × 10⁻⁹ m) = 6.00 × 10¹⁴ Hz, unchanged in water. E = hf = (6.63 × 10⁻³⁴ J·s)(6.00 × 10¹⁴ Hz) = 3.98 × 10⁻¹⁹ J.

CED 15.1.A.2.i · Read this in Fix

1. AIt travels along a wavy path, swinging from side to side.
   A student who pictures light's wave behavior as a wiggling path picks this. The photon's path is a straight line; its wave behavior shows up in interference and diffraction.
2. BIt travels in a straight line, slowing as it loses energy.
   A student who thinks light runs out of energy as it travels picks this. In a vacuum a photon keeps its speed c and its energy; distant sources look dim because their photons are spread over a larger area.
3. CIt travels in a straight line at a constant speed. Correct
   Photons travel in straight lines unless they interact with matter, and in free space every photon moves at the constant speed c.
4. DIt travels in a straight line at a speed set by its frequency.
   A student who thinks higher-frequency photons move faster picks this. Every photon in free space moves at the same speed, c, whatever its frequency.

CED 15.1.A.2.ii · Read this in Fix

1. Ahc/λ_g
   A student who thinks photons travel at c in glass picks this, taking f = c/λ_g. In the glass the speed is c/n, so f = c/(nλ_g).
2. Bnhc/λ_g
   A student who thinks light is faster in a medium of greater index picks this, taking v = nc. The speed in the glass is c/n, smaller than c.
3. Chc/(nλ_g) Correct
   In the glass the photon's speed is v = c/n, so its frequency is f = v/λ_g = c/(nλ_g), and E = hf = hc/(nλ_g). Equivalently, the wavelength in vacuum is nλ_g.
4. Dhc/(n³λ_g)
   A student who treats a photon's energy as kinetic energy picks this: the energy in vacuum, hc/(nλ_g), divided by n² because the photon's speed in the glass is c/n. A photon's energy is hf, and the frequency does not change in the glass, so the energy is hc/(nλ_g).

Working v = c/n in the glass; f = v/λ_g = c/(nλ_g); E = hf = hc/(nλ_g).

CED 15.1.A.3.ii · Read this in Fix

1. AGraph D Correct
   The speed of photons in a medium is inversely proportional to its index of refraction, v = c/n: c at n = 1, c/2 at n = 2 and c/3 at n = 3. That is a curve that falls steeply at first and then more gradually.
2. BGraph A
   A student who thinks light travels faster in a medium of greater index, as sound does in denser media, picks this. For light, v = c/n decreases as n increases.
3. CGraph B
   A student who pictures inverse proportion as a straight line sloping down picks this. Graph B gives 2c/3 at n = 2, but v = c/n gives c/2 there; the graph of an inverse proportion is a curve.
4. DGraph C
   A student who thinks photons travel at c in every medium picks this horizontal line. In a medium of index n, the speed is c/n.

Working v = c/n: n = 1 → c, n = 2 → c/2, n = 3 → c/3 — a curve, falling steeply then gradually (Graph D). Graph B, a straight line from c to c/3, gives 2c/3 at n = 2.

CED 15.1.A.3.ii · Read this in Fix

1. AThe bands form because electrons collide with one another as they pass the slits.
   A student who explains the pattern with colliding particles picks this. The same bands build up even when the beam is so weak that the electrons pass through one at a time, so the pattern comes from each electron's wave behavior.
2. BEach electron arrives at one point, but together they build up an interference pattern. Correct
   Every electron makes a single dot, which is particle-like behavior. As more electrons arrive, the dots build up several evenly spaced bands, an interference pattern, which is wave-like behavior: the electrons demonstrate wave properties in this version of Young's double-slit experiment.
3. CEach electron spreads out as a wave and arrives over the whole screen at once.
   A student who thinks an electron is spread out like a classical wave picks this. The images show each electron as a single dot at one point; the wave behavior appears only in the pattern made by many dots.
4. DThe electrons travel in straight lines, so the bands are merely shadows of the two slits.
   A student who treats the electrons as classical particles picks this. Straight-line particles would give two bands, one behind each slit; the images show several evenly spaced bands, an interference pattern.

CED 15.1.A.4 · Read this in Fix

1. AIt becomes three times as long. Correct
   λ = h/p = h/(mv). With the speed one-third as great, the momentum is one-third as great, so the wavelength is three times as long.
2. BIt becomes one-third as long.
   A student who thinks the wavelength grows with momentum picks this. λ = h/p is an inverse relationship: less momentum means a longer wavelength.
3. CIt becomes nine times as long.
   A student who uses the kinetic energy in place of the momentum picks this: K falls by a factor of 9, so h/K rises by 9. The wavelength depends on p = mv, which falls by a factor of 3 only.
4. DIt stays the same length.
   A student who uses the speed of light in the momentum picks this, giving h/(mc), which does not depend on the electron's speed. The momentum is mv, so the wavelength depends on v.

Working λ = h/(mv); v → v/3 gives λ → 3λ.

CED 15.1.A.4.i · Read this in Fix

1. AX > Z > Y > W
   A student who thinks a larger momentum gives a longer wavelength picks this, the momentum ranking itself. λ = h/p, so the particle with the largest momentum, X, has the shortest wavelength.
2. BW > Z > X > Y
   A student who ranks by speed alone, slowest first, picks this. The wavelength depends on the momentum mv: Y is the fastest, but its small mass gives it the second-smallest momentum and so the second-longest wavelength.
3. CW > Y > Z > X Correct
   λ = h/p = h/(mv), so the longest wavelength goes with the smallest momentum. Momenta, in units of m₀v₀: W 2 × 1 = 2, Y 1 × 5 = 5, Z 3 × 2 = 6, X 4 × 3 = 12. Wavelengths from longest to shortest: W > Y > Z > X.
4. DY > W > Z > X
   A student who ranks by mass alone, lightest first, picks this. The wavelength depends on the momentum mv: Y is the lightest, but it is so fast that its momentum (5) exceeds W's (2).

Working p = mv (units m₀v₀): W 2, X 12, Y 5, Z 6. λ ∝ 1/p: W > Y > Z > X.

CED 15.1.A.4.i · Read this in Fix

1. AWhen they pass through a slit 1.0 mm wide that is cut in a thin metal sheet
   A student who thinks any narrow slit produces wave effects picks this. A 1.0 mm slit is millions of times wider than the wavelength, so diffraction is negligible and the electrons pass straight through.
2. BWhen they pass between rows of atoms in a crystal, 2.0 × 10⁻¹⁰ m apart Correct
   Quantum theory is necessary when the de Broglie wavelength is comparable to the size of the system. The spacing of the atoms, 2.0 × 10⁻¹⁰ m, is comparable to 1.2 × 10⁻¹⁰ m, so the electrons diffract and a wave description is needed.
3. CWhen they pass a pollen grain 2.0 × 10⁻⁵ m across, visible only by microscope
   A student who thinks anything too small to see without a microscope needs quantum theory picks this. The grain is about 1.7 × 10⁵ times the electrons' wavelength (2.0 × 10⁻⁵ m ÷ 1.2 × 10⁻¹⁰ m), so the electrons' motion past it is described well by classical physics.
4. DWhen they move through empty space at their speed of 6.1 × 10⁶ m/s
   A student who thinks electrons' energies are always quantized, even when they move freely, picks this. An electron moving freely is not bound, so its energy is not quantized. In empty space it is not confined to any region comparable to its wavelength, and classical physics describes its motion.

Working Compare λ = 1.2 × 10⁻¹⁰ m with each size: crystal spacing 2.0 × 10⁻¹⁰ m (comparable, quantum needed); slit 1.0 × 10⁻³ m (about 10⁷ λ); pollen grain 2.0 × 10⁻⁵ m (about 1.7 × 10⁵ λ). Speed from λ: v = h/(mλ) = 6.1 × 10⁶ m/s.

CED 15.1.A.4.ii · Read this in Fix

1. Ah/(eΔV)
   A student who puts the kinetic energy eΔV in place of the momentum in λ = h/p picks this. p is the momentum, p = √(2mK), not K; h/(eΔV) has units of time, not of length.
2. Bhc/(eΔV)
   A student who uses the photon relationship λ = hc/E with the electron's kinetic energy eΔV picks this. That relationship holds only for photons, which travel at c; for the electron, λ = h/p with p = √(2meΔV).
3. Ch/√(2meΔV) Correct
   The electron gains kinetic energy K = eΔV. Since K = p²/(2m), its momentum is p = √(2meΔV), so its de Broglie wavelength is λ = h/p = h/√(2meΔV).
4. Dh/√(2mΔV)
   A student who takes the energy the electron gains to be ΔV itself, not eΔV, picks this. ΔV is energy per unit charge; the electron's kinetic energy is its charge times the potential difference, K = eΔV.

Working Energy: the electron gains kinetic energy K = eΔV. Momentum: K = p²/(2m), so p = √(2mK) = √(2meΔV). de Broglie: λ = h/p = h/√(2meΔV). Errors: K put in place of p → h/(eΔV) (units of time); photon relationship λ = hc/E → hc/(eΔV); K taken as ΔV → h/√(2mΔV).

CED 15.1.A.4.i · Read this in Fix

1. A√(2E/m)
   A student who finds the electron's wavelength from its kinetic energy with the photon relationship λ = hc/K picks this: equal wavelengths then mean K = E, and ½mv² = E. The relationship λ = hc/E holds only for photons; the electron's wavelength is h/(mv), so equal wavelengths fix the electron's momentum mv = E/c, not its energy.
2. BE/m
   A student who takes a photon's energy to be h/λ picks this: the photon's wavelength is then h/E, and h/(mv) = h/E gives v = E/m, which does not even have units of speed. The photon's energy is hc/λ, so its wavelength is hc/E.
3. CE/(mc) Correct
   The photon's wavelength is λ = c/f = hc/E, and the electron's de Broglie wavelength is h/(mv). Setting them equal gives mv = E/c, so v = E/(mc). Equal wavelengths do not mean equal energies: the electron's kinetic energy is not E.
4. D√(2E/(mc))
   A student who replaces the p in λ = h/p with the kinetic energy picks this: h/K = hc/E gives K = E/c, and then ½mv² = E/c. The p in λ = h/p is the momentum mv, and h/K does not have units of length.

Working Photon: E = hf and λ = c/f, so λ = hc/E. Electron: λ = h/p = h/(mv). Equal wavelengths: h/(mv) = hc/E, so mv = E/c and v = E/(mc). Equal wavelengths fix the electron's momentum, mv = E/c; they do not make its energy equal to E. Errors: the photon relationship λ = hc/K used for the electron → K = E → ½mv² = E → v = √(2E/m); λ = h/K for the electron → h/K = hc/E → ½mv² = E/c → v = √(2E/(mc)); photon energy taken as h/λ → photon wavelength h/E → mv = E → v = E/m.

CED 15.1.A.2.i · Read this in Fix

1. A1.89 × 10⁻¹⁵ J
   A student who uses the photon relationship E = hc/λ for the electron picks this. That relationship holds only for photons, which travel at c; for the electron, find p = h/λ first and then K = p²/(2m_e).
2. B2.19 × 10⁻¹⁷ J Correct
   The wavelength fixes the momentum: p = h/λ = 6.31 × 10⁻²⁴ kg·m/s. At a speed much less than c, K = p²/(2m_e) = 2.19 × 10⁻¹⁷ J.
3. C6.31 × 10⁻²⁴ J
   A student who treats the p in λ = h/p as the kinetic energy picks this, taking K = h/λ. That number is the electron's momentum in kg·m/s; the kinetic energy is p²/(2m_e).
4. D4.38 × 10⁻¹⁷ J
   A student who writes the kinetic energy as p²/m, leaving out the factor ½, picks this. Since K = ½m_ev² and p = m_ev, K = p²/(2m_e), half of this value.

Working p = h/λ = (6.63 × 10⁻³⁴ J·s)/(0.105 × 10⁻⁹ m) = 6.31 × 10⁻²⁴ kg·m/s. K = p²/(2m_e) = (6.31 × 10⁻²⁴ kg·m/s)²/(2 × 9.11 × 10⁻³¹ kg) = 2.19 × 10⁻¹⁷ J (about 137 eV). Errors: photon relationship K = hc/λ → 1.89 × 10⁻¹⁵ J; λ = h/K, so K = h/λ → 6.31 × 10⁻²⁴ (this has units of momentum, not energy); K = p²/m without the ½ → 4.38 × 10⁻¹⁷ J.

CED 15.1.A.4.i · Read this in Fix