University Physics IV · Midterm Examination

Cavity radiation at temperature 300 K is examined at wavelength 1.00 mm. By what factor does the Rayleigh–Jeans spectral energy density exceed the Planck spectral energy density at this wavelength? Take h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹ and k_B = 1.38 × 10⁻²³ J K⁻¹.

1. A1.00, because the two laws agree exactly at long wavelength
2. B1.02
3. C1.05
4. D20.8

A proton moves at a speed for which γ = 2.00. A force of magnitude F∥ applied parallel to its velocity gives it an acceleration of magnitude a. A force of magnitude F⊥ applied perpendicular to its velocity gives it an acceleration of the same magnitude a. What is F∥/F⊥ at that instant?

1. A0.250
2. B1.00
3. C4.00
4. D8.00

A wave packet for a free, non-relativistic electron is built from de Broglie waves with angular frequency ω(k) = E/ℏ, where E = ℏ²k²/(2m) + V₀ and V₀ is a constant potential energy whose value depends on where the zero of energy is chosen. The packet is centred on wavenumber k₀, and v = ℏk₀/m is the electron's classical speed. Which statement is correct?

1. AThe phase velocity is v/2 and the group velocity is v, whatever value of V₀ is chosen.
2. BThe phase velocity is v and the group velocity is v/2.
3. CBoth the phase velocity and the group velocity change when V₀ is changed.
4. DThe group velocity is v for every choice of V₀, but the phase velocity depends on V₀.

Two identical narrow wave packets, ψ_L centred at x = −d and ψ_R centred at x = +d, do not overlap anywhere. Many copies of a particle are prepared either (I) in the superposition (ψ_L + ψ_R)/√2, or (II) as a statistical mixture in which half the copies are in ψ_L and half are in ψ_R. Which measurement, repeated on many copies, distinguishes preparation I from preparation II?

1. AMomentum: for I the momentum distribution is modulated by fringes of spacing h/(2d), and for II it is not.
2. BPosition: for I the position distribution shows interference fringes between the two packets, and for II it does not.
3. CThe mean position ⟨x⟩, which is zero for I but not for II.
4. DNone, because in both preparations the particle is found near each centre with probability ½.

Ehrenfest's theorem gives d⟨x⟩/dt = ⟨p⟩/m and d⟨p⟩/dt = −⟨∂V/∂x⟩. For which potential does ⟨x⟩ obey exactly the classical equation m d²⟨x⟩/dt² = −V′(⟨x⟩) for every state of the particle?

1. AV(x) = αx⁴ with α > 0, because it is a symmetric confining well.
2. BAny potential, provided the wave packet is narrow at t = 0.
3. CV(x) = V₀ cos(qx), because the force it exerts is bounded.
4. DV(x) = ½mω²x² + Fx, with F a constant.

Electrons of kinetic energy 8.00 eV travel in the +x direction towards a potential step at which the potential energy rises abruptly from 0 to 6.00 eV. What fraction of the incident probability current is reflected?

1. A0
2. B0.111
3. C0.333
4. D0.360

For a particle of mass m in a harmonic oscillator of angular frequency ω, x̂ = √(ℏ/(2mω)) (â + â†), where â|n⟩ = √n |n − 1⟩ and â†|n⟩ = √(n + 1) |n + 1⟩. What is ⟨x²⟩ in the energy eigenstate |n⟩?

1. A(n + ½)ℏ/(mω)
2. Bnℏ/(mω)
3. C(2n + 1)ℏ/(mω)
4. DZero

An inspection gantry at rest in frame S has two laser gates: G₁ at x = −60.0 m and G₂ at x = +60.0 m, so the gantry's rest length is 120 m. A cargo shuttle of rest length 150 m moves in the +x direction at 0.800c. Frame S′ is the shuttle's rest frame, in standard configuration with S: the origins coincide at t = t′ = 0, when the centre of the shuttle passes x = 0. Each gate sits at the edge of the track and can fire a brief pulse straight across it, taking negligible time to cross; a pulse that meets the shuttle leaves a mark on its hull. The gantry controller fires both gates at t = 0 in S. Take c = 3.00 × 10⁸ m s⁻¹.

1. (a)

   Calculate Calculate the length of the shuttle measured in S, and the length of the gantry measured in S′.

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   1. 1

      γ = 1/√(1 − 0.800²) = 5/3, so the shuttle's length in S is 150 m/γ = 90.0 m; at t = 0 it spans x = −45.0 m to +45.0 m, clear of both gates

   2. 2

      The gantry's length in S′ is 120 m/γ = 72.0 m

2. (b)

   Calculate The firing of G₁ is event E₁ and the firing of G₂ is event E₂. Calculate the coordinates x′ and t′ of both events in S′, and state which gate fires first in S′ and by how much.

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   1. 1

      Uses x′ = γ(x − vt) and t′ = γ(t − vx/c²) with t = 0 for both events

   2. 2

      E₁: x′ = −100 m, t′ = +γ(0.800)(60.0 m)/c = +2.67 × 10⁻⁷ s; E₂: x′ = +100 m, t′ = −2.67 × 10⁻⁷ s

   3. 3

      G₂, the gate the shuttle is heading towards, fires first in S′, 5.33 × 10⁻⁷ s before G₁

3. (c)

   Explain In S′ the shuttle is 150 m long and the gantry is only 72.0 m long. Explain, using your results from (b), why neither pulse strikes the shuttle in S′ either.

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   1. 1

      In S′ the shuttle occupies −75.0 m ≤ x′ ≤ +75.0 m at all times; E₂ happens at x′ = +100 m, 25.0 m ahead of the nose, and E₁ at x′ = −100 m, 25.0 m behind the tail, so neither event is on the hull

   2. 2

      Between the two firings the gantry moves (0.800c)(5.33 × 10⁻⁷ s) = 128 m in the −x′ direction, so G₂ fires before it reaches the nose and G₁ fires after it has passed the tail

   3. 3

      Whether a pulse meets the hull is a coincidence of events at one place, on which every frame agrees; the statement that the shuttle is between the gates when they fire depends on E₁ and E₂ being simultaneous, which is true in S and false in S′, so there is no contradiction

4. (d)

   Determine A drone is launched forwards from the shuttle's nose at 0.600c relative to the shuttle. By composing rapidities, determine the drone's speed in S.

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   1. 1

      Rapidities φ = artanh β add along a line: artanh 0.800 = ln 3 = 1.099 and artanh 0.600 = ln 2 = 0.693, so φ = ln 6 = 1.792

   2. 2

      β = tanh(1.792) = 35/37 = 0.946, so the speed is 0.946c = 2.84 × 10⁸ m s⁻¹, in agreement with (0.800 + 0.600)/(1 + 0.480) and below c

A secondary beam of positive kaons, each with momentum 2.00 GeV/c, travels along an evacuated decay tunnel. A K⁺ at rest has mean lifetime τ = 1.238 × 10⁻⁸ s and decays mainly by K⁺ → μ⁺ + ν_μ. Take m_K c² = 493.7 MeV, m_μ c² = 105.7 MeV, m_p c² = 938.3 MeV, m_Λ c² = 1115.7 MeV and c = 3.00 × 10⁸ m s⁻¹, and treat the neutrino as massless.

1. (a)

   Calculate Calculate the mean decay length of the kaons in the laboratory.

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   1. 1

      βγ = pc/(m_K c²) = 2000 MeV/493.7 MeV = 4.05 (γ = 4.17, β = 0.971)

   2. 2

      Mean decay length = βγcτ = 4.05 × (3.00 × 10⁸ m s⁻¹)(1.238 × 10⁻⁸ s) = 15.0 m

2. (b)

   Derive Starting from conservation of four-momentum, derive expressions for the muon's energy E* and momentum p* in the kaon rest frame, and evaluate both.

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   1. 1

      Writes P_K = p_μ + p_ν and isolates the neutrino: p_ν² = (P_K − p_μ)² = 0 because the neutrino is massless

   2. 2

      Expands in the kaon rest frame: m_K²c⁴ + m_μ²c⁴ − 2m_K c² E* = 0, so E* = (m_K² + m_μ²)c²/(2m_K) = 258.2 MeV

   3. 3

      p*c = √(E*² − m_μ²c⁴) = (m_K² − m_μ²)c²/(2m_K) = 235.5 MeV, so p* = 235.5 MeV/c, equal to the neutrino's energy divided by c

3. (c)

   Determine Determine the largest energy a muon from this beam can have in the laboratory, and state the direction of emission in the kaon rest frame that gives it.

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   1. 1

      Boosting, E = γ(E* + βp*c cos θ*), which is greatest for emission along the beam direction, θ* = 0

   2. 2

      E_max = 4.173 × (258.2 MeV + 0.9709 × 235.5 MeV) = 2031 MeV ≈ 2.03 GeV (accept 2.02 to 2.04 GeV), just below the kaon's energy of 2060 MeV

4. (d)

   Calculate The kaons are produced when a proton beam strikes protons at rest, through p + p → p + Λ + K⁺. Calculate the minimum kinetic energy of a beam proton for this reaction.

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   1. 1

      At threshold the products are at rest in the centre-of-momentum frame, so the invariant (E_beam + m_p c²)² − (p_beam c)² equals (m_p + m_Λ + m_K)²c⁴ = (2547.7 MeV)²

   2. 2

      The invariant expands to 2m_p²c⁴ + 2E_beam m_p c², so E_beam = [(2547.7)² − 2(938.3)²]/(2 × 938.3) MeV = 2520 MeV

   3. 3

      K_th = E_beam − m_p c² = 1582 MeV ≈ 1.58 GeV, far more than the 671 MeV of rest energy created, because momentum conservation requires the products to move in the laboratory

Technetium-99m, used in medical gamma imaging, emits photons of energy 140.5 keV. Before reaching the camera, many of these photons Compton scatter in soft tissue from electrons that may be treated as free and at rest. The camera rejects scattered photons only by their energy: it accepts any photon whose energy is at least 90.0% of 140.5 keV. Take m_e c² = 511 keV, h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹ and e = 1.60 × 10⁻¹⁹ C.

1. (a)

   Derive A photon of wavelength λ scatters through angle θ from a free electron at rest. By conserving relativistic energy and both components of momentum, derive λ′ − λ = (h/(m_e c))(1 − cos θ) for the scattered wavelength λ′.

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   1. 1

      Energy: pc + m_e c² = p′c + E_e, with p = h/λ and p′ = h/λ′ for the incident and scattered photons

   2. 2

      Momentum as a vector: p_e = p − p′, so p_e² = p² + p′² − 2pp′ cos θ, which uses both components

   3. 3

      Eliminates the electron with E_e² = p_e²c² + m_e²c⁴: (pc − p′c + m_e c²)² = (p² + p′² − 2pp′ cos θ)c² + m_e²c⁴, which reduces to m_e c(p − p′) = pp′(1 − cos θ)

   4. 4

      Divides by m_e c pp′ to get 1/p′ − 1/p = (1 − cos θ)/(m_e c), and substitutes 1/p = λ/h to reach λ′ − λ = (h/(m_e c))(1 − cos θ)

2. (b)

   Calculate Calculate the largest angle through which a photon can be scattered once and still be accepted by the camera.

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   1. 1

      In energy form 1/E′ − 1/E = (1 − cos θ)/(m_e c²), with E′ = 0.900 × 140.5 keV = 126.45 keV

   2. 2

      1 − cos θ = 511 keV × (1/126.45 keV − 1/140.5 keV) = 0.404

   3. 3

      cos θ = 0.596, so θ_max = 53.4°

3. (c)

   Calculate Calculate the de Broglie wavelength of the recoil electron for the scattering in (b).

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   1. 1

      K = 140.5 keV − 126.45 keV = 14.05 keV, and pc = √(K² + 2K m_e c²) = 120.7 keV

   2. 2

      λ_e = hc/(pc) = (6.63 × 10⁻³⁴ J s × 3.00 × 10⁸ m s⁻¹)/(120.7 × 10³ × 1.60 × 10⁻¹⁹ J) = 1.03 × 10⁻¹¹ m = 10.3 pm

4. (d)

   Compare A PET camera detects 511 keV annihilation photons and uses the same fractional window, accepting photons of at least 90.0% of 511 keV. Compare its largest accepted single-scattering angle with your answer to (b), and explain the difference.

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   1. 1

      For 511 keV, 1 − cos θ = (511 keV/511 keV)(1/0.900 − 1) = 0.111, giving θ_max = 27.3°, about half the 53.4° accepted at 140.5 keV

   2. 2

      In general 1 − cos θ_max = (m_e c²/E)(1/0.900 − 1): at a given angle a photon with E well below m_e c² loses a smaller fraction of its energy, so it must scatter through a larger angle before falling below 90.0%, and the same fractional window accepts a much wider cone at 140.5 keV than at 511 keV

An electron in a GaAs quantum well is modelled as a particle of effective mass m* = 0.067m_e in a one-dimensional infinite square well of width L = 12.0 nm, between x = 0 and x = L. The energy eigenfunctions are ψ_n(x) = √(2/L) sin(nπx/L), with energies E_n = n²π²ℏ²/(2m*L²). At t = 0 the electron is prepared in the state Ψ(x, 0) = A[√3 ψ₁(x) + ψ₂(x)], with A real and positive. You may use ∫₀ᴸ x sin(πx/L) sin(2πx/L) dx = −8L²/(9π²). Take m_e = 9.11 × 10⁻³¹ kg, ℏ = 1.055 × 10⁻³⁴ J s and h = 6.63 × 10⁻³⁴ J s.

1. (a)

   Determine Determine A, and the probabilities that a measurement of the energy at t = 0 returns E₁ and returns E₂.

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   1. 1

      Orthonormality of ψ₁ and ψ₂ gives |A|²(3 + 1) = 1, so A = 1/2

   2. 2

      P(E₁) = |A√3|² = 3/4 and P(E₂) = |A|² = 1/4

2. (b)

   Calculate The mean position ⟨x⟩ of the electron oscillates. Calculate the period T of this oscillation.

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   1. 1

      E₁ = π²ℏ²/(2m*L²) = 6.25 × 10⁻²¹ J (39.1 meV), so E₂ − E₁ = 3E₁ = 1.87 × 10⁻²⁰ J

   2. 2

      T = 2πℏ/(E₂ − E₁) = h/(E₂ − E₁) = 3.54 × 10⁻¹⁴ s

3. (c)

   Show that Show that ⟨x⟩(t) = L/2 − (8√3L/(9π²)) cos ωt, where ω = (E₂ − E₁)/ℏ.

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   1. 1

      Ψ(x, t) = ½[√3 ψ₁ e^(−iE₁t/ℏ) + ψ₂ e^(−iE₂t/ℏ)], with each stationary state carrying its own phase

   2. 2

      ⟨x⟩ = ¾⟨ψ₁|x|ψ₁⟩ + ¼⟨ψ₂|x|ψ₂⟩ + (√3/2)⟨ψ₁|x|ψ₂⟩ cos ωt, and both diagonal terms equal L/2 because each |ψ_n|² is symmetric about the centre of the well

   3. 3

      ⟨ψ₁|x|ψ₂⟩ = (2/L)(−8L²/(9π²)) = −16L/(9π²), so ⟨x⟩ = L/2 − (8√3L/(9π²)) cos ωt, an oscillation of amplitude 0.156L = 1.87 nm

4. (d)

   Sketch On one set of axes, sketch |Ψ(x, t)|² for 0 ≤ x ≤ L at t = 0 and at t = T/2.

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   1. 1

      t = 0: zero at both walls, with a single main peak left of centre near x ≈ 0.32L, falling to zero at x = 5L/6 and staying very small between there and x = L

   2. 2

      t = T/2: the mirror image of the t = 0 curve about x = L/2, with its main peak near x ≈ 0.68L and a zero at x = L/6; the two curves enclose equal areas

5. (e)

   Predict At t = T/2 the energy of the electron is measured and the result is E₂. State the probability of this result, and predict ⟨x⟩ at all later times.

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   1. 1

      The probability is 1/4, the same at every time, since the modulus of each coefficient c_n e^(−iE_nt/ℏ) does not change

   2. 2

      The state is updated to ψ₂ (up to a phase), which is stationary, so there is no cross term and ⟨x⟩ = L/2 at all later times: the oscillation stops

A vacuum photocell has a cathode whose surface work function is quoted by its manufacturer as 2.00 eV, and an anode made of a different metal. Monochromatic light of each wavelength in the table falls on the cathode. An external supply applies a retarding potential difference between anode and cathode, which is increased until the photocurrent just falls to zero; that reading is the stopping potential V_s. Each V_s reading has an uncertainty of ±0.02 V, and the wavelengths may be treated as exact. The frequencies were calculated with c = 3.00 × 10⁸ m s⁻¹. Take e = 1.60 × 10⁻¹⁹ C.

1. (a)

   State State the relation between V_s and f predicted by Einstein's photon model, and what the gradient and the vertical intercept of a graph of V_s against f represent.

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   1. 1

      eV_s = hf − φ, so V_s = (h/e)f − φ/e

   2. 2

      The gradient is h/e, and the intercept on the V_s axis is −φ/e (equivalently, the intercept on the f axis is the threshold frequency φ/h)

2. (b)

   Determine Using a straight-line fit of V_s against f, determine the Planck constant h and the work function φ indicated by the data.

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   1. 1

      Plots or fits V_s against the tabulated f; all six points lie on one straight line within ±0.02 V

   2. 2

      Gradient = 4.12 × 10⁻¹⁵ V s (accept 4.05 × 10⁻¹⁵ to 4.20 × 10⁻¹⁵ V s)

   3. 3

      h = e × gradient = 6.6 × 10⁻³⁴ J s (accept 6.5 × 10⁻³⁴ to 6.7 × 10⁻³⁴ J s)

   4. 4

      Intercept on the V_s axis = −2.29 V (accept −2.19 V to −2.35 V), so φ = 2.29 eV; or threshold frequency 5.55 × 10¹⁴ Hz with φ = hf₀ = 2.29 eV

3. (c)

   Estimate Assuming the ±0.02 V reading uncertainty is the only significant source of error, estimate the uncertainty in your value of h.

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   1. 1

      Steepest and shallowest lines through the ends of the ±0.02 V error bars at 300 nm and 500 nm have gradients 4.20 × 10⁻¹⁵ V s and 4.00 × 10⁻¹⁵ V s

   2. 2

      Δh ≈ ½(6.72 − 6.40) × 10⁻³⁴ J s = 0.16 × 10⁻³⁴ J s, so h = (6.6 ± 0.2) × 10⁻³⁴ J s, a 2 to 3% uncertainty (a justified least-squares standard error is also accepted: about 0.10 × 10⁻³⁴ J s from σ = 0.02 V on each reading, or about 0.05 × 10⁻³⁴ J s from the scatter of the residuals)

4. (d)

   Justify A student concludes that the cathode's true work function is the value found in (b), and that the manufacturer's 2.00 eV is wrong. Justify the alternative conclusion that the intercept measures the work function of the anode, not the cathode.

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   1. 1

      An electron leaves the cathode with at most hf − φ_C, but between two different metals there is also a contact potential difference (φ_A − φ_C)/e, so the external stopping potential satisfies eV_s = (hf − φ_C) − (φ_A − φ_C) = hf − φ_A

   2. 2

      The intercept range 2.19 to 2.35 eV excludes 2.00 eV, so the 0.3 eV difference is significant and is accounted for as the contact potential; this does not depend on f, so the gradient and h are unaffected and the manufacturer's value can stand

A particle of mass m moves in the potential V(x) = ½mω²x², so Ĥ = p̂²/(2m) + ½mω²x̂², with [x̂, p̂] = iℏ. Define â = √(mω/(2ℏ)) (x̂ + ip̂/(mω)) and ↠= √(mω/(2ℏ)) (x̂ − ip̂/(mω)). In parts (d) and (e) the stretching vibration of a nitric oxide (NO) molecule is modelled as such an oscillator, with reduced mass μ = 1.24 × 10⁻²⁶ kg, vibrational frequency f = 5.64 × 10¹³ Hz and equilibrium bond length 115 pm. Take ℏ = 1.055 × 10⁻³⁴ J s, h = 6.63 × 10⁻³⁴ J s and k_B = 1.38 × 10⁻²³ J K⁻¹. You may use ∫ exp(−αx²) dx = √(π/α) and ∫ x² exp(−αx²) dx = √π/(2α^(3/2)), both taken over all x.

1. (a)

   Show that Show that [â, â†] = 1 and that Ĥ = ℏω(â†â + ½).

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   1. 1

      â↠= (mω/(2ℏ))[x̂² + p̂²/(m²ω²)] − (i/(2ℏ))[x̂, p̂] = (mω/(2ℏ))[x̂² + p̂²/(m²ω²)] + ½

   2. 2

      Similarly â†â = (mω/(2ℏ))[x̂² + p̂²/(m²ω²)] − ½, so [â, â†] = â↠− â†â = 1

   3. 3

      (mω/(2ℏ))[x̂² + p̂²/(m²ω²)] = Ĥ/(ℏω), so Ĥ = ℏω(â†â + ½)

2. (b)

   Derive Using only the two results of (a), derive the energy spectrum E_n = (n + ½)ℏω, n = 0, 1, 2, …

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   1. 1

      [Ĥ, â] = ℏω[â†â, â] = ℏω[â†, â]â = −ℏωâ, so if Ĥ|E⟩ = E|E⟩ then Ĥ(â|E⟩) = (E − ℏω)â|E⟩: â lowers the energy by ℏω, and likewise ↠raises it by ℏω

   2. 2

      E = ⟨E|Ĥ|E⟩ = ℏω(‖â|E⟩‖² + ½) ≥ ½ℏω, so repeated lowering cannot continue and must end on a state |0⟩ with â|0⟩ = 0

   3. 3

      Then Ĥ|0⟩ = ℏω(â†â + ½)|0⟩ = ½ℏω|0⟩; every eigenvalue lies a whole number of steps ℏω above this, and raising |0⟩ n times reaches each one, so E_n = (n + ½)ℏω

3. (c)

   Derive Derive the normalised ground-state wavefunction ψ₀(x) from the condition â|0⟩ = 0.

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   1. 1

      In the position representation p̂ = −iℏ d/dx, so â|0⟩ = 0 becomes (x + (ℏ/(mω)) d/dx)ψ₀ = 0

   2. 2

      Separating, dψ₀/ψ₀ = −(mω/ℏ)x dx, so ψ₀ = N exp(−mωx²/(2ℏ))

   3. 3

      ∫|ψ₀|² dx = N²√(πℏ/(mω)) = 1, so N = (mω/(πℏ))^(1/4)

4. (d)

   Calculate For the NO ground state, calculate σ_x and σ_p, and compare their product with the lower bound set by the commutator [x̂, p̂].

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   1. 1

      |ψ₀|² ∝ exp(−μωx²/ℏ) is a Gaussian of variance ℏ/(2μω); with ω = 2πf = 3.54 × 10¹⁴ rad s⁻¹, σ_x = √(ℏ/(2μω)) = 3.46 × 10⁻¹² m = 3.46 pm, about 3% of the bond length

   2. 2

      ⟨p⟩ = 0 and ⟨p²⟩/(2μ) = ½⟨Ĥ⟩ = ¼ℏω (equal kinetic and potential shares), so σ_p = √(μℏω/2) = 1.52 × 10⁻²³ kg m s⁻¹

   3. 3

      σ_xσ_p = 5.27 × 10⁻³⁵ J s = ℏ/2, equal to the Robertson bound ½|⟨[x̂, p̂]⟩| = ℏ/2, so the Gaussian ground state saturates it

5. (e)

   Evaluate Evaluate whether the harmonic model can be trusted for (i) the fraction of NO molecules in the first excited vibrational level at 300 K, and (ii) the energies of vibrational levels close to dissociation.

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   Mark-by-mark answer

   1. 1

      (i) hf/(k_BT) = (6.63 × 10⁻³⁴ J s × 5.64 × 10¹³ Hz)/(1.38 × 10⁻²³ J K⁻¹ × 300 K) = 9.03, so N₁/N₀ = e^(−9.03) = 1.2 × 10⁻⁴, which is also the fraction in n = 1 because almost every molecule is in n = 0; this depends only on the lowest spacing, and with σ_x only about 3% of the bond length the lowest levels sample just the bottom of the well, where the quadratic approximation holds, so the model can be trusted

   2. 2

      (ii) High levels have large amplitudes that reach the anharmonic part of the potential, which flattens towards dissociation; the real levels crowd together and end at a finite dissociation energy, while the harmonic ladder stays evenly spaced for ever, so the model cannot be trusted there