University Physics IV · Final Examination

A clean tungsten surface of work function φ = 4.50 eV is illuminated in vacuum by femtosecond laser pulses of wavelength 800 nm, and electrons are emitted even though each photon carries less energy than the work function. An electron spectrometer shows that the most energetic electrons leave with 0.15 eV at every pulse intensity used, and doubling the peak intensity I multiplies the number of electrons emitted per pulse by 7.9 ± 0.4. Take hc = 1240 eV nm. Which conclusion do these measurements justify?

1. AThe intense pulses heat the surface and the electrons are emitted thermally, so these data say nothing about how light exchanges energy with electrons.
2. BEinstein's photoelectric equation is refuted, because it forbids emission whenever the photon energy is below the work function, however intense the light.
3. CEach emitted electron absorbs three quanta of 1.55 eV, since 3hν − φ = 0.15 eV and an emission rate proportional to I³ grows by 2³ = 8 when I doubles; energy is exchanged with the light in quanta of hν, but these data do not by themselves prove that the light field is quantised.
4. DEach emitted electron absorbs three quanta, which proves that the light field consists of photons, because a classical wave cannot deliver energy to an electron in whole multiples of hν.

At a pulsed neutron source, each pulse releases neutrons with a wide spread of speeds from the moderator at the same instant. They travel a total flight path of 12.0 m, by way of a powder sample, to a detector that records neutrons scattered elastically through 90.0° by the sample. A Bragg peak reaches the detector 6.00 ms after the pulse. Treating it as a first-order reflection, what is the spacing of the lattice planes that produce it? Take h = 6.626 × 10⁻³⁴ J s and mₙ = 1.675 × 10⁻²⁷ kg.

1. A0.140 nm
2. B0.0989 nm
3. C0.280 nm
4. D0.0223 nm

An electron is held in a one-dimensional trap whose potential energy is not symmetric about any point. It is in the trap's ground state, described by a real, normalised wavefunction u(x) whose probability density is lopsided. A uniform force then acts for a time so short that u(x) does not change while it acts, delivering an impulse ℏk₀ in the +x direction, so that immediately afterwards the wavefunction is u(x)e^(ik₀x). With φ(p) the momentum-space wavefunction, which statement is correct?

1. ABefore the impulse ⟨p⟩ is not zero, because the lopsided density |u(x)|² makes the momentum distribution lopsided too; the impulse then adds ℏk₀ to ⟨p⟩.
2. BThe factor e^(ik₀x) has modulus 1 and leaves the position density unchanged, so no measurement made immediately afterwards, of position or of momentum, can detect the impulse.
3. CBefore the impulse |φ(p)|² is symmetric in p only if u(x) is an even function; immediately afterwards the momentum distribution is shifted by ℏk₀ and also broadened, because the factor e^(ik₀x) adds momenta near ℏk₀ to those already present.
4. DBefore the impulse |φ(p)|² = |φ(−p)|², so ⟨p⟩ = 0 even though |u(x)|² is lopsided; immediately afterwards the momentum distribution is the same curve shifted by ℏk₀, while the position density is unchanged.

A single ion held in a radio-frequency trap oscillates along the trap axis as a quantum harmonic oscillator of frequency f = 4.00 MHz, with energy levels Eₙ = (n + ½)hf. After laser cooling, its axial motion is in thermal equilibrium at T = 0.600 mK. What is the probability that the axial motion is found in its ground state, n = 0? Take h = 6.626 × 10⁻³⁴ J s and k_B = 1.381 × 10⁻²³ J K⁻¹.

1. A0.726
2. B0.274
3. C0.579
4. D0.852

A particle is in the simultaneous eigenstate of J² and J_z with quantum numbers j = 3/2 and m = 3/2. What is the expectation value ⟨J_x²⟩?

1. A3ℏ²/4
2. B0
3. C3ℏ²/2
4. D15ℏ²/4

A free ion has the ground configuration [Ar]3d³. Applying all three of Hund's rules, what is its ground-state term symbol?

1. A⁴F₉/₂
2. B²H₉/₂
3. C⁴P₁/₂
4. D⁴F₃/₂

In a one-dimensional tight-binding model an electron band has dispersion E(k) = E₀ − 2t cos(ka), with hopping energy t = 0.50 eV and lattice spacing a = 0.30 nm. What is the effective mass of an electron near the bottom of the band, at k = 0? Take ℏ = 1.055 × 10⁻³⁴ J s, e = 1.602 × 10⁻¹⁹ C and mₑ = 9.109 × 10⁻³¹ kg.

1. A0.42mₑ
2. B0.85mₑ
3. C1.69mₑ
4. D−0.85mₑ

In the spin-orbit shell model the proton single-particle levels fill in the order 1s₁/₂, 1p₃/₂, 1p₁/₂, 1d₅/₂, 2s₁/₂, 1d₃/₂, 1f₇/₂, each holding 2j + 1 protons, and neutrons fill the same sequence. What spin and parity does the model predict for the ground state of ³⁹K, with Z = 19 and N = 20?

1. A3/2⁺
2. B1/2⁺
3. C3/2⁻
4. D7/2⁻

A neutral meson decays in flight into two photons, which an electromagnetic calorimeter records with energies 1.50 GeV and 0.500 GeV and an angle of 36.9° between their directions of travel. Treating the photons as massless, what is the rest energy of the meson?

1. A2.00 GeV
2. B1.64 GeV
3. C0.388 GeV
4. D0.548 GeV

Entangled photon pairs are measured at two stations whose analyser settings are chosen randomly and independently, with the detection events at the two stations spacelike separated and the detection efficiency high enough that no fair-sampling assumption is needed. The measured CHSH combination is S = 2.70 ± 0.02. Which statement correctly describes what this result establishes?

1. AThe choice of setting at one station changes the outcomes recorded at the other, so a message could be sent faster than light.
2. BEvery hidden-variable theory, local or non-local, is ruled out, which proves that quantum mechanics is complete.
3. CAll models in which each outcome is fixed by variables carried locally from the source, independent of the settings, are excluded; the correlations still cannot be used to send a signal faster than light, and non-local hidden-variable models are not excluded.
4. DThe source emitted a statistical mixture, since a pure entangled state would give S exactly equal to 2.

A handheld X-ray fluorescence analyser is aimed at a corroded brass fitting. Its primary X-rays eject inner-shell electrons from atoms in the sample, and the detector records two strong lines at 8.04 keV and 8.63 keV, together with weaker lines, one of which lies at 8.90 keV. Model each K line with hydrogen-like levels and a screening constant σ: a transition from shell n to the K shell (n = 1) emits a photon of energy E = (13.6 eV)(Z − σ)²(1 − 1/n²), so that the Kα line (n = 2) follows Moseley's law, E = (3/4)(13.6 eV)(Z − σ)².

1. (a)

   Describe Describe the sequence of events, beginning with the absorption of a primary X-ray photon, that leads to the emission of a Kα photon, and say why the primary photons must exceed a threshold energy.

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   1. 1

      1 mark: a primary photon with energy above the K-shell binding energy is absorbed and ejects a 1s electron, leaving a K-shell vacancy; a photon below this K edge cannot create the vacancy, hence the threshold

   2. 2

      1 mark: an electron from the n = 2 (L) shell falls into the vacancy, and the energy difference between the two inner-shell levels is carried away by the Kα photon (the competing Auger process ejects an electron instead)

2. (b)

   Calculate Taking σ = 1, calculate the atomic numbers of the elements that produce the 8.04 keV and 8.63 keV lines, and identify the elements.

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   1. 1

      1 mark: rearranges Moseley's law to Z = 1 + √[E/((3/4)(13.6 eV))] = 1 + √(E/10.2 eV), with E in eV

   2. 2

      1 mark: 8.04 keV gives Z = 1 + √788 = 29.1, so Z = 29, copper

   3. 3

      1 mark: 8.63 keV gives Z = 1 + √846 = 30.1, so Z = 30, zinc, the two constituents of brass

3. (c)

   Explain Explain why σ ≈ 1 is a reasonable screening constant for a Kα line.

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   1. 1

      1 mark: with one 1s electron ejected, one 1s electron remains; it lies almost entirely inside the orbit of the n = 2 electron that makes the jump, and so screens one unit of the nuclear charge

   2. 2

      1 mark: the electrons of the n = 2 shell and beyond are spread at the same or larger radii, so by Gauss's law they screen little, leaving an effective charge close to Z − 1 in a hydrogen-like energy

4. (d)

   Evaluate A colleague suggests that the 8.90 keV line is the Kα line of a third element. Evaluate this suggestion against the alternative that it is the Kβ line (n = 3 → n = 1) of copper, and state what your conclusion implies about σ for a Kβ line.

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   1. 1

      1 mark: as a Kα line with σ = 1, 8.90 keV gives Z = 1 + √(8900/10.2) = 30.5, not an integer, whereas the two strong lines gave Z within 0.1 of an integer; the neighbouring Kα lines, zinc (observed at 8.63 keV) and gallium (predicted 9.18 keV), bracket it, so the suggestion is rejected

   2. 2

      1 mark: copper Kβ with σ = 1 is predicted at (13.6 eV)(28)²(1 − 1/9) = 9.48 keV, within 7% of 8.90 keV, and copper is known to be present, so the line is copper Kβ

   3. 3

      1 mark: matching 8.90 keV exactly needs σ ≈ 1.9, because the n = 3 electron is screened by the remaining 1s electron and also, in part, by the eight n = 2 electrons; σ = 1 does not carry over from Kα to Kβ

A single lithium-7 atom, of mass m = 1.165 × 10⁻²⁶ kg, is confined along one direction between two sheets of laser light that act as impenetrable walls at x = 0 and x = L, with L = 10.0 μm. Its stationary states are ψₙ(x) = √(2/L) sin(nπx/L) for 0 ≤ x ≤ L, with energies Eₙ = n²π²ℏ²/(2mL²), and it is prepared in the ground state ψ₁. At t = 0 a pulse of light lasting far less than any timescale of the atom's motion illuminates only the half of the box 0 < x < L/2, multiplying the wavefunction there by e^(iπ) = −1 and leaving the half L/2 < x < L untouched. Take ℏ = 1.055 × 10⁻³⁴ J s and k_B = 1.381 × 10⁻²³ J K⁻¹. You may use ∫₀^(L/2) sin(nπx/L) sin(πx/L) dx = (L/(2π))[sin((n − 1)π/2)/(n − 1) − sin((n + 1)π/2)/(n + 1)] for n ≥ 2.

1. (a)

   Calculate Calculate E₁, and the temperature T at which k_BT equals E₂ − E₁, the temperature scale that the atom must be cooled well below for its ground state to be prepared reliably.

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   1. 1

      1 mark: E₁ = π²ℏ²/(2mL²) = π²(1.055 × 10⁻³⁴ J s)²/[2(1.165 × 10⁻²⁶ kg)(1.00 × 10⁻⁵ m)²] = 4.71 × 10⁻³² J

   2. 2

      1 mark: E₂ − E₁ = 3E₁ = 1.41 × 10⁻³¹ J, so T = 3E₁/k_B = 1.02 × 10⁻⁸ K, about 10 nK

2. (b)

   Derive Starting from the expansion of Ψ(x, 0) in the stationary states and the Born rule, derive the probability Pₙ that a measurement of the energy made just after the pulse returns Eₙ. Find every n for which Pₙ = 0, and evaluate P₂ and P₄.

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   1. 1

      1 mark: Ψ(x, 0) = Σ cₙψₙ(x), and orthonormality of the ψₙ gives cₙ = ∫₀ᴸ ψₙ(x)Ψ(x, 0) dx, with Pₙ = |cₙ|² by the Born rule

   2. 2

      1 mark: Ψ(x, 0) = −ψ₁ in the left half and +ψ₁ in the right half, so cₙ is the integral of ψₙψ₁ over the right half minus its integral over the left half; since ψₙ(L − x) = (−1)^(n+1)ψₙ(x), the product ψₙψ₁ is symmetric about x = L/2 for odd n, the two integrals are equal, and Pₙ = 0 for every odd n, the ground state included

   3. 3

      1 mark: for even n the product is antisymmetric about x = L/2, so cₙ = −2∫₀^(L/2) ψₙψ₁ dx = −(2/π)[sin((n − 1)π/2)/(n − 1) − sin((n + 1)π/2)/(n + 1)], which gives Pₙ = 16n²/[π²(n² − 1)²] for even n

   4. 4

      1 mark: P₂ = 64/(9π²) = 0.721 and P₄ = 256/(225π²) = 0.115

3. (c)

   Explain The pulse leaves the probability density |Ψ(x, 0)|² equal to |ψ₁(x)|². Explain why the density nevertheless changes after the pulse, why it remains symmetric about the centre of the box at all later times, and why it is zero at x = L/2 at every t > 0.

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   1. 1

      1 mark: the pulse changes the relative phase of the two halves, so Ψ(x, 0) is no longer a stationary state but a superposition of ψ₂, ψ₄, ψ₆, …; each term acquires its own phase e^(−iEₙt/ℏ), so the cross terms in |Ψ|² change with time

   2. 2

      1 mark: every component present is antisymmetric about x = L/2 and keeps that symmetry as its phase evolves, so Ψ(L − x, t) = −Ψ(x, t) and |Ψ(x, t)|² = |Ψ(L − x, t)|² at all times

   3. 3

      1 mark: antisymmetry requires Ψ(L/2, t) = −Ψ(L/2, t), so Ψ(L/2, t) = 0, as every even ψₙ has a node there; a notch opens at the centre, where the density had its maximum before the pulse

4. (d)

   Predict In the real trap the light sheets form barriers of finite height V₀ = 400E₁, with E₁ the value found in (a), and the same pulse, which reverses the sign of the wavefunction everywhere to the left of x = L/2, is applied to the ground state of this finite well. Predict, with reasons, whether a measurement of the energy just after the pulse could return the ground-state energy of the finite well, and whether the atom is certain to remain in the trap.

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   1. 1

      1 mark: the finite well is still symmetric about x = L/2, so its bound states still have definite parity and its ground state is symmetric; the pulsed wavefunction is antisymmetric about the centre, so its overlap with the ground state is zero and the ground-state energy cannot be returned

   2. 2

      1 mark: the sign jump makes the amplitudes fall off only as 1/n, so part of the probability lies in unbound states above the barrier top (in the infinite-well estimate the states with n > 20, whose energies exceed 400E₁, carry about 4%), and the atom is not certain to remain trapped

The resonance lines of the alkali atoms come from an np → ns transition that spin-orbit coupling splits into D₁ (²P₁/₂ → ²S₁/₂) and D₂ (²P₃/₂ → ²S₁/₂). The fine-structure splitting between ²P₃/₂ and ²P₁/₂ is 7.16 × 10⁻³ eV for the 4p level of potassium and 4.16 × 10⁻⁵ eV for the 2p level of lithium. Both vapours are placed in a uniform magnetic field of 0.500 T. Ignore hyperfine structure throughout. Take μ_B = 5.788 × 10⁻⁵ eV T⁻¹ and g_s = 2, so that g_J = 1 + [j(j + 1) + s(s + 1) − l(l + 1)]/[2j(j + 1)].

1. (a)

   Calculate Calculate g_J for the ²S₁/₂, ²P₁/₂ and ²P₃/₂ levels.

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   1. 1

      1 mark: ²S₁/₂: g_J = 1 + (3/4 + 3/4)/(3/2) = 2, the moment being pure spin

   2. 2

      1 mark: ²P₁/₂: g_J = 1 + (3/4 + 3/4 − 2)/(3/2) = 2/3; ²P₃/₂: g_J = 1 + (15/4 + 3/4 − 2)/(15/2) = 4/3

2. (b)

   Determine In the weak-field limit each sublevel shifts by g_J m_j μ_B B. Using the selection rule Δm_j = 0, ±1, determine the number of components into which the D₁ line splits and the shift of each component from the zero-field line, in units of μ_B B.

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   1. 1

      1 mark: the ²P₁/₂ sublevels shift by ±(1/3)μ_B B and the ²S₁/₂ sublevels by ±μ_B B

   2. 2

      1 mark: all four pairings of m_j = ±1/2 in the upper level with m_j = ±1/2 in the lower level satisfy Δm_j = 0, ±1

   3. 3

      1 mark: four components, at ±(2/3)μ_B B for Δm_j = 0 and ±(4/3)μ_B B for Δm_j = ±1

3. (c)

   Compare Compare the fine-structure splitting with μ_B B at 0.500 T for each atom.

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   1. 1

      1 mark: μ_B B = (5.788 × 10⁻⁵ eV T⁻¹)(0.500 T) = 2.89 × 10⁻⁵ eV

   2. 2

      1 mark: potassium: 7.16 × 10⁻³ eV/2.89 × 10⁻⁵ eV ≈ 250; lithium: 4.16 × 10⁻⁵ eV/2.89 × 10⁻⁵ eV ≈ 1.4, so the same field is a small perturbation for potassium but comparable with the spin-orbit energy for lithium

4. (d)

   Justify Justify, for each atom, whether the D₁ pattern you found in (b) describes its spectrum at 0.500 T.

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   1. 1

      1 mark: potassium: the Zeeman energy is about 250 times smaller than the spin-orbit energy, so j and m_j remain good quantum numbers, ²P₁/₂ and ²P₃/₂ do not mix, and the four-component pattern of (b) applies

   2. 2

      1 mark: lithium: the two energies are comparable, so the field mixes ²P₁/₂ and ²P₃/₂ sublevels of the same m_j; j is no longer a good quantum number, the shifts are not linear in B, and the weak-field pattern does not apply (the intermediate regime)

5. (e)

   Predict Predict the appearance of the lithium resonance line at 5.00 T, naming the good quantum numbers.

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   1. 1

      1 mark: at 5.00 T, μ_B B = 2.89 × 10⁻⁴ eV, about seven times the fine-structure splitting, so the Paschen–Back limit is approached: spin and orbit decouple, and m_l and m_s, not j and m_j, are the good quantum numbers

   2. 2

      1 mark: each state shifts by (m_l + 2m_s)μ_B B; with Δm_s = 0 and Δm_l = 0, ±1 the line becomes a triplet at 0 and ±μ_B B = ±2.89 × 10⁻⁴ eV, like the normal Zeeman pattern, with each component slightly split by the residual spin-orbit energy

The muon decays by μ⁻ → e⁻ + ν̄ₑ + ν_μ with a branching fraction of essentially 1 and a mean lifetime τ_μ = 2.197 × 10⁻⁶ s. The τ⁻ lepton decays through the same charged weak current into e⁻ + ν̄ₑ + ν_τ with branching fraction B_e = 0.178, and also into μ⁻ + ν̄_μ + ν_τ and into hadrons. Take m_μc² = 105.66 MeV, m_τc² = 1776.9 MeV and the Fermi constant G_F/(ℏc)³ = 1.166 × 10⁻⁵ GeV⁻². In natural units, with ℏ = c = 1, a decay width Γ = ℏ/τ has the dimensions of energy.

1. (a)

   State State the values of Lₑ, L_μ and L_τ before and after the decay τ⁻ → e⁻ + ν̄ₑ + ν_τ, and state which conservation law forbids τ⁻ → e⁻ + γ.

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   1. 1

      1 mark: before, (Lₑ, L_μ, L_τ) = (0, 0, 1); after, e⁻ (1, 0, 0) + ν̄ₑ (−1, 0, 0) + ν_τ (0, 0, 1) = (0, 0, 1), so each lepton-family number is conserved

   2. 2

      1 mark: τ⁻ → e⁻ + γ changes L_τ from 1 to 0 and Lₑ from 0 to 1, so it violates conservation of lepton-family number, although charge, energy and total lepton number all balance

2. (b)

   Describe Describe the lowest-order Feynman diagram for τ⁻ → e⁻ + ν̄ₑ + ν_τ, and give the power of the weak coupling g_W in its amplitude and in its rate.

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   1. 1

      1 mark: at one vertex the τ⁻ becomes a ν_τ by emitting a virtual W⁻, which propagates to a second vertex where it becomes e⁻ and ν̄ₑ

   2. 2

      1 mark: two vertices give an amplitude proportional to g_W² and a rate proportional to its square, g_W⁴; at energies far below M_W the propagator is about 1/M_W², so the rate goes as (g_W²/M_W²)², that is as G_F²

3. (c)

   Show that Show that if the width depends only on G_F and on the mass m of the decaying lepton, the result of (b) requires Γ ∝ G_F²m⁵.

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   1. 1

      1 mark: in natural units G_F has dimensions of energy⁻² and Γ of energy, so writing Γ ∝ G_F^a m^b requires −2a + b = 1

   2. 2

      1 mark: the rate goes as G_F², so a = 2 and b = 5, giving Γ ∝ G_F²m⁵

4. (d)

   Estimate Estimate the mean lifetime of the τ, stating the assumptions you make.

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   1. 1

      1 mark: assumes lepton universality, so the W couples equally to τ and μ, and neglects the electron and neutrino masses, so Γ(τ → eν̄ν)/Γ(μ → eν̄ν) = (m_τ/m_μ)⁵ = (16.82)⁵ = 1.35 × 10⁶

   2. 2

      1 mark: the partial width is Γ(τ → eν̄ν) = B_eℏ/τ_τ, so τ_τ = B_eτ_μ(m_μ/m_τ)⁵

   3. 3

      1 mark: τ_τ = 0.178 × 2.197 × 10⁻⁶ s × 7.43 × 10⁻⁷ = 2.91 × 10⁻¹³ s

5. (e)

   Evaluate The measured τ lifetime is 2.903 × 10⁻¹³ s. Evaluate what the agreement with your estimate tests, and use a count of the final states available to the virtual W⁻ to explain why B_e is close to 0.2.

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   1. 1

      1 mark: the estimate agrees with the measurement to about 0.2%, so the charged weak coupling of the τ equals that of the μ to within the few-tenths-of-a-percent precision of the inputs: lepton universality holds across generations

   2. 2

      1 mark: the W⁻ from a τ can materialise as e⁻ν̄ₑ, μ⁻ν̄_μ or dū, and the quark pair counts three times, once for each colour, giving five channels of nearly equal weight and B_e ≈ 1/5

   3. 3

      1 mark: the measured 0.178 lies below 1/5 because strong-interaction (QCD) corrections raise the hadronic width: 1/B_e = 5.6 requires the quark channels to count as about 3.6 instead of 3, an enhancement of about 20%

²¹⁰Po decays by α emission to ²⁰⁶Pb with a half-life of 138.4 days. The atomic masses are ²¹⁰Po 209.982874 u, ²⁰⁶Pb 205.974465 u and ⁴He 4.002603 u, with 1 u = 931.494 MeV/c². In the Gamow model the α particle, of reduced mass μ, meets a Coulomb barrier V(r) = Z_dZ_ααℏc/r that begins at the nuclear radius R, and tunnels out to the classical turning point b at which V(b) = Q. The WKB exponent is 2G = 2αZ_dZ_α√(2μc²/Q)[arccos√(R/b) − √((R/b)(1 − R/b))], and the half-life is proportional to e^(2G) divided by an assault frequency. Take αℏc = 1.440 MeV fm, α = 1/137.0, μc² = 3.66 × 10³ MeV, R = 9.0 fm, Z_d = 82 and Z_α = 2.

1. (a)

   Calculate Calculate the Q value of the decay.

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   1. 1

      1 mark: Q = [m(²¹⁰Po) − m(²⁰⁶Pb) − m(⁴He)]c² = (0.005806 u)(931.494 MeV/u), the electron masses cancelling when atomic masses are used throughout

   2. 2

      1 mark: Q = 5.41 MeV

2. (b)

   Calculate Calculate the outer turning point b.

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   1. 1

      1 mark: b = Z_dZ_ααℏc/Q = (82)(2)(1.440 MeV fm)/(5.41 MeV) = 43.7 fm, nearly five times R, so R/b = 0.21

3. (c)

   Show that Using arccos√x ≈ π/2 − √x and √(x(1 − x)) ≈ √x for x = R/b much less than 1, show that 2G ≈ παZ_dZ_α√(2μc²/Q) − 4√(2αZ_dZ_αμc²R/ℏc), and hence that the second term does not depend on Q.

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   1. 1

      1 mark: the bracket becomes π/2 − 2√(R/b), so the first term is 2αZ_dZ_α√(2μc²/Q)(π/2) = παZ_dZ_α√(2μc²/Q)

   2. 2

      1 mark: the second term is 4αZ_dZ_α√(2μc²/Q)√(R/b), and with b = Z_dZ_ααℏc/Q, √(R/b) = √(RQ/(Z_dZ_ααℏc))

   3. 3

      1 mark: multiplying out, the factors of Q cancel: 4√[α²Z_d²Z_α²(2μc²/Q)(RQ/(Z_dZ_ααℏc))] = 4√(2αZ_dZ_αμc²R/ℏc), independent of Q

4. (d)

   Predict ²¹⁴Po also decays by α emission to a lead isotope (Z_d = 82), with Q = 7.833 MeV. Assuming the same R, μ and assault frequency for both nuclei, predict the half-life of ²¹⁴Po from that of ²¹⁰Po using the result of (c).

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   1. 1

      1 mark: with R, μ and the assault frequency unchanged only the first term differs, so ln[t½(²¹⁰Po)/t½(²¹⁴Po)] = παZ_dZ_α√(2μc²)[1/√Q(²¹⁰Po) − 1/√Q(²¹⁴Po)]

   2. 2

      1 mark: = (3.761)(85.56 MeV^(1/2))(0.4300 − 0.3573) MeV^(−1/2) = 23.4, the first terms being 138.4 and 115.0

   3. 3

      1 mark: the ratio is e^23.4 = 1.4 × 10¹⁰, so t½(²¹⁴Po) = (138.4 × 86 400 s)/(1.4 × 10¹⁰) = 8.3 × 10⁻⁴ s

5. (e)

   Evaluate The measured half-life of ²¹⁴Po is 164 μs. Evaluate the model in the light of this comparison.

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   1. 1

      1 mark: the prediction, 0.83 ms, is about five times the measured 0.164 ms, yet the model spans a half-life ratio of order 10¹⁰: the measured ln ratio is 25.0 against the predicted 23.4, so the Q dependence of the barrier accounts for nearly all of the ten orders of magnitude

   2. 2

      1 mark: agreement to within a factor of a few is all the model can claim, because the half-life depends exponentially on R and on the approximations made; it also omits nuclear structure, and ²¹⁰Po, with the closed neutron shell N = 126, has a reduced probability of preforming an α cluster, lengthening its half-life relative to ²¹⁴Po (N = 130), in the direction of the discrepancy

A sample-return probe leaves a space station, which defines an inertial frame S, and travels in a straight line at 0.600c to a relay beacon at rest in S at a distance of 3.00 light-years. At the beacon it reverses in a negligibly short time and returns at 0.800c. Work in units in which c = 1 light-year per year, label events by their S coordinates (t, x), and take the departure as the event (0, 0). Let S′ be the inertial frame moving with the probe on the outbound leg and S″ the inertial frame moving with it on the return leg.

1. (a)

   Calculate Calculate the time that elapses on the station clock, and the proper time that elapses on the probe's clock, for the round trip.

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   1. 1

      1 mark: station: 3.00/0.600 = 5.00 y out and 3.00/0.800 = 3.75 y back, a total of 8.75 y

   2. 2

      1 mark: probe, leg by leg with Δτ = Δt√(1 − v²/c²): 5.00 y × 0.800 = 4.00 y and 3.75 y × 0.600 = 2.25 y

   3. 3

      1 mark: total proper time 6.25 y, so the probe returns 2.50 y younger than the station clock reads

2. (b)

   Determine Determine the reading of the station clock that is simultaneous with the turnaround event, first in S′ and then in S″.

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   1. 1

      1 mark: the turnaround is the event (5.00 y, 3.00 ly); in S′, with v = +0.600c and γ = 1.25, t′ = γ(t − vx/c²) = 1.25(5.00 − 1.80) = 4.00 y, the probe's own clock reading

   2. 2

      1 mark: a station event (t_s, 0) has t′ = γt_s, and setting this equal to 4.00 y gives t_s = 3.20 y

   3. 3

      1 mark: in S″, with v = −0.800c and γ = 5/3, t″ = γ(t + 0.800x) = (5/3)(5.00 + 2.40) = 12.3 y

   4. 4

      1 mark: setting γt_s = 12.3 y gives t_s = 7.40 y, so the station reading simultaneous with the turnaround jumps from 3.20 y to 7.40 y

3. (c)

   Show that Show that the probe's account, with the station clock running slow on each leg plus the jump found in (b), reproduces the station's total of 8.75 y.

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   1. 1

      1 mark: in S′ the station clock runs slow by γ = 1.25, advancing 4.00 y/1.25 = 3.20 y; in S″ it runs slow by γ = 5/3, advancing 2.25 y × 0.600 = 1.35 y

   2. 2

      1 mark: 3.20 y + (7.40 − 3.20) y + 1.35 y = 8.75 y, agreeing with the station's own clock

4. (d)

   Evaluate Evaluate the claim: “The probe ages less because it accelerated, and the harder it accelerates at the turnaround, the more time it loses.”

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   1. 1

      1 mark: the ageing difference is fixed by the speeds and distances of the two inertial legs and is 2.50 y however brief the turnaround; the size of the acceleration does not appear in it, so the second half of the claim is false

   2. 2

      1 mark: the first half is right only in a limited sense: the acceleration makes the probe's worldline non-inertial and breaks the symmetry, since the probe changes inertial frame and with it its simultaneity (the 4.20 y jump), while the station stays in one frame, and of all worldlines joining departure and return the inertial one has the greatest proper time

The intrinsic carrier density nᵢ of an undoped semiconductor bar is found from Hall-effect and conductivity measurements at six temperatures, and each value of nᵢ carries a random uncertainty of ±3%. For a non-degenerate intrinsic semiconductor with a temperature-independent band gap E_g, nᵢ = CT^(3/2)exp(−E_g/2kT), where C is a constant. Take k = 8.617 × 10⁻⁵ eV K⁻¹.

1. (a)

   Explain Explain why a graph of ln(nᵢ/T^(3/2)) against 1/T is expected to be a straight line, and what its gradient gives.

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   1. 1

      1 mark: taking logarithms, ln(nᵢ/T^(3/2)) = ln C − (E_g/2k)(1/T), which is linear in 1/T with the constant intercept ln C

   2. 2

      1 mark: the gradient is −E_g/2k, so E_g = −2k × gradient; the T^(3/2) comes from the effective densities of states of the two bands and must be divided out, or it curves the line

2. (b)

   Determine Determine E_g from the data.

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   1. 1

      1 mark: computes 1/T = 3.333, 2.941, 2.632, 2.381, 2.174 and 2.000 (× 10⁻³ K⁻¹) and ln(nᵢ/T^(3/2)) = 35.90, 37.64, 39.07, 40.17, 41.15 and 41.91

   2. 2

      1 mark: plots or fits a straight line, the points lying on it within the ±0.03 expected from ±3% uncertainties, with gradient −4.5 × 10³ K (least-squares −4.52 × 10³ K; end points −4.51 × 10³ K)

   3. 3

      1 mark: E_g = 2(8.617 × 10⁻⁵ eV K⁻¹)(4.52 × 10³ K) = 0.78 eV (accept 0.77 eV to 0.79 eV)

3. (c)

   Estimate Using the first and last rows only, estimate the uncertainty in E_g that follows from the ±3% uncertainty in nᵢ, stating the assumption you make about the two uncertainties.

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   1. 1

      1 mark: assuming the two uncertainties are independent, the difference of the two logarithms is uncertain by √2 × 0.03 = 0.042, over a span in 1/T of 1.333 × 10⁻³ K⁻¹, so the gradient is uncertain by about 32 K

   2. 2

      1 mark: δE_g = 2(8.617 × 10⁻⁵ eV K⁻¹)(32 K) = 0.0055 eV, so E_g = 0.779 ± 0.005 eV, an uncertainty of about 0.7%

4. (d)

   Compare A student instead plots ln nᵢ against 1/T and takes E_g = −2k × gradient. Compare the gap this gives with your value from (b), and account for the difference.

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   1. 1

      1 mark: ln nᵢ = 44.46, 46.39, 47.98, 49.23, 50.34 and 51.23 gives a gradient of about −5.09 × 10³ K and an apparent gap of 0.88 eV, about 0.10 eV (13%) larger than 0.78 eV

   2. 2

      1 mark: ln nᵢ contains (3/2)ln T, which rises with T and so steepens the line; its contribution to the apparent gap is about 3kT at the mid-range temperature of about 390 K, 3(8.6 × 10⁻⁵ eV K⁻¹)(390 K) ≈ 0.10 eV, some twenty times the random uncertainty from (c)

5. (e)

   Evaluate Optical absorption measured on the same bar at 400 K shows an absorption edge at 0.66 eV. Semiconductor gaps typically narrow as the temperature rises. Evaluate whether this contradicts your value from (b).

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   1. 1

      1 mark: if E_g(T) = E_g0 − βT, then exp(−E_g/2kT) = exp(β/2k)exp(−E_g0/2kT), so the linear term changes only the constant C, and the gradient in (b) returns E_g0, the gap extrapolated to 0 K, not the gap at the measurement temperatures

   2. 2

      1 mark: there is no contradiction: both results fit E_g0 = 0.78 eV with β = (0.78 − 0.66) eV/(400 K) = 3.0 × 10⁻⁴ eV K⁻¹, a typical narrowing, so the transport fit and the optical edge together determine both E_g0 and β

An electron is bound to a nucleus of charge Ze. With μ the reduced mass, the radial function u(r) = rR(r) of a state of orbital quantum number l satisfies the radial Schrödinger equation −(ℏ²/2μ)(d²u/dr²) + [−Ze²/(4πε₀r) + l(l + 1)ℏ²/(2μr²)]u = Eu, with E < 0 for a bound state. Write κ = √(−2μE)/ℏ and a_μ = 4πε₀ℏ²/(μe²). You may use ∫₀^∞ rᵏe^(−βr) dr = k!/β^(k+1) for integer k ≥ 0. For the numerical parts take a₀ = 5.292 × 10⁻¹¹ m and 13.606 eV as the Bohr radius and the ionisation energy of hydrogen for an infinitely heavy nucleus, m_p/mₑ = 1836 and h = 4.136 × 10⁻¹⁵ eV s.

1. (a)

   Show that Show that a normalisable solution must behave as e^(−κr) at large r and as r^(l+1) near r = 0.

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   1. 1

      1 mark: as r → ∞ both potential terms vanish, leaving d²u/dr² = κ²u, with solutions e^(±κr); normalisability rejects e^(+κr), so u ∝ e^(−κr)

   2. 2

      1 mark: as r → 0 the centrifugal term dominates, d²u/dr² ≈ l(l + 1)u/r², and u = r^s requires s(s − 1) = l(l + 1), so s = l + 1 or s = −l; u ∝ r^(−l) makes R = u/r diverge, non-normalisably for l ≥ 1 and, for l = 0, as 1/r, which fails the Schrödinger equation at the origin, so u ∝ r^(l+1)

2. (b)

   Derive By substituting u = r^(l+1)e^(−κr) into the radial Schrödinger equation, derive the conditions under which it is an exact solution, and hence derive κ and the energy E of this state. Identify its principal quantum number n.

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   1. 1

      1 mark: u′ = [(l + 1)/r − κ]u and u″ = [l(l + 1)/r² − 2κ(l + 1)/r + κ²]u

   2. 2

      1 mark: on substitution the 1/r² terms, −(ℏ²/2μ)l(l + 1)/r² and +l(l + 1)ℏ²/(2μr²), cancel identically

   3. 3

      1 mark: the 1/r terms cancel only if ℏ²κ(l + 1)/μ = Ze²/(4πε₀), so κ = Z/[(l + 1)a_μ]

   4. 4

      1 mark: the constant terms give E = −ℏ²κ²/(2μ) = −Z²ℏ²/[2μa_μ²(l + 1)²] = −μZ²e⁴/[2(4πε₀)²ℏ²(l + 1)²]

   5. 5

      1 mark: u has no radial nodes, so n = n_r + l + 1 = l + 1, and E_n = −Z²e²/(8πε₀a_μn²), the Bohr energy with the reduced mass, for the state with l = n − 1

3. (c)

   Determine For this state, determine the most probable radius r_mp and the mean radius ⟨r⟩ in terms of n, a_μ and Z.

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   1. 1

      1 mark: the radial probability density is P(r) ∝ |u|² = r^(2n)e^(−2Zr/(na_μ)), the r² Jacobian being already contained in u = rR

   2. 2

      1 mark: dP/dr = 0 gives 2n/r = 2Z/(na_μ), so r_mp = n²a_μ/Z

   3. 3

      1 mark: ⟨r⟩ = ∫r^(2n+1)e^(−βr) dr/∫r^(2n)e^(−βr) dr = (2n + 1)/β with β = 2Z/(na_μ), so ⟨r⟩ = n(n + ½)a_μ/Z

4. (d)

   Calculate Circular Rydberg states, with l = n − 1, are used in microwave cavity experiments. For the n = 50 circular state of hydrogen, calculate the most probable radius and the frequency of the photon emitted in the transition to n = 49.

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   1. 1

      1 mark: a_μ = a₀(1 + mₑ/m_p) = (5.292 × 10⁻¹¹ m)(1.000545) = 5.295 × 10⁻¹¹ m

   2. 2

      1 mark: r_mp = 2500a_μ = 1.32 × 10⁻⁷ m (132 nm)

   3. 3

      1 mark: ΔE = (13.606 eV/1.000545)(1/49² − 1/50²) = 2.24 × 10⁻⁴ eV, so f = ΔE/h = 5.42 × 10¹⁰ Hz (54.2 GHz)

5. (e)

   Justify Justify, using the electric-dipole selection rules, the claim that this state can decay by only one electric-dipole transition.

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   1. 1

      1 mark: electric-dipole transitions require Δl = ±1 (with Δm = 0, ±1), and spontaneous emission requires a final level of lower energy, n′ < 50

   2. 2

      1 mark: Δl = +1 needs l′ = 50 and so n′ ≥ 51, which is higher in energy; Δl = −1 needs l′ = 48 and so n′ ≥ 49, leaving only n′ = 49, l′ = 48, the next circular state

Model the universe as a uniform, pressureless dust of mass density ρ(t) in which every separation grows in proportion to a scale factor a(t), with a = 1 today. A particle of mass m moving with the dust lies at distance r = a(t)x from an arbitrary origin, where x is constant, and by the shell theorem it is attracted only by the mass inside the sphere of radius r. Take G = 6.674 × 10⁻¹¹ N m² kg⁻², the present expansion rate H₀ = ȧ/a = 67.4 km s⁻¹ Mpc⁻¹, 1 Mpc = 3.086 × 10²² m, the mass of a hydrogen atom as 1.674 × 10⁻²⁷ kg and 1 yr = 3.156 × 10⁷ s.

1. (a)

   Derive Starting from conservation of energy for the particle, derive the Friedmann equation (ȧ/a)² = (8πG/3)ρ − K/a², where K is a constant, and state what the sign of K means for the future of the expansion in this Newtonian picture.

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   1. 1

      1 mark: the mass inside the sphere, M = (4π/3)r³ρ, is fixed because the dust moves with the sphere, so ρa³ is constant

   2. 2

      1 mark: energy conservation for the particle: ½mṙ² − GMm/r = E, with E constant

   3. 3

      1 mark: substituting r = ax, ṙ = ȧx and M = (4π/3)a³x³ρ and dividing by ½ma²x² gives (ȧ/a)² = (8πG/3)ρ + 2E/(mx²a²), so K = −2E/(mx²), a constant

   4. 4

      1 mark: K > 0 (E < 0): the expansion halts and reverses; K < 0 (E > 0): it continues for ever; K = 0: the expansion rate falls to zero only as t → ∞

2. (b)

   Calculate The critical density ρ_c is the density for which K = 0. Calculate ρ_c today, and express it as a number of hydrogen atoms per cubic metre.

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   1. 1

      1 mark: K = 0 at a = 1 gives ρ_c = 3H₀²/(8πG), with H₀ = (67.4 × 10³ m s⁻¹)/(3.086 × 10²² m) = 2.18 × 10⁻¹⁸ s⁻¹

   2. 2

      1 mark: ρ_c = 3(2.18 × 10⁻¹⁸ s⁻¹)²/(8π × 6.674 × 10⁻¹¹ N m² kg⁻²) = 8.5 × 10⁻²⁷ kg m⁻³

   3. 3

      1 mark: ρ_c/m_H = (8.53 × 10⁻²⁷ kg m⁻³)/(1.674 × 10⁻²⁷ kg) = 5.1 hydrogen atoms per cubic metre

3. (c)

   Derive For K = 0 and ρ ∝ a⁻³, derive a(t) from the Friedmann equation, taking a = 0 at t = 0, and hence obtain the age t₀ of the universe in terms of H₀ and in gigayears.

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   1. 1

      1 mark: ρ = ρ_c a⁻³ = (3H₀²/8πG)a⁻³ turns the Friedmann equation into ȧ² = H₀²/a, so a^(1/2) da = H₀ dt

   2. 2

      1 mark: integrating from a = 0 at t = 0 gives (2/3)a^(3/2) = H₀t, so a(t) = (3H₀t/2)^(2/3) ∝ t^(2/3)

   3. 3

      1 mark: a = 1 today gives t₀ = 2/(3H₀) = 3.05 × 10¹⁷ s = 9.7 Gyr

4. (d)

   Evaluate The oldest globular-cluster stars in the Milky Way are dated at about 12.5 Gyr. Evaluate the flat, matter-only model in the light of this, including the value of H₀ the model would need.

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   1. 1

      1 mark: 9.7 Gyr is younger than stars that formed within the universe, so the model with the measured H₀ is excluded; an age of 12.5 Gyr would need H₀ = 2/(3t₀) ≈ 52 km s⁻¹ Mpc⁻¹, well below the measured value of about 67 km s⁻¹ Mpc⁻¹

   2. 2

      1 mark: the model must change, not H₀: a component that makes the expansion accelerate, such as a cosmological constant, means the expansion was slower in the past and the universe older for the same H₀ (about 13.8 Gyr for matter and dark-energy fractions of 0.315 and 0.685 with H₀ = 67.4 km s⁻¹ Mpc⁻¹), and pressureless dust alone cannot supply it

5. (e)

   Calculate The cosmic microwave background was released at redshift z = 1090. Calculate the scale factor at that time and the factor by which the matter density then exceeded its present value.

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   1. 1

      1 mark: a = 1/(1 + z) = 1/1091 = 9.17 × 10⁻⁴

   2. 2

      1 mark: ρ/ρ₀ = a⁻³ = 1091³ = 1.30 × 10⁹