University Physics III · Final Examination

A particle of mass 0.150 kg moves along the positive x-axis in the potential U(x) = A/x² − B/x, where A = 0.360 J·m² and B = 1.20 J·m. What is the angular frequency of small oscillations about its stable equilibrium position?

1. A0.969 rad/s
2. B6.09 rad/s
3. C10.5 rad/s
4. D13.6 rad/s

The one-dimensional wave equation for waves of speed v is ∂²y/∂x² = (1/v²) ∂²y/∂t². In each function below, A, k and b are positive constants and ω = kv. Which function is NOT a solution of this equation?

1. Ay = A cos(kx) sin(ωt)
2. By = A/[1 + (x + vt)²/b²]
3. Cy = A sin(kx − ωt) + A sin(kx + ωt)
4. Dy = A exp[−(x² + v²t²)/b²]

A Galilean beam expander widens a collimated laser beam of diameter 2.00 mm. The beam passes first through a thin diverging lens of focal length −25.0 mm and then through a thin converging lens of focal length +100 mm on the same axis. Which lens separation makes the beam emerge collimated, and what is its diameter then?

1. ASeparation 125 mm; diameter 8.00 mm
2. BSeparation 75.0 mm; diameter 0.500 mm
3. CSeparation 75.0 mm; diameter 8.00 mm
4. DSeparation 75.0 mm; diameter 2.00 mm

Sunlight is reflected from the calm surface of a lake, of refractive index 1.33, towards an angler whose sunglasses have polarising lenses with vertical transmission axes. At what angle below the horizontal must the angler look at the water for the reflected glare to be completely polarised, so that the sunglasses block it entirely?

1. A53.1°
2. B36.9°
3. C41.2°
4. D48.8°

A sample of argon (molar mass 39.9 g/mol) is at 320 K. At what temperature would a sample of helium (molar mass 4.00 g/mol) have the same root-mean-square atomic speed? Treat both as ideal monatomic gases.

1. A32.1 K
2. B101 K
3. C320 K
4. D3190 K

A rigid, thermally insulated container is divided by a partition into two equal halves. One half holds 1.00 mol of an ideal gas and the other is evacuated. The partition is punctured and the gas fills the whole container. Take R = 8.314 J/(mol·K). Which statement is correct?

1. ANo heat enters the gas, so its entropy change is zero.
2. BThe gas does work as it expands, so its temperature falls.
3. CThe internal energy of the gas is unchanged, and its entropy rises by 5.76 J/K.
4. DThe entropy of the gas rises by 5.76 J/K, but the entropy of the universe is unchanged because the container is insulated.

A bunch of protons travels at constant speed 0.800c along a straight section of a linear accelerator, between two beam-position monitors that are fixed in the laboratory 36.0 m apart. Take c = 3.00 × 10⁸ m/s. Which pair gives the distance between the monitors and the time the bunch takes to travel from one to the other, both as measured in the rest frame of the bunch?

1. A21.6 m and 90.0 ns
2. B36.0 m and 90.0 ns
3. C21.6 m and 150 ns
4. D60.0 m and 250 ns

Light of wavelength 400 nm ejects photoelectrons from a metal surface whose work function is 2.15 eV. The wavelength is then changed to 200 nm while the intensity is kept the same. Take hc = 1240 eV·nm. What is the stopping potential with the 200 nm light?

1. A0.95 V
2. B1.90 V
3. C4.05 V
4. D6.20 V

An electron in a hydrogen atom has orbital angular momentum quantum number l = 3. What is the smallest angle that its orbital angular momentum vector can make with the z-axis?

1. A0°
2. B30.0°
3. C54.7°
4. D60.0°

Rest energies: proton 938.3 MeV, neutron 939.6 MeV, electron 0.511 MeV, muon 105.7 MeV, charged pion 139.6 MeV; neutrino masses are negligible. Which of the following processes is allowed by every conservation law?

1. An → p + e⁻ + ν_e
2. Bp → n + e⁺ + ν_e, for a free proton at rest
3. Cμ⁻ → e⁻ + γ
4. Dπ⁺ → μ⁺ + ν_μ

An aerial-mapping camera photographs a field from a height of 120 m. Its lens is a thin converging lens of focal length 8.80 mm and aperture diameter 3.14 mm, and its image sensor has square pixels 2.40 μm wide. Take the wavelength of the light as 550 nm in air. To reduce reflections, the lens glass, of refractive index 1.52, carries a single coating of magnesium fluoride, of refractive index 1.38, designed for 550 nm at normal incidence.

1. (a)

   Calculate Calculate how far behind the lens the sensor must be placed to focus the field, and how far this is from the lens's focal plane.

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   1. 1

      1/s′ = 1/f − 1/s = 1/(8.80 × 10⁻³ m) − 1/(120 m), so s′ = 8.800 65 × 10⁻³ m behind the lens

   2. 2

      s′ − f = 6.5 × 10⁻⁷ m = 0.65 μm, about a quarter of a pixel width, so the field is effectively at infinity and the sensor sits in the focal plane

2. (b)

   Calculate Calculate the width of the strip of ground that is imaged onto one pixel.

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   1. 1

      The magnification has magnitude s′/s = 8.8006 × 10⁻³ m/120 m = 7.334 × 10⁻⁵

   2. 2

      Ground width = 2.40 × 10⁻⁶ m/(7.334 × 10⁻⁵) = 3.27 × 10⁻² m = 32.7 mm

3. (c)

   Justify Justify whether diffraction at the lens aperture or the size of the pixels sets the smallest ground detail this camera can resolve at 550 nm.

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   1. 1

      Rayleigh criterion for a circular aperture: θ_min = 1.22λ/D = 1.22 × 550 × 10⁻⁹ m/(3.14 × 10⁻³ m) = 2.14 × 10⁻⁴ rad

   2. 2

      On the ground this is 120 m × 2.14 × 10⁻⁴ rad = 25.6 mm; equivalently fθ_min = 1.88 μm on the sensor

   3. 3

      One pixel covers 32.7 mm of ground (2.40 μm on the sensor), coarser than the 25.6 mm diffraction limit, so the pixels limit the resolution: two features need at least two pixels, about 65 mm, to be separated

4. (d)

   Determine Determine the minimum thickness of the coating that minimises the reflection of 550 nm light at normal incidence.

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   1. 1

      Both reflections, air to coating (1.00 to 1.38) and coating to glass (1.38 to 1.52), go from lower to higher index, so both waves are shifted by π and cancellation needs a path difference 2n_c t = λ/2

   2. 2

      t = λ/(4n_c) = 550 nm/(4 × 1.38) = 99.6 nm

5. (e)

   Explain When white light is reflected from the coated lens, the reflection looks purple. Explain why.

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   1. 1

      The path difference 2n_c t = 275 nm is half of 550 nm, but only 0.39 of a wavelength at 700 nm and 0.69 of a wavelength at 400 nm, so the two reflected waves cancel best near the middle of the visible spectrum and only partly at its ends

   2. 2

      The reflected light is therefore depleted in green and yellow and relatively strong at the red and violet ends, which together look purple

The pendulum of a longcase clock is a light rigid rod carrying a solid brass sphere of mass 2.30 kg and radius r = 0.0400 m, whose centre is L = 0.990 m below the pivot. The moment of inertia of a solid sphere about an axis through its centre is (2/5)mr². Take g = 9.81 m/s². The clock's escapement gives the pendulum a small push once each swing to keep its amplitude at 4.00°.

1. (a)

   Derive Derive, from the rotational form of Newton's second law, an expression for the period of small oscillations of this pendulum in terms of L, r and g, stating the approximation you make.

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   1. 1

      Parallel-axis theorem: the moment of inertia about the pivot is I = (2/5)mr² + mL²

   2. 2

      Gravity exerts a torque −mgL sin θ about the pivot, so I d²θ/dt² = −mgL sin θ ≈ −mgLθ, using sin θ ≈ θ (θ in radians) for small swings

   3. 3

      This is simple harmonic motion with ω₀² = mgL/I, so T₀ = 2π√(I/(mgL)) = 2π√((L² + 2r²/5)/(gL))

2. (b)

   Calculate Calculate the period, and the fraction by which it exceeds the period of a simple pendulum of length 0.990 m.

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   1. 1

      T₀ = 2π√((0.990² + 0.4 × 0.0400²) m²/(9.81 m/s² × 0.990 m)) = 1.9967 s, against 2π√(0.990 m/9.81 m/s²) = 1.9960 s for a simple pendulum

   2. 2

      Fractional excess = √(1 + 2r²/(5L²)) − 1 ≈ r²/(5L²) = 3.3 × 10⁻⁴, about 28 s per day

3. (c)

   Determine With the escapement disconnected, the amplitude falls from 4.00° to 2.00° in 12.0 minutes. Determine the average power the escapement must supply to keep the amplitude at 4.00°, stating the assumption about the damping that your method requires.

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   1. 1

      For light damping proportional to angular velocity the amplitude decays exponentially, so the energy, proportional to the square of the amplitude, falls by a factor of 4 with time constant τ_E = 720 s/ln 4 = 519 s

   2. 2

      Energy of the swing at 4.00° (0.0698 rad): E = ½Iω₀²θ₀² = ½mgLθ₀² = ½ × 2.30 kg × 9.81 m/s² × 0.990 m × (0.0698 rad)² = 5.44 × 10⁻² J

   3. 3

      P = E/τ_E = 5.44 × 10⁻² J/519 s = 1.05 × 10⁻⁴ W, replacing the 0.38% of the energy lost in each cycle

4. (d)

   Calculate For a finite amplitude θ₀ the period is T ≈ T₀(1 + θ₀²/16), with θ₀ in radians. The clock is adjusted to keep perfect time at an amplitude of 4.00°, but friction in its gear train lets the amplitude wander between 3.50° and 4.50°. Calculate the largest time the clock gains or loses in one day as a result.

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   1. 1

      Relative to 4.00° (0.06981 rad), the period at 4.50° (0.07854 rad) is longer by the fraction (0.07854² − 0.06981²)/16 = 8.09 × 10⁻⁵, and at 3.50° (0.06109 rad) it is shorter by 7.14 × 10⁻⁵

   2. 2

      The largest error is 8.09 × 10⁻⁵ × 86 400 s = 7.0 s lost per day at 4.50° (the clock gains 6.2 s per day at 3.50°)

5. (e)

   Evaluate The clock must keep time to within 1 s per day. Evaluate which idealisation of the simple-pendulum model, the point-mass bob or the small-angle limit, limits this clock's timekeeping.

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   1. 1

      The point-mass error, 3.3 × 10⁻⁴ or about 28 s per day, is constant, so it is removed once when the clock is regulated by raising or lowering the bob, and it does not limit the timekeeping

   2. 2

      The amplitude-dependent error cannot be regulated out: it varies by up to 7 s per day as the amplitude wanders, seven times the tolerance, so the small-angle idealisation is the one that fails, and the remedy is a steadier or smaller amplitude rather than a smaller bob

A pair of extraction fans of the same model stand 6.0 m apart on a flat roof. Each has 8 blades and radiates a steady tone at its blade-passing frequency, the number of blades multiplied by the fan's rotation rate in revolutions per second, and the controller of fan A shows a rotation rate of 24.50 revolutions per second. With both fans running, a technician's sound level meter, filtered to reject everything except a narrow band around this tone and quick enough to follow any change in the tone's amplitude, records a level that rises and falls smoothly and regularly between a maximum of 79.5 dB and a minimum of 71.9 dB, with successive maxima 5.00 s apart. The meter responds to the pressure at its microphone and reports the sound intensity level of a single plane wave with the same pressure amplitude. Treat the fans as small sources in open air and ignore reflections. Take the reference intensity as 1.00 × 10⁻¹² W/m², the density of air as 1.20 kg/m³, the speed of sound as 343 m/s and atmospheric pressure as 1.01 × 10⁵ Pa.

1. (a)

   Calculate Calculate the pressure amplitude of the sound at the meter when the level is at a maximum and when it is at a minimum.

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   1. 1

      At the maximum, I = 1.00 × 10⁻¹² W/m² × 10^(79.5/10) = 8.91 × 10⁻⁵ W/m², so p₀ = √(2ρvI) = √(2 × 1.20 kg/m³ × 343 m/s × 8.91 × 10⁻⁵ W/m²) = 0.271 Pa

   2. 2

      At the minimum, I = 1.00 × 10⁻¹² W/m² × 10^(71.9/10) = 1.55 × 10⁻⁵ W/m², so p₀ = √(2ρvI) = 0.113 Pa, a pressure-amplitude ratio of 2.40 between maximum and minimum

2. (b)

   Explain The sound from the two fans reaches the meter travelling in different directions. Explain why the excess pressures of the two waves at the meter can be added as numbers, whereas the displacements of the air in the two waves cannot, and why the superposition principle holds for these waves.

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   1. 1

      Excess pressure is a scalar, so the two waves' pressures add algebraically at each instant; sound is longitudinal, so each wave displaces the air along its own direction of travel, and displacements arriving from different directions add as vectors, which is why a pressure-sensing meter, like the ear, responds to the summed pressure

   2. 2

      Superposition needs a linear medium: even the largest pressure amplitude, 0.271 Pa, is only 2.7 × 10⁻⁶ of atmospheric pressure, so the air responds linearly and each wave passes through the other unchanged

3. (c)

   Determine Determine the sound intensity level that the louder fan alone would produce at the meter, the level that the quieter fan alone would produce, and the difference between the two fans' rotation rates.

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   1. 1

      The two tones differ slightly in frequency, so their relative phase advances steadily: the pressures add in step at a maximum and in opposition at a minimum, p_max = p₁ + p₂ and p_min = p₁ − p₂

   2. 2

      p₁ = (0.271 Pa + 0.113 Pa)/2 = 0.192 Pa and p₂ = (0.271 Pa − 0.113 Pa)/2 = 0.0790 Pa, so the quieter fan's pressure amplitude is about 0.41 of the louder fan's

   3. 3

      I₁ = p₁²/(2ρv) = 4.47 × 10⁻⁵ W/m², a level of 76.5 dB, and I₂ = p₂²/(2ρv) = 7.58 × 10⁻⁶ W/m², a level of 68.8 dB

   4. 4

      The level peaks once for each 2π of relative phase, so the blade-passing frequencies differ by 1/(5.00 s) = 0.200 Hz; dividing by the 8 blades, the rotation rates differ by 0.0250 revolutions per second (the beats alone cannot show which fan is faster)

4. (d)

   Describe The technician adjusts the controller of fan B until the level at the meter stops rising and falling. Describe how the steady level now depends on the fans and on where the meter is placed, giving the range of steady levels the meter could read at its present position, and the change in the difference between its distances from the two fans that takes the level from its highest to its lowest value, ignoring any change in each fan's own amplitude.

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   1. 1

      With equal frequencies the two tones keep a fixed phase difference, set by the relative angular positions of the two fans' blades and by the difference between the meter's distances from the fans, so the level is steady at some value between 71.9 dB (antiphase) and 79.5 dB (in phase), not necessarily at the 77.2 dB of the two intensities added

   2. 2

      Changing the difference in distance by half a wavelength, v/(2f) = 343 m/s/(2 × 196 Hz) = 0.875 m, turns in-phase arrival into antiphase and takes the level from its highest to its lowest value, so the roof now carries a fixed pattern of loud and quiet places, and the 77.2 dB sum of intensities is only the average over them

In the laboratory frame S, three events are recorded on the x-axis. Event A: a detector at x = 0 fires at t = 0. Event B: a detector at x = 900 m fires at t = 2.00 μs. Event C: the same detector at x = 900 m fires again at t = 4.00 μs. Frame S′ moves at +0.800c along the x-axis relative to S, and its origin coincides with that of S at event A. Take c = 3.00 × 10⁸ m/s.

1. (a)

   Calculate Calculate the invariant interval s² = (cΔt)² − (Δx)² between events A and B, and classify the separation.

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   1. 1

      cΔt = 3.00 × 10⁸ m/s × 2.00 × 10⁻⁶ s = 600 m, so s² = (600 m)² − (900 m)² = −4.50 × 10⁵ m²

   2. 2

      s² < 0, so the separation is spacelike

2. (b)

   Determine Determine the time interval Δt′ and the spatial separation Δx′ between A and B measured in S′, and state which event occurs first in S′.

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   1. 1

      γ = 1/√(1 − 0.800²) = 1.667

   2. 2

      Δt′ = γ(Δt − vΔx/c²) = 1.667 × (2.00 μs − 2.40 μs) = −0.667 μs, so B occurs before A in S′

   3. 3

      Δx′ = γ(Δx − vΔt) = 1.667 × (900 m − 480 m) = 700 m, consistent with (200 m)² − (700 m)² = −4.50 × 10⁵ m²

3. (c)

   Justify Justify the claim that event A cannot have caused event B, giving two separate arguments.

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   1. 1

      A signal from A would have to cover 900 m in 2.00 μs, a speed of 4.50 × 10⁸ m/s = 1.5c, faster than light; equivalently, the interval is spacelike

   2. 2

      The time order of A and B depends on the frame (B precedes A in S′), whereas a cause must precede its effect in every inertial frame

4. (d)

   Determine Determine the velocity, relative to S, of the inertial frame in which events A and B are simultaneous.

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   1. 1

      Setting Δt′ = γ(Δt − vΔx/c²) = 0 gives v = c²Δt/Δx

   2. 2

      v = c × (600 m/900 m) = 0.667c = 2.00 × 10⁸ m/s, in the +x direction

5. (e)

   Predict Predict the order of events A and C in S′ and in every other inertial frame, giving the reason.

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   1. 1

      For A and C, cΔt = 1200 m exceeds Δx = 900 m, so s² = +6.30 × 10⁵ m² and the separation is timelike

   2. 2

      C follows A in every inertial frame: in S′, Δt′ = 1.667 × (4.00 μs − 2.40 μs) = +2.67 μs, and no v < c can make Δt − vΔx/c² negative when cΔt > Δx

A radioisotope power source for a deep-space probe contains 3.50 kg of plutonium-238 (²³⁸Pu, Z = 94), which decays by alpha emission to uranium-234 with a half-life of 87.7 years. Atomic masses: ²³⁸Pu 238.049 560 u, ²³⁴U 234.040 952 u, ⁴He 4.002 603 u. Take 1 u = 931.494 MeV/c², 1 MeV = 1.602 18 × 10⁻¹³ J, N_A = 6.022 × 10²³ mol⁻¹, 1 year = 3.156 × 10⁷ s, and the molar mass of ²³⁸Pu as 238.05 g/mol.

1. (a)

   State State the decay equation, showing the nucleon number and proton number of every particle.

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   1. 1

      ²³⁸₉₄Pu → ²³⁴₉₂U + ⁴₂He, with nucleon numbers 238 = 234 + 4 and proton numbers 94 = 92 + 2

2. (b)

   Explain Explain why the Q value of this decay can be calculated directly from atomic masses, even though the decay is a nuclear process.

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   1. 1

      An atomic mass includes the atom's electrons: the ²³⁸Pu atom carries 94 electron masses, and the products carry 92 in the uranium atom plus 2 in the helium atom, also 94

   2. 2

      The electron masses therefore cancel in the mass difference, and the change in the atoms' total electron binding energy, at most a few tens of keV, is under 1% of a Q value of several MeV

3. (c)

   Calculate Calculate the Q value of the decay in MeV.

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   1. 1

      Δm = 238.049 560 u − 234.040 952 u − 4.002 603 u = 0.006 005 u

   2. 2

      Q = 0.006 005 u × 931.494 MeV/u = 5.59 MeV

4. (d)

   Show that Show that, for a parent nucleus at rest, the alpha particle carries about 98% of the Q value as kinetic energy.

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   1. 1

      Momentum conservation gives the alpha particle and the daughter equal and opposite momenta, so with K = p²/2m their kinetic energies are inversely proportional to their masses

   2. 2

      K_α = Q m_U/(m_U + m_α) = 5.59 MeV × 234.04/238.04 = 5.50 MeV, which is 98.3% of Q

5. (e)

   Calculate Calculate the initial thermal power of the source, assuming that all of the decay energy is absorbed within it.

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   1. 1

      N = (3500 g/238.05 g/mol) × 6.022 × 10²³ mol⁻¹ = 8.85 × 10²⁴ nuclei

   2. 2

      λ = ln 2/(87.7 × 3.156 × 10⁷ s) = 2.50 × 10⁻¹⁰ s⁻¹, so the activity is A = λN = 2.22 × 10¹⁵ Bq

   3. 3

      P = AQ = 2.22 × 10¹⁵ s⁻¹ × 5.59 MeV × 1.602 × 10⁻¹³ J/MeV = 1.99 × 10³ W

6. (f)

   Determine Determine how long after the source is made its thermal power falls to 1500 W.

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   1. 1

      P(t) = P₀e^(−λt), so t = (T½/ln 2) ln(P₀/P) = 87.7 years × log₂(1987 W/1500 W)

   2. 2

      t = 87.7 years × 0.406 = 35.6 years

An engine takes 0.500 mol of an ideal diatomic gas (C_V = 5R/2, γ = 1.40) around a three-stage cycle. A → B: reversible isothermal compression at 300 K, in contact with a cold reservoir at 300 K, until the volume is one quarter of its value at A. B → C: heating at constant volume, by contact with a hot reservoir at 600 K, until the gas reaches temperature T_C, when contact is broken. C → A: reversible adiabatic expansion back to state A. Take R = 8.314 J/(mol·K).

1. (a)

   Show that Show that T_C = 522 K.

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   1. 1

      Along the reversible adiabat C → A, TV^(γ−1) is constant, and V_C = V_B = V_A/4

   2. 2

      T_C = T_A(V_A/V_C)^(γ−1) = 300 K × 4^0.40 = 522 K

2. (b)

   Calculate Calculate the heat Q absorbed by the gas and the work W done by the gas in each of the three stages, and the net work done by the gas per cycle.

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   1. 1

      A → B: ΔU = 0, so Q = W = nRT ln(V_B/V_A) = 0.500 mol × 8.314 J/(mol·K) × 300 K × ln(1/4) = −1.73 × 10³ J

   2. 2

      B → C: W = 0, so Q = nC_V(T_C − T_B) = 0.500 mol × 20.79 J/(mol·K) × 222 K = +2.31 × 10³ J

   3. 3

      C → A: Q = 0, so W = −ΔU = nC_V(T_C − T_A) = +2.31 × 10³ J

   4. 4

      Net work = −1729 J + 0 + 2311 J = +582 J per cycle, equal to the net heat absorbed

3. (c)

   Compare Compare the thermal efficiency of this cycle with the Carnot efficiency of an engine operating between the same two reservoirs.

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   1. 1

      η = W/Q_in = 582 J/2311 J = 0.252 (25.2%), where only the B → C heat is heat input

   2. 2

      η_Carnot = 1 − 300 K/600 K = 0.500, so the cycle achieves about half of the Carnot bound

4. (d)

   Determine Determine the total entropy change of the gas and the two reservoirs per cycle, and identify the stage in which entropy is generated.

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   1. 1

      Over a complete cycle ΔS_gas = 0, because entropy is a state function

   2. 2

      ΔS_hot = −2311 J/600 K = −3.85 J/K and ΔS_cold = +1729 J/300 K = +5.76 J/K, so ΔS_total = +1.91 J/K

   3. 3

      Entropy is generated in B → C, where heat crosses a finite temperature difference from the 600 K reservoir into gas at 300 K to 522 K; A → B and C → A are reversible

5. (e)

   Show that Show that the work this cycle loses, compared with a Carnot engine that absorbs the same heat Q_BC from the 600 K reservoir, equals T_cold × S_gen, where T_cold = 300 K and S_gen is the entropy generated per cycle.

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   1. 1

      Over a cycle W = Q_BC − |Q_AB| and S_gen = |Q_AB|/T_cold − Q_BC/T_hot; eliminating |Q_AB| gives W = Q_BC(1 − T_cold/T_hot) − T_cold S_gen

   2. 2

      Numerically, η_Carnot Q_BC − W = 1155 J − 582 J = 574 J, and T_cold S_gen = 300 K × 1.91 J/K = 574 J

In a low-energy electron diffraction experiment, electrons accelerated from rest through a potential difference V strike a clean copper crystal surface along its normal. The rows of surface atoms, a distance a apart, act as a reflection grating: first-order diffracted beams leave the surface at an angle φ to the normal, where a sin φ = λ and λ is the de Broglie wavelength of the electrons. The angle φ is measured at several accelerating potential differences, and the energy scale of the electron gun is calibrated to ±2%. Take h = 6.626 × 10⁻³⁴ J·s, m_e = 9.109 × 10⁻³¹ kg and e = 1.602 × 10⁻¹⁹ C. In bulk copper the atoms of a plane of this orientation form a square array of side 0.2556 nm.

1. (a)

   Show that Show that the model predicts a straight line through the origin when sin φ is plotted against 1/√V, and give its gradient in terms of h, m_e, e and a.

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   1. 1

      The kinetic energy eV = p²/(2m_e) gives the de Broglie wavelength λ = h/p = h/√(2m_e eV)

   2. 2

      Then sin φ = λ/a = [h/(a√(2m_e e))] × (1/√V), a straight line through the origin with gradient G = h/(a√(2m_e e))

2. (b)

   Determine Using all of the readings, determine the gradient of the graph of sin φ against 1/√V.

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   1. 1

      Values of 1/√V (in V^−½) and sin φ: 0.1414 and 0.681, 0.1155 and 0.550, 0.1000 and 0.483, 0.0894 and 0.426, 0.0816 and 0.394, 0.0756 and 0.360, 0.0707 and 0.340

   2. 2

      A best-fit straight line through all seven points and the origin (or a least-squares fit), not a gradient taken from a single reading

   3. 3

      Gradient G = 4.80 V^½ (least-squares line through the origin 4.798 V^½; accept 4.75 V^½ to 4.85 V^½)

3. (c)

   Calculate Calculate the row spacing a, with its uncertainty arising from the calibration of the energy scale.

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   1. 1

      a = h/(G√(2m_e e)), with √(2m_e e) = √(2 × 9.109 × 10⁻³¹ kg × 1.602 × 10⁻¹⁹ C) = 5.402 × 10⁻²⁵ kg^½ C^½

   2. 2

      a = 6.626 × 10⁻³⁴ J·s/(4.80 V^½ × 5.402 × 10⁻²⁵ kg^½ C^½) = 2.56 × 10⁻¹⁰ m = 0.256 nm (accept 0.253 nm to 0.258 nm)

   3. 3

      A ±2% error in V changes 1/√V, and so G and a, by ±1%, so a = 0.256 ± 0.003 nm

4. (d)

   Evaluate A calculation predicts that the outermost layer of copper atoms relaxes towards the layer beneath it, reducing the spacing between those two layers by about 2%. Evaluate whether these data support a surface whose atoms keep their bulk arrangement within the layer, and whether this experiment could detect the predicted relaxation.

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   1. 1

      The measured a = 0.256 ± 0.003 nm agrees with the bulk value of 0.2556 nm within its uncertainty, so the data support a surface layer whose atoms keep the bulk spacing within the plane

   2. 2

      The predicted relaxation changes only the spacing perpendicular to the surface, which does not enter a sin φ = λ, so the beam angles cannot detect it at any precision; it would have to be sought in how the beam intensities vary with electron energy

In muonic hydrogen a negative muon, of mass 1.8835 × 10⁻²⁸ kg, takes the place of the electron and is bound to a proton of mass 1.6726 × 10⁻²⁷ kg; its energy levels are measured to find the size of the proton. The relative motion obeys the hydrogen-atom Schrödinger equation with the reduced mass μ = m_μm_p/(m_μ + m_p) in place of the electron mass. For a state with no angular dependence (l = 0) the equation reads −(ℏ²/2μ)(1/r²) d/dr(r² dψ/dr) − (e²/(4πε₀r))ψ = Eψ. Take ℏ = 1.0546 × 10⁻³⁴ J·s, e = 1.602 × 10⁻¹⁹ C, ε₀ = 8.854 × 10⁻¹² F/m, the proton radius as 0.84 fm and the Bohr radius of ordinary hydrogen as 5.29 × 10⁻¹¹ m. You may use ∫₀^∞ rⁿe^(−βr) dr = n!/β^(n+1).

1. (a)

   Derive By substituting the trial solution ψ(r) = Ce^(−r/a) into the Schrödinger equation, derive expressions for the length a and the energy E, and evaluate both for muonic hydrogen.

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   1. 1

      dψ/dr = −ψ/a, so r² dψ/dr = −(r²/a)ψ

   2. 2

      (1/r²) d/dr(r² dψ/dr) = (1/a² − 2/(ar))ψ

   3. 3

      Substituting and dividing by ψ: −(ℏ²/2μ)(1/a² − 2/(ar)) − e²/(4πε₀r) = E, which must hold for every r, so the 1/r terms must cancel: a = 4πε₀ℏ²/(μe²)

   4. 4

      The remaining constant terms give E = −ℏ²/(2μa²)

   5. 5

      μ = 1.693 × 10⁻²⁸ kg (185.8 electron masses), a = 2.85 × 10⁻¹³ m and E = −4.05 × 10⁻¹⁶ J = −2.53 keV

2. (b)

   Show that Show that the normalised wavefunction has C = 1/√(πa³).

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   1. 1

      Normalisation over all space: ∫₀^∞ |ψ|² 4πr² dr = 4πC² ∫₀^∞ r²e^(−2r/a) dr = 1

   2. 2

      The integral is 2!/(2/a)³ = a³/4, so πa³C² = 1 and C = 1/√(πa³)

3. (c)

   Determine Determine the most probable distance of the muon from the proton, in metres.

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   1. 1

      dP/dr ∝ (2r − 2r²/a)e^(−2r/a) = 0 gives r = a, the maximum (r = 0 is a minimum)

   2. 2

      r = a = 2.85 × 10⁻¹³ m, about 186 times smaller than the corresponding distance in ordinary hydrogen

4. (d)

   Sketch Sketch the radial probability density P(r) = 4πr²|ψ(r)|² against r/a, marking the most probable radius found in (c) and the mean radius ⟨r⟩ = 1.5a.

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   1. 1

      Axes labelled P(r) and r/a; the curve is zero at r = 0, rises initially as r², and has a single maximum marked at r/a = 1

   2. 2

      A long exponential tail towards large r, so the curve is asymmetric, with ⟨r⟩ = 1.5a marked to the right of the peak

5. (e)

   Estimate Estimate the probability that the muon is found inside the proton, stating the assumption you make, and compare it with the corresponding probability for the electron in ordinary hydrogen.

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   1. 1

      Because the proton radius R = 0.84 fm is far smaller than a, take |ψ|² as constant at its value at r = 0 across the proton: P ≈ (4/3)πR³|ψ(0)|² = (4/3)(R/a)³

   2. 2

      For muonic hydrogen P ≈ (4/3)(0.84 × 10⁻¹⁵ m/2.85 × 10⁻¹³ m)³ = 3.4 × 10⁻⁸

   3. 3

      For ordinary hydrogen P ≈ (4/3)(0.84 × 10⁻¹⁵ m/5.29 × 10⁻¹¹ m)³ = 5.3 × 10⁻¹⁵, so the muon is about 6 × 10⁶ times more likely to be inside the proton, which is why muonic energy levels are far more sensitive to the proton's size

The neutron's mass is determined most precisely from the gamma ray emitted when a slow neutron is captured by a proton at rest: n + p → d + γ. A crystal-diffraction spectrometer measures the photon energy as E_γ = 2.223 25 MeV, with an uncertainty of ±0.4 eV. Take the rest energies m_pc² = 938.2721 MeV and m_dc² = 1875.6129 MeV, and treat the neutron's kinetic energy before capture as negligible. Write M = m_n + m_p for the total rest mass before capture, and B = (M − m_d)c² for the binding energy of the deuteron.

1. (a)

   Derive Derive, from conservation of energy and momentum, an exact expression for E_γ in terms of M, m_d and c.

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   1. 1

      The initial system is at rest, so the deuteron recoils with momentum equal and opposite to the photon's, of magnitude E_γ/c

   2. 2

      Energy conservation: Mc² = E_γ + √((m_dc²)² + E_γ²)

   3. 3

      Isolating the square root and squaring: (Mc² − E_γ)² = (m_dc²)² + E_γ², so M²c⁴ − 2Mc²E_γ = m_d²c⁴

   4. 4

      E_γ = (M² − m_d²)c²/(2M)

2. (b)

   Show that Show that the result of (a) can be written exactly as E_γ = B − B²/(2Mc²), and explain the physical meaning of the term B²/(2Mc²).

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   1. 1

      M² − m_d² = (M − m_d)(M + m_d), with (M − m_d)c² = B and M + m_d = 2M − B/c²

   2. 2

      E_γ = B(2Mc² − B)/(2Mc²) = B − B²/(2Mc²), with no approximation

   3. 3

      By energy conservation Mc² = E_γ + m_dc² + K_d, so B²/(2Mc²) = B − E_γ is exactly the kinetic energy of the recoiling deuteron, about E_γ²/(2m_dc²) = 1.32 keV

3. (c)

   Determine Determine the rest energy of the neutron, m_nc², from the measured photon energy.

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   1. 1

      B = E_γ + B²/(2Mc²); the correction is small, so evaluate it with B ≈ E_γ and Mc² ≈ 1877.8 MeV: (2.2233 MeV)²/(2 × 1877.8 MeV) = 1.32 keV

   2. 2

      B = 2.223 25 MeV + 0.001 32 MeV = 2.224 57 MeV

   3. 3

      m_nc² = m_dc² + B − m_pc² = 1875.6129 MeV + 2.2246 MeV − 938.2721 MeV = 939.5654 MeV

4. (d)

   Evaluate Evaluate whether the recoil term must be included when the neutron's rest energy is quoted from this measurement.

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   1. 1

      Neglecting recoil gives m_nc² = 939.5641 MeV, lower by 1.32 keV, which is about 3 × 10³ times the ±0.4 eV uncertainty in E_γ and larger than the 0.1 keV rounding of the rest energies given

   2. 2

      So the recoil term is essential: without it the result is wrong by thousands of standard deviations of the photon measurement, and once it is included the precision of m_n is limited by the proton and deuteron rest energies rather than by the spectrometer

5. (e)

   Compare In the reverse process, photodisintegration γ + d → n + p, a photon strikes a deuteron at rest. Compare the minimum photon energy for this process with the photon energy emitted in capture, and explain the difference.

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   1. 1

      At threshold the neutron and proton move off together with the photon's momentum, so the invariant mass must just reach M: (E_th + m_dc²)² − E_th² = M²c⁴, giving E_th = (M² − m_d²)c²/(2m_d) = B + B²/(2m_dc²)

   2. 2

      E_th = 2.224 57 MeV + 0.001 32 MeV = 2.225 89 MeV, which is 2.64 keV more than the 2.223 25 MeV emitted in capture

   3. 3

      In capture the photon carries the binding energy less the deuteron's recoil energy; in photodisintegration the photon must supply the binding energy plus the kinetic energy that momentum conservation forces the n–p pair to carry, so the two energies differ by about two recoil energies, B²/(m_dc²)