20 questions in 35 minutes, like Section I, marked with worked explanations.
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HSC Physics · Module 7 flashcards
84 cards. Work out each answer first, then open the card to check it and see the working.
Answer: the electric and magnetic constants of free space
Maxwell showed c = 1/√(μ0ε0), which gave a value matching the measured speed of light.
Answer: 5.6 × 1014 Hz
f0 = W/h = (2.3 × 1.602 × 10-19)/(6.626 × 10-34) = 5.6 × 1014 Hz.
Answer: 11 μs
γ = 1/√(1 − 0.98²) = 5.03. t = γt0 = 5.03 × 2.2 = 11 μs.
Answer: could be reflected, refracted and polarised like light
Hertz generated and detected radio waves and showed they had the same wave properties as light, confirming Maxwell's prediction of electromagnetic waves.
Answer: 80 m
γ = 1/√(1 − 0.36) = 1.25. L = L0/γ = 100/1.25 = 80 m.
Answer: provided no evidence for the existence of the aether
The experiment looked for a change in the speed of light due to Earth's motion through the aether. No shift was found, so no aether motion was detected.
Answer: timing variations in the eclipses of Jupiter's moon Io
Rømer noticed Io's eclipses appeared late when Earth was far from Jupiter and early when near, and attributed the difference to light's travel time across Earth's orbit.
Answer: 36.9°
d = 1/500 000 = 2.0 × 10-6 m. sinθ = mλ/d = 2 × 6.0 × 10-7/2.0 × 10-6 = 0.60, so θ = 36.9°.
Answer: 1.3 eV
Ephoton = hf = 6.626 × 10-34 × 8.0 × 1014 = 5.3 × 10-19 J = 3.3 eV. KEmax = 3.3 − 2.0 = 1.3 eV.
Answer: 16.7 years
γ = 1/√(1 − 0.64) = 1.67. The crew measure proper time, so Earth measures t = γt0 = 16.7 years.
Answer: 5.0 × 10-19 J
E = hc/λ = 6.626 × 10-34 × 3.0 × 108/4.0 × 10-7 = 5.0 × 10-19 J (about 3.1 eV).
Answer: 2.4 × 10-19 J
KEmax = eVstop = 1.602 × 10-19 × 1.5 = 2.4 × 10-19 J (that is, 1.5 eV).
Answer: a rotating toothed wheel and a distant mirror
Light passed through a gap in a spinning wheel, reflected from a mirror kilometres away, and was blocked by the next tooth at a particular rotation rate, giving the travel time.
Answer: a continuous spectrum
Closely packed atoms in a hot solid interact so strongly that all wavelengths are emitted. Line spectra come from low-pressure gases.
Answer: may not be simultaneous in a frame moving relative to the first
Because c is the same for all observers, observers in relative motion can disagree about whether spatially separated events occurred at the same time.
Answer: unify electricity, magnetism and light as electromagnetic waves
Maxwell's equations predicted self-propagating electric and magnetic waves at a speed equal to that of light, identifying light as an electromagnetic wave.
Answer: the lines are broadened as one edge approaches and the other recedes
Light from the approaching limb is blue-shifted and from the receding limb red-shifted, smearing each line over a range of wavelengths.
Answer: 3.9 × 1026 W
P = (Δm/Δt)c² = 4.3 × 109 × 9.0 × 1016 = 3.9 × 1026 W.
Answer: electromagnetic waves had travelled from the transmitter to the receiver
The receiving loop was not connected to the transmitter. The spark showed that an oscillating electromagnetic wave had induced a current in it.
Answer: an emission spectrum of discrete bright lines
Isolated atoms emit only the wavelengths matching their energy level differences.
Answer: a transverse wave
Only transverse waves have an oscillation direction perpendicular to travel that can be filtered by a polariser.
Answer: an observer for whom the events happen at the same place
Proper time is the shortest time interval, measured by the clock that is present at both events. Every other observer measures a longer, dilated interval.
Answer: perpendicular to each other and both perpendicular to the direction of travel
E, B and the direction of propagation are mutually perpendicular, and E and B oscillate in phase.
Answer: sharper, brighter maxima
More slits means more waves adding constructively at the maxima and cancelling more completely between them, giving narrow bright lines.
Answer: the frequency of the light and the work function of the metal
KEmax = hf − W. Intensity affects how many electrons leave, not their energy.
Answer: in the frame in which the object is at rest
Proper length is the longest length. Any observer moving relative to the object measures a contracted length L = L0/γ.
Answer: moving towards Earth
A shift to shorter wavelengths indicates a source approaching the observer.
Answer: 600 nm
λ = dΔy/L = 5.0 × 10-4 × 2.4 × 10-3/2.0 = 6.0 × 10-7 m = 600 nm.
Answer: 5.0 × 1018
Ephoton = hc/λ = 3.97 × 10-19 J. N = 2.0/3.97 × 10-19 = 5.0 × 1018 per second.
Answer: The energy required becomes infinite as v approaches c
γ → ∞ as v → c, so the kinetic energy (γ − 1)mc² and the required work become unbounded.
Answer: setting up standing waves and measuring their wavelength, then using v = fλ
Reflecting the waves created standing waves. The node spacing gave the wavelength, and the known oscillator frequency gave the speed, which equalled c.
Answer: 290 nm
λmax = b/T = 2.898 × 10-3/10 000 = 2.9 × 10-7 m = 290 nm (ultraviolet).
Answer: I0/2
Malus' law: I = I0 cos²45° = I0 × 0.5. (For unpolarised light there would be an extra factor of ½.)
Answer: the travelling twin is younger
The travelling twin's clock runs slow relative to Earth. The situation is not symmetric because the traveller changes frames when turning around.
Answer: the elements present in the Sun's outer atmosphere
Each element absorbs at its own set of wavelengths, so the pattern of dark lines identifies which elements are present in the cooler outer layers.
Answer: d sinθ = mλ
Constructive interference needs a path difference d sinθ equal to a whole number of wavelengths.
Answer: 1.8 V
Ephoton = 1240/300 = 4.13 eV. KEmax = 4.13 − 2.3 = 1.8 eV, so Vstop = 1.8 V.
Answer: 6.0 years
Earth frame: t = 8.0/0.80 = 10 years. γ = 1.67, so crew time = 10/1.67 = 6.0 years. (Equivalently the distance contracts to 4.8 ly.)
Answer: 3.0 × 108 m s-1
c = 1/√(μ0ε0) ≈ 3.0 × 108 m s-1, matching the known speed of light.
Answer: 9 × 104 m s-1
v = c Δλ/λ = 3.0 × 108 × 0.2/656.3 = 9.1 × 104 m s-1, receding (red shift).
Answer: a larger angle
More lines per mm means smaller slit spacing d. sinθ = λ/d increases.
Answer: running slow
Time dilation is symmetric between inertial frames: each sees the other's clock as slow, because each sees the other as moving.
Answer: the Doppler shift of the lines
The amount of shift Δλ/λ = v/c gives the speed towards or away from the observer.
Answer: the ultraviolet catastrophe
Planck resolved it by proposing that energy is emitted in packets E = hf, which suppresses high-frequency emission.
Answer: half a wavelength
A half-wavelength path difference puts the waves out of phase so they cancel.
Answer: 1.0 m
Length contraction occurs only along the direction of motion. Perpendicular dimensions are unchanged.
Answer: rotating the receiving loop and observing the spark strength change
The spark in the detector depended on the orientation of the receiver relative to the transmitter, which is the signature of a polarised transverse wave.
Answer: hotter
Wien's law: hotter objects peak at shorter wavelengths. Blue stars have surface temperatures above about 10 000 K; red stars below about 3500 K.
Answer: one half
Averaging cos²θ over all orientations gives ½. Malus' law applies only after the light is already polarised.
Answer: 6600 m
Lab lifetime = 10 × 2.2 μs = 22 μs. Distance = 0.995 × 3.0 × 108 × 22 × 10-6 = 6.6 × 10³ m.
Answer: varied with the Earth–Jupiter distance
The eclipses lagged when Earth was further away because the light had further to travel. The lag gave a finite speed for light.
Answer: the Balmer formula
Balmer found an empirical formula for the visible lines; Rydberg generalised it and Bohr later derived it.
Answer: 11°
d = 1/300 000 = 3.33 × 10-6 m. sinθ = 6.5 × 10-7/3.33 × 10-6 = 0.195, θ = 11°.
Answer: 8.2 × 10-14 J (0.51 MeV)
E = mc² = 9.109 × 10-31 × 9.0 × 1016 = 8.2 × 10-14 J ≈ 0.51 MeV.
Answer: produces a magnetic field
This is the key addition Maxwell made to Ampère's law. Together with Faraday's law it allows E and B to regenerate each other as a travelling wave.
Answer: white, with coloured fringes either side
All wavelengths have zero path difference at the centre, so they all add constructively. Away from centre, different wavelengths peak at different positions.
Answer: the speed of light in a vacuum is the same for all observers
The constancy of c for all inertial observers, regardless of the motion of the source, leads directly to time dilation and length contraction.
Answer: travelled through air at a finite speed, later measured as c
No conductor linked transmitter and receiver. The waves propagated through space, as Maxwell's theory required.
Answer: 4100 K
T = 2.898 × 10-3/7.0 × 10-7 = 4100 K — a cool orange-red star.
Answer: closer together
Δy = λL/d. Blue light has a shorter wavelength, so the spacing decreases.
Answer: no electrons emitted, however bright the light
Each photon has less energy than the work function. No single photon can free an electron, and intensity does not change the energy per photon.
Answer: 69 minutes
γ = 1/√(1 − 0.25) = 1.155. t = 1.155 × 60 = 69 minutes.
Answer: radio, microwave, infrared, visible, ultraviolet, X-ray, gamma
Radio has the longest wavelength and lowest frequency; gamma the shortest wavelength and highest frequency.
Answer: 5800 K
Its spectrum peaks near 500 nm, which by Wien's law corresponds to about 5800 K.
Answer: inversely proportional to its wavelength
E = hf = hc/λ.
Answer: the motion of Earth through the aether via a shift in interference fringes
If light moved relative to a stationary aether, rotating the apparatus should have shifted the fringes. No shift was seen.
Answer: dark lines on a continuous spectrum
A cooler gas in front of a continuous source removes specific wavelengths, leaving gaps.
Answer: 1.67 × 10-6 m
d = 1 mm/600 = 1.0 × 10-3/600 = 1.67 × 10-6 m.
Answer: the work function and the electron's kinetic energy
hf = W + KEmax. The work function is the minimum energy to free the electron; the rest is kinetic energy.
Answer: 1.8 × 1014 J
E = mc² = 2.0 × 10-3 × 9.0 × 1016 = 1.8 × 1014 J.
Answer: zero
After the first polariser the light is polarised. Malus' law: I = I1 cos²90° = 0.
Answer: the shape of the black body spectrum
Treating energy as emitted in multiples of hf gave a curve matching experiment and removed the ultraviolet catastrophe.
Answer: 970 nm
λmax = 2.898 × 10-3/3000 = 9.7 × 10-7 m = 970 nm (infrared).
Answer: KEmax = hf − W
The electron gains hf from one photon and spends W escaping the metal; the rest is kinetic energy.
20 questions in 35 minutes, like Section I, marked with worked explanations.
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317 cards across Modules 5–8.
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