Edexcel International A Level Physics · Unit 5 · Thermodynamics, Radiation, Oscillations and Cosmology
Stars: radiation, distances and the HR diagram
40 free, high-quality past-paper-style questions on Stars: radiation, distances and the HR diagram for Edexcel IAL Physics, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Physics notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.40 new this month · 149 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1IALP-U5-8-A-01Unit 5 · A2Easy1 mark
Which statement describes a black body?
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Answer
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Notes
A black body is an idealised object that absorbs all the electromagnetic radiation falling on it. It is also the best possible emitter, with a continuous spectrum whose shape depends only on its temperature.
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Answer C
Answer: C
Exam tip: A black body is a perfect absorber and a perfect emitter. Its spectrum is continuous, not a single wavelength: only the position of the peak is fixed by the temperature.
Why can trigonometric parallax be used to measure the distance only to relatively nearby stars?
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The parallax angle is inversely proportional to the distance. For distant stars it becomes too small to measure compared with the uncertainty in the angle measurement.
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Answer C
Answer: C
Exam tip: Link the limit to the measurement: the angle shrinks as 1/d, so its percentage uncertainty grows until the measurement is meaningless.
The radiation emitted by stars can be modelled as black-body radiation. Star A has a surface temperature of 8000 K and star B has a surface temperature of 4000 K.
(a)
State what is meant by a black body.
(1)
(b)
Describe two ways in which the black-body radiation curve (intensity against wavelength) of star B differs from that of star A.
(2)
(c)
Calculate the wavelength at which the radiation from star A has its maximum intensity.
(1)
(d)
State the region of the electromagnetic spectrum in which this wavelength lies.
(1)
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Part (a)
an (idealised) object that absorbs all the electromagnetic radiation falling on it
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Part (b)
the peak of B is at a longer wavelength (twice that of A)
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B's curve is lower at every wavelength / the area under it is smaller (1/16 of A's per unit area)
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Part (c)
λmax = 2.898 × 10−3 ÷ 8000 = 3.62 × 10−7 m
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Part (d)
ultraviolet (362 nm is shorter than about 400 nm)
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Model sketch
Figure: black-body curves; B's peak is at twice A's wavelength and far lower (dashed: B scaled up ×10)
Exam tip: When comparing black-body curves, describe both the position of the peak (Wien) and the height or area (Stefan–Boltzmann).
A rover on the surface of Mars is powered by solar panels. Mars orbits the Sun at a mean distance of 2.28 × 1011 m. The luminosity of the Sun is 3.83 × 1026 W.
(a)
Calculate the intensity of the Sun's radiation at the mean distance of Mars.
(2)
(b)
The solar panels have a total area of 1.2 m2 and convert 22% of the radiation falling on them into electrical energy. Calculate the maximum electrical power output when the Sun's radiation is perpendicular to the panels.
(2)
(c)
State one reason why the actual power output is usually less than this value.
(1)
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Part (a)
I = L/4πd2 = 3.83 × 1026 ÷ (4π × (2.28 × 1011)2)
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= 586 W m−2
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Part (b)
power incident = 586 × 1.2 = 703 W
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electrical power = 0.22 × 703 = 155 W
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Part (c)
any one: some radiation is absorbed or scattered by the Martian atmosphere or dust / dust settles on the panels / the radiation is not perpendicular to the panels for most of the day
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Exam tip: Intensity is power per unit area: multiply by the area to get the power arriving, then by the efficiency for the useful output.
The graph shows the black-body radiation curves for three stars, X, Y and Z, drawn to the same scale. Which statement is correct?
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Z's curve peaks at the longest wavelength, so by Wien's law Z is the coolest. Its curve is also lowest at every wavelength, so the area under it (the power emitted per unit surface area, ∝ T4) is the smallest: Z has the lowest temperature and emits the least power per unit area.
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Answer A
Answer: A
Exam tip: Read two things from black-body curves: the peak position gives the temperature (Wien), and the area under the curve gives the power per unit area (∝ T4).
The diagram shows the Earth at two positions, E1 and E2, on opposite sides of its orbit around the Sun, and a nearby star. The angle θ at the star between the lines from E1 and E2 is shown. What is the distance from the Sun to the star?
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θ is subtended by the diameter of the orbit (2 AU). The parallax angle p is subtended by the radius (1 AU), so p = θ/2 = 0.25 arcseconds. d = 1/p = 1 ÷ 0.25 = 4.0 pc.
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Answer B
Answer: B
Exam tip: The parsec is defined with a baseline of 1 AU, the radius of the orbit. If an angle is drawn across the whole orbit, halve it before using d = 1/p.
The flow chart shows the later stages in the life of a star whose main-sequence mass is about 25 times the mass of the Sun. Three stages are labelled X, Y and Z. Which row identifies X, Y and Z?
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A star this massive swells into a red supergiant (X). When fusion in its core can no longer support it, the core collapses and the star explodes as a supernova (Y). A remnant core of more than about 3 M☉ collapses to a black hole (Z).
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Answer C
Answer: C
Exam tip: The route depends on mass: low-mass stars end with a planetary nebula and a white dwarf; high-mass stars end with a supernova, then a neutron star or a black hole.
Two images of the same small region of sky are taken six months apart, in January and July. The diagram shows the two images overlaid on a grid. The background stars are in the same place in both images, but a nearby star S has moved. 1 parsec = 3.09 × 1016 m.
(a)
Explain why the background stars do not appear to move.
(1)
(b)
Use the diagram to determine the total angular shift of star S between the two images.
(1)
(c)
Calculate the distance to star S in parsecs.
(2)
(d)
Express this distance in metres.
(1)
(e)
The smallest shift of a star between two images that this telescope can detect is one grid square. Calculate the greatest distance that can be measured with this telescope by this method.
(1)
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Part (a)
they are so far away that their parallax angles are far too small to detect
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Part (b)
12 squares × 0.010 arcsecond = 0.120 arcsecond
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Part (c)
parallax angle p = ½ × 0.120 = 0.060 arcsecond
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d = 1/p = 1 ÷ 0.060 = 17 pc (16.7 pc)
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Part (d)
16.7 × 3.09 × 1016 = 5.2 × 1017 m
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Part (e)
smallest shift 0.010 arcsecond, so smallest parallax angle 0.005 arcsecond; d = 1 ÷ 0.005 = 2.0 × 102 pc
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Exam tip: With p in arcseconds, d = 1/p gives parsecs directly: no conversion to radians is needed unless the answer is wanted in metres by trigonometry.
Two Cepheid variable stars X and Y have the same period of pulsation and therefore the same mean luminosity. The mean intensity of the radiation from X measured at the Earth is 1/9 of that from Y. What is the ratio distance to X/distance to Y?
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I = L/4πd2 with L the same, so d ∝ 1/√I. dX/dY = √9 = 3.
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Answer C
Answer: C
Exam tip: For standard candles with the same luminosity, intensity ∝ 1/d²: a ninth of the intensity means three times the distance.
Which sequence describes the stages of a star of about the same mass as the Sun after it leaves the main sequence?
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A low-mass star swells into a red giant, sheds its outer layers as a planetary nebula and leaves its core as a white dwarf. Ending in a core-collapse supernova, a neutron star or a black hole needs a much more massive star.
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Answer A
Answer: A
Exam tip: Learn the two routes separately: low mass → red giant → planetary nebula → white dwarf; high mass → red supergiant → supernova → neutron star or black hole.
An astronomer photographs a star against a background of very distant stars on two dates six months apart. Between the two photographs the star's position shifts by a total angle of 0.62 arcseconds. 1 AU = 1.50 × 1011 m. 1 arcsecond = 1/3600 degree.
Figure: the star observed from opposite sides of the Earth's orbit (not to scale)
(a)
Explain why the photographs are taken six months apart.
(1)
(b)
Calculate the distance to the star in metres.
(3)
(c)
Calculate the distance to the star in parsecs.
(1)
(d)
Define the parsec.
(1)
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Part (a)
the Earth is then on opposite sides of its orbit, giving the largest baseline (2 AU) and so the largest shift
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Part (b)
parallax angle p = 0.62 ÷ 2 = 0.31 arcseconds = 0.31/3600 degrees = 1.50 × 10−6 rad
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d = r/tan p = 1.50 × 1011 ÷ tan(0.31/3600 degrees)
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= 1.0 × 1017 m
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Part (c)
d = 1 ÷ 0.31 = 3.2 pc
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Part (d)
the distance at which a radius of 1 AU subtends an angle of 1 arcsecond / the distance to a star whose parallax angle is 1 arcsecond
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Common error: Using the total shift as the parallax angle. The parallax angle is half the shift seen over six months, because it is defined using the orbit's radius, not its diameter.
A Cepheid variable star is observed in a small galaxy near our own. From its period of pulsation, its mean luminosity is found to be 1.1 × 1030 W. The mean intensity of its radiation at the Earth is 3.9 × 10−14 W m−2. 1 parsec = 3.09 × 1016 m.
(a)
Explain why a Cepheid variable star can be used as a standard candle.
(1)
(b)
Calculate the distance to the galaxy in metres.
(2)
(c)
Express this distance in parsecs.
(1)
(d)
Dust between the galaxy and the Earth absorbs some of the star's radiation. Explain the effect this has on the calculated distance.
(2)
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Part (a)
its luminosity can be found (from its period), so its distance can be calculated from the measured intensity
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Part (b)
d = √(L/4πI) = √(1.1 × 1030 ÷ (4π × 3.9 × 10−14))
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= 1.5 × 1021 m
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Part (c)
1.5 × 1021 ÷ 3.09 × 1016 = 4.8 × 104 pc
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Part (d)
the measured intensity is lower than it would be without the dust
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so the calculated distance (∝ 1/√I) is too large / an overestimate
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Exam tip: Rearrange I = L/4πd2 in symbols first, d = √(L/4πI), then substitute. It makes forgetting the square root much less likely.
The data below are for four stars. The Sun's surface temperature is 5780 K and its luminosity is L☉. Star P: surface temperature 3400 K, luminosity 0.010 L☉ Star Q: surface temperature 4100 K, luminosity 300 L☉ Star R: surface temperature 12 000 K, luminosity 0.0030 L☉ Star S: surface temperature 3700 K, luminosity 1.5 × 105L☉
(a)
Using a Hertzsprung–Russell diagram, identify the type of each star.
(2)
(b)
Calculate the radius of star Q as a multiple of the Sun's radius.
(3)
(c)
Explain why star R has a very low luminosity even though it is hotter than the Sun.
(1)
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Part (a)
P main sequence (red dwarf) and S supergiant
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Q red giant and R white dwarf
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Part (b)
L = 4πr2σT4, so r/r☉ = √(L/L☉) × (T☉/T)2
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= √300 × (5780/4100)2
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= 34
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Part (c)
its radius (surface area) must be very small, since L ∝ r2T4
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Model sketch
Figure: the four stars placed on a Hertzsprung–Russell diagram (dashed line: main sequence)
Exam tip: Work in solar units with ratios: the constants 4π and σ cancel, and the numbers stay small.
The graph shows the black-body radiation curves for two stars, P and Q, drawn to the same scale. What is the ratio power emitted per unit surface area of P/power emitted per unit surface area of Q?
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From the graph the peaks are at 450 nm (P) and 600 nm (Q). Wien's law gives T ∝ 1/λmax, so TP/TQ = 600/450 = 1.33. Power per unit area = σT4, so the ratio = 1.334 = 3.2.
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Answer D
Answer: D
Exam tip: You do not need the temperatures themselves: since λmax ∝ 1/T, the temperature ratio is the inverse of the ratio of the peak wavelengths.
The graph shows how the wavelength λmax at which a black body emits most intensely varies with 1/T, where T is the temperature of the black body. One point on the line is labelled. What is the gradient of the line in SI units?
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Gradient = λmax ÷ (1/T) = 580 × 10−9 m ÷ 2.00 × 10−4 K−1 = 2.9 × 10−3 m K: the line is Wien's law, λmax = (2.898 × 10−3 m K) × 1/T.
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Answer B
Answer: B
Common error: Ignoring the powers of ten in the axis labels. Here the wavelength is in nm (× 10−9) and 1/T is in units of 10−4 K−1; convert both before dividing.
A student measures the intensity I of the radiation from a small lamp at different distances d from it. The lamp radiates uniformly in all directions. The graph shows I against 1/d2. What is the total power radiated by the lamp?
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I = L/4πd2, so the gradient of I against 1/d2 is L/4π. Gradient = 10.0 ÷ 50 = 0.20 W, so L = 4π × 0.20 = 2.5 W.
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Answer D
Answer: D
Exam tip: Write the equation in the form y = mx before reading the gradient: I = (L/4π) × 1/d2, so the gradient is L/4π, not L.
The bar charts show the radius, in units of the Sun's radius R☉, and the surface temperature of four stars, W, X, Y and Z. Which star has the greatest luminosity?
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L = 4πr2σT4, so compare r2T4 (radius in R☉, T in 103 K): W: 402 × 3.04 = 1.3 × 105; X: 2.02 × 124 = 8.3 × 104; Y: 8.02 × 5.04 = 4.0 × 104; Z: 0.502 × 254 = 9.8 × 104. Star W is the most luminous: its huge surface area outweighs its low temperature.
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Answer A
Answer: A
Common error: Picking the hottest star. Luminosity depends on both factors: the radius squared and the temperature to the fourth power.
The graph shows how the intensity at the Earth of the radiation from a type Ia supernova varies with time. Type Ia supernovae all reach about the same peak luminosity, 2.0 × 1036 W. What is the distance from the Earth to the supernova?
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Peak intensity from the graph = 4.0 × 10−14 W m−2. d = √(L/4πI) = √(2.0 × 1036 ÷ (4π × 4.0 × 10−14)) = 2.0 × 1024 m.
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Answer A
Answer: A
Exam tip: A standard candle gives L; the measurement gives I. Use the peak value on both sides: the known luminosity is the peak luminosity, so read the peak intensity.
The Hertzsprung–Russell diagram shows the path of a star as it changes from A, on the main sequence, to B. Which statement about the star is correct?
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From A to B the luminosity stays at about 104L☉ while T falls from 20 000 K to 3500 K. With L constant, r ∝ 1/T2, so rB/rA = (20 000/3500)2 ≈ 33: the radius increases by a factor of about 30. The star is massive and becomes a red supergiant.
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Answer D
Answer: D
Exam tip: A horizontal move to the right on an HR diagram means constant luminosity and falling temperature, so the star must be growing: r ∝ 1/T2 at constant L.
A computer model calculates how the luminosity L of a star would change if its surface temperature T changed while its radius stayed the same. The graph shows the model's results as L against T4.
(a)
Explain why the graph is a straight line through the origin.
(1)
(b)
Determine the gradient of the graph.
(2)
(c)
Use your answer to (b) to calculate the radius of the model star.
(2)
(d)
Use the graph to find the surface temperature at which the luminosity of the model star is 2.8 × 1026 W.
(1)
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Part (a)
L = 4πr2σT4 with r (and σ) constant, so L ∝ T4
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Part (b)
use a large triangle, e.g. from (0, 0) to (1.6 × 1015 K4, 5.6 × 1026 W)
Figure 1(a) shows how the intensity at the Earth of the radiation from a Cepheid variable star varies with time. Figure 1(b) shows how the mean luminosity of Cepheid variables depends on their period, in units of the Sun's luminosity L☉ = 3.83 × 1026 W.
(a)
Use Figure 1(a) to determine the period of the star.
(2)
(b)
Use Figure 1(b) to determine the mean luminosity of the star in watts.
(2)
(c)
Calculate the distance from the Earth to the star.
(2)
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Part (a)
maxima at t = 1, 7, 13 and 19 days: three cycles in 18 days
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period = 6.0 days
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Part (b)
L = 2.5 × 103L☉ (allow 2.4–2.6 × 103)
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= 2.5 × 103 × 3.83 × 1026 = 9.6 × 1029 W
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Part (c)
mean I = 2.0 × 10−13 W m−2 from Figure 1(a); d = √(L/4πI) = √(9.6 × 1029 ÷ (4π × 2.0 × 10−13))
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= 6.2 × 1020 m — allow error carried forward
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Exam tip: Time several cycles and divide by the number of cycles: the period is then far more precise than one cycle read on its own.
A red dwarf star has a radius of 1.4 × 108 m and a surface temperature of 3200 K. The diagram shows the orbits of three of its planets, b, c and d.
(a)
Calculate the luminosity of the star.
(2)
(b)
Calculate the intensity of the star's radiation at planet c.
(2)
(c)
The intensity of the Sun's radiation at the Earth is 1.36 × 103 W m−2. Show that planet c is the only one of the three planets whose intensity is within a factor of 2 of this value.
(2)
(d)
Calculate the wavelength at which the star's radiation is most intense, and state the region of the electromagnetic spectrum in which it lies.
A star has a parallax angle of 0.125 arcseconds. The intensity of its radiation at the Earth is 3.0 × 10−8 W m−2 and its spectrum has a peak at a wavelength of 610 nm. 1 parsec = 3.09 × 1016 m. The radius of the Sun is 6.96 × 108 m.
The Sun has been on the main sequence for about 4.6 × 109 years and will remain there for several billion years more.
(a)
Describe how the Sun will change after it leaves the main sequence, referring to its position on a Hertzsprung–Russell diagram, and explain why a star of 20 solar masses would end its life differently.
(6)
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Indicative content:
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when the hydrogen in its core runs out, the core contracts and heats while the outer layers expand and cool
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the Sun becomes a red giant, moving up and to the right on the HR diagram (more luminous, cooler, much larger)
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its outer layers drift away as a planetary nebula, leaving the hot core
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the core becomes a white dwarf at the bottom left of the diagram, with no fusion, which cools and fades (moving down and to the right)
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a 20 solar mass star has enough mass to become a red supergiant and fuse heavier elements; when fusion in its core can no longer supply energy the core collapses and the star explodes as a supernova
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leaving a neutron star or, if the remaining core is more than about 3 solar masses, a black hole, rather than a white dwarf
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Model sketch
Figure: sketch of the Sun's path on a Hertzsprung–Russell diagram (dashed line: main sequence; not to scale)
Exam tip: For a life-cycle explanation, give the stages in order, say where each one is on the HR diagram, and state what decides the route: the mass of the star.
An astronomer estimates the radius of a star from the following measurements. peak wavelength of the spectrum: λmax = (520 ± 10) nm intensity at the Earth: I = (8.4 ± 0.3) × 10−9 W m−2 distance from the Earth: d = 1.9 × 1017 m, with a percentage uncertainty of 5%
Figure: spectrum of the star (not to scale)
(a)
Calculate the surface temperature of the star.
(1)
(b)
Calculate the percentage uncertainty in the temperature.
(1)
(c)
Calculate the luminosity of the star.
(2)
(d)
Calculate the percentage uncertainty in the luminosity.
(1)
(e)
Calculate the radius of the star.
(2)
(f)
Calculate the percentage uncertainty in the radius.
(1)
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Answer
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Notes
Part (a)
T = 2.898 × 10−3 ÷ 520 × 10−9 = 5570 K
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Part (b)
10 ÷ 520 × 100 = 1.9% (the same as the percentage uncertainty in λmax)
Models suggest that, as a red giant, the Sun will reach a surface temperature of about 3500 K and a luminosity of about 2000 L☉. Today the Sun's surface temperature is 5780 K and its radius is 6.96 × 108 m. 1 AU = 1.50 × 1011 m.
(a)
Calculate the radius of the Sun as a red giant, in metres.
(2)
(b)
Express this radius as a fraction of the radius of the Earth's orbit.
(1)
(c)
Later, the Sun's core becomes a white dwarf with a radius of 0.012 times the Sun's present radius and a surface temperature of 30 000 K. Calculate its luminosity in units of L☉.
(2)
(d)
Explain why the white dwarf moves down and to the right on a Hertzsprung–Russell diagram as time passes.
(2)
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Part (a)
r/r☉ = √2000 × (5780/3500)2 = 122
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r = 122 × 6.96 × 108 = 8.5 × 1010 m
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Part (b)
8.5 × 1010 ÷ 1.50 × 1011 = 0.57
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Part (c)
L/L☉ = (0.012)2 × (30 000/5780)4
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= 0.10
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Part (d)
there is no fusion, so it radiates away its stored thermal energy and its temperature falls (moves right)
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its radius stays (almost) constant, so L ∝ T4 falls too (moves down)
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Model sketch
Figure: the white dwarf moves down and to the right as it cools (dashed line: main sequence; not to scale)
Exam tip: In solar units, L/L☉ = (r/r☉)2(T/T☉)4: a quick way to compare any star with the Sun.
A space probe orbits the Sun in a circular orbit of radius 1.52 AU. It observes a star from positions P1 and P2 on opposite sides of its orbit and measures the parallax angle p shown in the diagram. 1 AU = 1.50 × 1011 m; 1 parsec = 3.09 × 1016 m.
(a)
Explain why the probe can measure the distances to stars further away than a telescope on the Earth with the same angular precision can.
(1)
(b)
Calculate the distance to the star in metres.
(3)
(c)
Show that this distance is about 40 pc.
(1)
(d)
The intensity of the star's radiation at the Earth is 2.6 × 10−10 W m−2. Calculate the luminosity of the star.
(2)
(e)
The luminosity of the Sun is 3.83 × 1026 W. Calculate the luminosity of the star as a multiple of the Sun's luminosity.
(1)
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Answer
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Notes
Part (a)
its baseline (orbit radius) is larger, so a star at a given distance gives a larger parallax angle, and the smallest measurable angle is reached at a greater distance
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Part (b)
radius of the probe's orbit = 1.52 × 1.50 × 1011 = 2.28 × 1011 m
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p = 0.0380/3600 degree = 1.84 × 10−7 rad
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d = 2.28 × 1011 ÷ tan p = 1.24 × 1018 m
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Part (c)
1.24 × 1018 ÷ 3.09 × 1016 = 40.1 pc ≈ 40 pc
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Part (d)
L = 4πd2I = 4π × (1.24 × 1018)2 × 2.6 × 10−10
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= 5.0 × 1027 W
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Part (e)
5.0 × 1027 ÷ 3.83 × 1026 = 13
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Common error: Using 1 AU as the baseline. Parallax measured from the probe uses the radius of the probe's own orbit, 1.52 AU, so d in parsecs = 1.52 ÷ p in arcseconds.
The graph shows how the luminosity L of main-sequence stars depends on their mass M. Both scales are logarithmic. P and Q are points on the line.
(a)
Show that the graph is consistent with L ∝ Mn where n is about 3.5.
(2)
(b)
The time a star spends on the main sequence is proportional to M/L. The Sun's main-sequence lifetime is 1.0 × 1010 years. Calculate the main-sequence lifetime of a star of mass 10 M☉.
(2)
(c)
Describe the stages in the life of the 10 M☉ star after it leaves the main sequence.
(3)
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Answer
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Part (a)
a straight line on log–log axes has gradient n = Δlog10L ÷ Δlog10M
The chart shows, on a logarithmic scale, the range of distances over which four methods are used to find distances to astronomical objects. 1 parsec = 3.09 × 1016 m; L☉ = 3.83 × 1026 W.
(a)
Use the chart to calculate the smallest parallax angle that the space telescope must be able to measure.
(2)
(b)
Explain why it is important that the range of distances for Cepheid variables overlaps the range for parallax.
(2)
(c)
A Cepheid at the greatest distance shown for this method has a mean luminosity of 3.0 × 104L☉. Calculate the mean intensity of its radiation at the Earth.
(2)
(d)
Suggest why type Ia supernovae can be used to find much greater distances than Cepheid variables.
(1)
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Answer
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Notes
Part (a)
greatest distance for space-telescope parallax = 104 pc
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p = 1/d = 1 ÷ 104 = 1.0 × 10−4 arcsecond
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Part (b)
the period–luminosity relationship is calibrated using Cepheids whose distances are known from parallax: L = 4πd2I
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then for more distant Cepheids the luminosity found from the period, with the measured intensity, gives the distance
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Part (c)
d = 4 × 107 × 3.09 × 1016 = 1.24 × 1024 m and L = 3.0 × 104 × 3.83 × 1026 = 1.15 × 1031 W
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I = L/4πd2 = 6.0 × 10−19 W m−2
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Part (d)
they are far more luminous, so their radiation is still intense enough to measure at much greater distances
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Exam tip: Each method is calibrated with the one that reaches less far: that is why the ranges on a distance ladder must overlap.
The diagram shows radiation from the Sun reaching the Earth. Model the Earth as a sphere of radius R = 6.37 × 106 m that reflects 30% of the radiation falling on it and radiates as a black body, at the same temperature over its whole surface.
(a)
Calculate the intensity of the Sun's radiation at the Earth.
(2)
(b)
Show that the power absorbed by the Earth is about 1.2 × 1017 W.
(2)
(c)
The Earth emits radiation at the same rate as it absorbs it. Calculate the Earth's surface temperature in this model.
(2)
(d)
Calculate the wavelength at which the Earth's emission is most intense, and state the region of the electromagnetic spectrum in which it lies.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
I = L/4πd2 = 3.83 × 1026 ÷ (4π × (1.50 × 1011)2)
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= 1.35 × 103 W m−2
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Part (b)
power intercepted = I × πR2 = 1355 × π × (6.37 × 106)2 = 1.73 × 1017 W
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power absorbed = 0.70 × 1.73 × 1017 = 1.21 × 1017 W
Common error: Using 4πR2 for the area that catches the Sun's radiation. The Earth intercepts the beam over its cross-section, πR2, but radiates from its whole surface, 4πR2.
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