A Level Physics 25.2 Stellar Radii MCQs – 30 Practice Questions with Answers (Cambridge 9702)
This page gives you 30 original practice MCQs on Section 25.2, Stellar Radii, from the Cambridge AS & A Level Physics 9702 syllabus. This is an A Level (A2) topic: you will model stars as black bodies, use Wien’s displacement law and the Stefan-Boltzmann law, and combine them to calculate the radius of a star from its luminosity and surface temperature.
Formula recap
- Wien’s displacement law: λmaxT = 2.9 × 10−3 m K, where λmax is the wavelength at which the star’s radiation is most intense.
- Stefan-Boltzmann law: L = 4πr2σT4, where L is the luminosity (total power radiated), r is the star’s radius and σ = 5.67 × 10−8 W m−2 K−4.
- The quantity σT4 is the power radiated per unit area of the star’s surface.
- Rearranged for radius: r = √(L / (4πσT4)).
- Data for the Sun: L = 3.85 × 1026 W, T = 5800 K, R = 7.0 × 108 m.
See the full A Level Physics Formula Sheet for every equation you need for Cambridge 9702.
Black-Body Radiation & Wien’s Law
Q1. A black body is best described as an object that
- (A) reflects all the radiation incident on it
- (B) absorbs all the radiation incident on it and is a perfect emitter of radiation
- (C) emits radiation at only one wavelength
- (D) stays at a temperature of absolute zero
Show Answer
Answer: B – A black body absorbs every wavelength incident on it and, when hot, emits a continuous spectrum that depends only on its temperature. No real object is perfect, but a star is a very good approximation.
Q2. Stars are modelled as black bodies mainly because
- (A) they look perfectly black in the night sky
- (B) they absorb all the light that reaches them, so none is seen
- (C) their emitted spectrum is approximately the continuous black-body spectrum, whose shape depends only on surface temperature
- (D) they emit no radiation at visible wavelengths
Show Answer
Answer: C – The intensity-wavelength curve of a star matches the black-body curve for its surface temperature. This is what allows the temperature, and then the radius, to be found from the observed spectrum.
Q3. The radiation from a star is most intense at a wavelength of 500 nm. Use λmaxT = 2.9 × 10−3 m K to find the star’s surface temperature.
- (A) 5800 K
- (B) 1450 K
- (C) 11 600 K
- (D) 290 K
Show Answer
Answer: A – T = 2.9 × 10−3 / λmax = 2.9 × 10−3 / (500 × 10−9) = 2.9 × 10−3 / 5.00 × 10−7 = 5800 K, the same as the Sun’s surface temperature.
Q4. A hot star’s emission peaks at a wavelength of 290 nm. Its surface temperature is
- (A) 1000 K
- (B) 2900 K
- (C) 5800 K
- (D) 10 000 K
Show Answer
Answer: D – T = 2.9 × 10−3 / (290 × 10−9) = 2.9 × 10−3 / 2.9 × 10−7 = 10 000 K. The peak lies in the ultraviolet, so the star appears blue-white.
Q5. A cool red star emits most intensely at a wavelength of 1.0 μm. Its surface temperature is approximately
- (A) 290 K
- (B) 2900 K
- (C) 5800 K
- (D) 29 000 K
Show Answer
Answer: B – T = 2.9 × 10−3 / (1.0 × 10−6) = 2900 K. The peak is in the infrared, and the visible light it does emit is mostly at the red end, so the star looks red.
Q6. Four stars appear blue-white, yellow, orange and red. Which has the highest surface temperature?
- (A) The blue-white star
- (B) The yellow star
- (C) The orange star
- (D) The red star
Show Answer
Answer: A – By Wien’s law, a hotter black body peaks at a shorter wavelength. Blue-white light peaks at the shortest wavelength of the four, so the blue-white star is hottest. The order from hottest to coolest is blue-white, yellow, orange, red.
Q7. Star A peaks at 400 nm and star B peaks at 800 nm. Which statement about their surface temperatures is correct?
- (A) B is twice as hot as A
- (B) A and B have the same temperature
- (C) A is twice as hot as B
- (D) A is half as hot as B
Show Answer
Answer: C – T is inversely proportional to λmax. TA = 2.9 × 10−3 / 4.0 × 10−7 = 7250 K and TB = 2.9 × 10−3 / 8.0 × 10−7 = 3625 K, so A is twice as hot as B.
Q8. A star has a surface temperature of 7250 K. At what wavelength is its radiation most intense?
- (A) 200 nm
- (B) 290 nm
- (C) 580 nm
- (D) 400 nm
Show Answer
Answer: D – λmax = 2.9 × 10−3 / T = 2.9 × 10−3 / 7250 = 4.0 × 10−7 m = 400 nm, at the violet edge of the visible spectrum.
Q9. The surface temperature of a star doubles. What happens to the wavelength at which its radiation is most intense?
- (A) It doubles
- (B) It halves
- (C) It quarters
- (D) It stays the same
Show Answer
Answer: B – λmaxT is constant, so λmax is inversely proportional to T. Doubling T halves λmax, shifting the peak towards the blue/ultraviolet end.
Q10. As the temperature of a black body increases, its emission spectrum changes so that
- (A) the total power radiated increases and the peak moves to a shorter wavelength
- (B) the total power radiated increases and the peak moves to a longer wavelength
- (C) the total power radiated decreases and the peak moves to a shorter wavelength
- (D) the peak moves to a longer wavelength but the total power stays the same
Show Answer
Answer: A – Wien’s law moves the peak to shorter wavelengths, and the Stefan-Boltzmann law (next section) shows the total power rises as T4. A hotter star is both bluer and much more luminous at the same size.
The Stefan-Boltzmann Law
Q11. In the Stefan-Boltzmann law, L = 4πr2σT4, the factor 4πr2 represents
- (A) the volume of the star
- (B) the cross-sectional area of the star
- (C) the total surface area of the star
- (D) the diameter of the star
Show Answer
Answer: C – 4πr2 is the surface area of a sphere. The luminosity is this area multiplied by the power emitted per unit area, σT4.
Q12. A star’s surface temperature doubles while its radius stays the same. Its luminosity increases by a factor of
- (A) 2
- (B) 4
- (C) 8
- (D) 16
Show Answer
Answer: D – L ∝ T4 at fixed radius, so doubling T multiplies L by 24 = 16. Temperature has a far stronger effect than size.
Q13. A star’s radius doubles while its surface temperature stays the same. Its luminosity increases by a factor of
- (A) 4
- (B) 2
- (C) 8
- (D) 16
Show Answer
Answer: A – L ∝ r2 at fixed temperature, so doubling r multiplies L by 22 = 4.
Q14. A star has twice the radius of the Sun but half the Sun’s surface temperature. Its luminosity, compared with the Sun’s, is
- (A) 4 times greater
- (B) one quarter as great
- (C) the same
- (D) 16 times greater
Show Answer
Answer: B – L/LSun = (r/rSun)2 × (T/TSun)4 = 22 × (1/2)4 = 4 × (1/16) = 1/4. The T4 effect wins over the area effect.
Q15. In the Stefan-Boltzmann law, the quantity σT4 represents
- (A) the total power radiated by the whole star
- (B) the power radiated per unit volume of the star
- (C) the power radiated per unit area of the star’s surface
- (D) the energy carried by one photon
Show Answer
Answer: C – For the Sun, σT4 = 5.67 × 10−8 × 58004 = 6.4 × 107 W m−2: every square metre of the Sun’s surface radiates about 64 MW. Multiplying by the full surface area 4πr2 gives the luminosity.
Q16. A star’s surface temperature triples while its radius is unchanged. Its luminosity increases by a factor of
- (A) 81
- (B) 27
- (C) 9
- (D) 3
Show Answer
Answer: A – L ∝ T4, so the factor is 34 = 81.
Q17. A star has the same radius as the Sun (7.0 × 108 m) but a surface temperature of 2900 K, half the Sun’s. Taking the Sun’s luminosity as 3.85 × 1026 W, the star’s luminosity is
- (A) 3.85 × 1026 W
- (B) 9.6 × 1025 W
- (C) 4.8 × 1025 W
- (D) 2.4 × 1025 W
Show Answer
Answer: D – At fixed radius, L ∝ T4. Halving T divides L by 24 = 16, giving 3.85 × 1026 / 16 = 2.4 × 1025 W.
Q18. A star has three times the radius of the Sun and the same surface temperature, 5800 K. Its luminosity is approximately (Sun: L = 3.85 × 1026 W)
- (A) 1.2 × 1027 W
- (B) 3.5 × 1027 W
- (C) 1.0 × 1028 W
- (D) 3.1 × 1028 W
Show Answer
Answer: B – At fixed temperature, L ∝ r2, so L = 32 × 3.85 × 1026 = 9 × 3.85 × 1026 = 3.5 × 1027 W.
Q19. Two stars have the same surface temperature. Star X has radius R and star Y has radius 2R. The ratio luminosity of X : luminosity of Y is
- (A) 1 : 1
- (B) 1 : 2
- (C) 1 : 4
- (D) 1 : 16
Show Answer
Answer: C – At the same temperature, L ∝ r2. LX/LY = (R/2R)2 = 1/4, so the ratio is 1 : 4.
Q20. A star contracts so that its surface temperature doubles, but its luminosity stays the same. Its new radius is
- (A) one quarter of the original radius
- (B) one half of the original radius
- (C) the same as the original radius
- (D) four times the original radius
Show Answer
Answer: A – With L fixed, r = √(L / (4πσT4)) so r ∝ 1/T2. Doubling T multiplies T4 by 16, so r2 falls by 16 and r falls by a factor of 4.
Calculating Stellar Radii
Q21. A star has a luminosity of 1.0 × 1028 W and a surface temperature of 10 000 K. Its radius is approximately
- (A) 1.2 × 108 m
- (B) 1.2 × 1010 m
- (C) 1.4 × 1018 m
- (D) 1.2 × 109 m
Show Answer
Answer: D – r = √(L / (4πσT4)). First, 4πσT4 = 4π × 5.67 × 10−8 × (10 000)4 = 7.1 × 109. Then r2 = 1.0 × 1028 / 7.1 × 109 = 1.4 × 1018 m2, so r = √(1.4 × 1018) = 1.2 × 109 m. (Option C is r2 before taking the square root – a common slip.)
Q22. The spectrum of a star peaks at 290 nm, and its luminosity is 4.0 × 1028 W. The radius of the star is approximately
- (A) 1.2 × 109 m
- (B) 2.4 × 109 m
- (C) 4.7 × 109 m
- (D) 5.6 × 1018 m
Show Answer
Answer: B – Step 1 (Wien): T = 2.9 × 10−3 / 2.9 × 10−7 = 10 000 K. Step 2 (Stefan-Boltzmann): 4πσT4 = 7.1 × 109, so r2 = 4.0 × 1028 / 7.1 × 109 = 5.6 × 1018 m2 and r = 2.4 × 109 m, about 3.4 times the Sun’s radius.
Q23. A star has the same luminosity as the Sun but a surface temperature of 2900 K, half the Sun’s 5800 K. Its radius, in solar radii, is
- (A) 4
- (B) 2
- (C) 16
- (D) 0.25
Show Answer
Answer: A – From L = 4πr2σT4, r ∝ √L / T2. With L unchanged and T halved, r changes by 1 / (1/2)2 = 4, so the radius is 4 solar radii (about 2.8 × 109 m). A cool star must be big to radiate the same total power.
Q24. A star has a luminosity of 4 times the Sun’s and the same surface temperature as the Sun, 5800 K. Its radius is
- (A) 7.0 × 108 m
- (B) 2.8 × 109 m
- (C) 1.4 × 109 m
- (D) 5.6 × 109 m
Show Answer
Answer: C – At the same temperature, r ∝ √L, so r = √4 × RSun = 2 × 7.0 × 108 = 1.4 × 109 m.
Q25. A red giant has a cool surface, around 3000 K, yet its luminosity is thousands of times greater than the Sun’s. This is mainly because
- (A) its core is much hotter than the Sun’s core
- (B) its radius is enormous, so its emitting surface area is huge
- (C) it is much closer to the Earth than the Sun
- (D) it reflects light from neighbouring stars
Show Answer
Answer: B – Each square metre of a red giant radiates far less than the Sun’s surface (low T, so small σT4), but the star may be hundreds of times wider, and area grows as r2. The huge area more than compensates, giving a very large total luminosity.
Q26. A white dwarf has a very high surface temperature, around 11 600 K, yet a luminosity well below the Sun’s. This is because
- (A) it is much further away than most stars
- (B) it absorbs most of the radiation it produces
- (C) its surface temperature is actually very low
- (D) its radius is extremely small, so its emitting surface area is tiny
Show Answer
Answer: D – A white dwarf is about the size of the Earth, roughly 0.01-0.03 solar radii. Even though σT4 is large (T is twice the Sun’s, so 16 times the power per unit area), the area 4πr2 is so small that the total luminosity stays low.
Q27. A star has twice the surface temperature of the Sun and 16 times the Sun’s luminosity. Its radius compared with the Sun’s radius is
- (A) the same
- (B) 4 times greater
- (C) 2 times greater
- (D) one quarter as great
Show Answer
Answer: A – r/rSun = √(L/LSun) / (T/TSun)2 = √16 / 22 = 4 / 4 = 1. The star is the same size as the Sun; its extra luminosity comes entirely from its higher temperature.
Q28. A star’s radiation peaks at 580 nm and its luminosity is 1.0 × 1026 W. Its radius is approximately
- (A) 4.7 × 107 m
- (B) 2.2 × 1017 m
- (C) 4.7 × 108 m
- (D) 4.7 × 109 m
Show Answer
Answer: C – Step 1 (Wien): T = 2.9 × 10−3 / 5.8 × 10−7 = 5000 K. Step 2: 4πσT4 = 4π × 5.67 × 10−8 × 50004 = 4.5 × 108. Then r2 = 1.0 × 1026 / 4.5 × 108 = 2.2 × 1017 m2, so r = 4.7 × 108 m, slightly smaller than the Sun.
Q29. A star has the same surface temperature as the Sun but a luminosity 100 times greater. Its radius, in solar radii, is
- (A) 100
- (B) 10
- (C) 10 000
- (D) 5
Show Answer
Answer: B – At fixed temperature, r ∝ √L, so r = √100 = 10 solar radii.
Q30. To determine the radius of a distant star, astronomers combine which two measured quantities using the Stefan-Boltzmann law?
- (A) The star’s colour and its distance from Earth
- (B) The star’s apparent brightness and its mass
- (C) The star’s mass and its surface temperature
- (D) The star’s luminosity (from its brightness and distance) and its surface temperature (from the peak wavelength of its spectrum)
Show Answer
Answer: D – The peak wavelength gives T through Wien’s law, and the brightness together with the distance gives L. Then r = √(L / (4πσT4)) gives the radius – the method used throughout this section.
Keep practising Physics
Browse hundreds more original practice questions on the Physics MCQs hub, and keep every equation for this topic in one place with the A Level Physics Formula Sheet.
These are original practice questions written for Quaidianpoint.com. They are not Cambridge past-paper questions, and this page is not affiliated with or endorsed by Cambridge Assessment International Education.