E.5 Fusion & stars: IB Physics questions with answers
Free E.5 Fusion & stars questions for IB Physics (SL & HL), part of E Nuclear & quantum physics. Each question has the answer and the marking points an examiner would look for.
A main-sequence star much more massive than the Sun reaches the end of its main-sequence life. Which row describes its next stage and final state?
next stage
final state
A
red giant
white dwarf
B
red supergiant
white dwarf
C
red supergiant
neutron star or black hole
D
white dwarf
black hole
Show answer
Answer: C
Question 2Multiple choice · 1 mark
A star of luminosity 7.6 × 1026 W is 9.5 × 1016 m from the Earth. What is its apparent brightness?
A 6.7 × 10−9 W m−2
B 1.3 × 10−8 W m−2
C 8.4 × 10−8 W m−2
D 6.4 × 108 W m−2
Show answer
Answer: A
Question 3Structured · 8 marksA* challenge
The Sun has luminosity 3.83 × 1026 W and surface temperature 5800 K. Star X has luminosity 40 times that of the Sun and surface temperature 10 000 K.
(a)
Calculate the radius of X in terms of the radius of the Sun.
[3]
(b)
Calculate the peak wavelength of the light from X.
[1]
(c)
X is 25 pc from Earth (1 pc = 3.09 × 1016 m). Calculate its apparent brightness.
[2]
(d)
Explain how X remains stable while on the main sequence.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
2.1 R☉
L ∝ R2T4✓ RX/R☉ = √40 × (5800/10 000)2✓ = 2.1 ✓
3
(b)
290 nm
λ = 2.9 × 10−3/10 000 = 2.9 × 10−7 m ✓
1
(c)
2.0 × 10−9 W m−2
b = L/4πd2✓ = 2.0 × 10−9 W m−2✓
2
(d)
pressure balances gravity
gravitational pressure acts inwards ✓ balanced by radiation and gas pressure from fusion outwards ✓
2
Question 4Structured · 5 marks
This question is about the evolution of stars.
(a)
State the main source of energy for a main-sequence star.
[1]
(b)
Describe the evolution of a star of one solar mass after it leaves the main sequence.
[3]
(c)
State what is meant by the Chandrasekhar limit.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
hydrogen fusion
fusion of hydrogen into helium in the core ✓
1
(b)
red giant → PN → white dwarf
becomes a red giant ✓ outer layers ejected as a planetary nebula ✓ core becomes a white dwarf ✓
3
(c)
1.4 M☉
the maximum mass of a white dwarf, about 1.4 solar masses ✓
1
Question 5Structured · 4 marks
A star has a surface temperature of 4000 K and a luminosity 100 times that of the Sun (Sun: 5800 K).
(a)
Calculate the peak wavelength of the star.
[1]
(b)
Calculate the ratio of the star’s radius to the Sun’s radius.
[2]
(c)
Identify the type of star.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
725 nm
λ = 2.9 × 10−3/4000 = 725 nm ✓
1
(b)
21
L ∝ R2T4✓ R/R☉ = 10 × (5800/4000)2 = 21 ✓
2
(c)
red giant
red giant ✓
1
Question 6Structured · 14 marksA* challenge
A star cluster has a parallax angle of 4.0 × 10−3 arc-seconds. The most luminous stars in the cluster that are still on the main sequence have an apparent brightness of 2.05 × 10−11 W m−2 and a peak emission wavelength of 290 nm. For main-sequence stars, L ∝ M3.5. The Sun’s luminosity is 3.83 × 1026 W and its main-sequence lifetime is 1.0 × 1010 years. 1 pc = 3.09 × 1016 m, Wien constant = 2.9 × 10−3 m K, σ = 5.67 × 10−8 W m−2 K−4.
(a)
Determine the luminosity of one of these stars in terms of the Sun’s luminosity.
[3]
(b)
Estimate the age of the cluster.
[4]
(c)
A red giant in the cluster has the same apparent brightness as these stars but a peak wavelength of 580 nm. Determine the radius of the red giant.
[3]
(d)
A student suggests that a white dwarf found in the cluster formed from a star of mass 1.0 solar masses. Explain why this is not possible.
[2]
(e)
Predict the luminosity of the most luminous main-sequence stars when the cluster is four times its present age.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
≈ 40 L☉
d = 1/p = 250 pc = 7.7 × 1018 m ✓ L = 4πd2b = 1.5 × 1028 W ✓ L ≈ 40L☉✓
3
(b)
≈ 7 × 10⁸ years
M = 401/3.5 = 2.9M☉✓ lifetime ∝ (fuel)/(rate of use) ∝ M/L ∝ M−2.5✓ turn-off stars are just ending their main-sequence life, so cluster age ≈ their lifetime ✓ age ≈ 1.0 × 1010 × 2.87−2.5 ≈ 7 × 108 years ✓
4
(c)
5.9 × 10⁹ m
same distance and brightness, so same luminosity ✓ T = 2.9 × 10−3/580 × 10−9 = 5000 K ✓ R = √(L/4πσT4) = 5.9 × 109 m ✓
3
(d)
1 M☉ star still on main sequence
a 1.0 M☉ star has a main-sequence lifetime of 1.0 × 1010 years ✓ this exceeds the cluster age, so it cannot yet have left the main sequence; the progenitor must have been more massive than the turn-off mass ✓
2
(e)
≈ 5.7 L☉
age ∝ M−2.5 and L ∝ M3.5, so turn-off L ∝ age−1.4✓ L = 40 × 4−1.4 ≈ 5.7L☉✓
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Original IB-style practice questions written for mentorriyaz.in. These are not IB past paper questions, and this site is not affiliated with or endorsed by the International Baccalaureate Organization. Updated 2026-11-04. Paper generator · mentorriyaz.in