IB · Physics SL & HL · E Nuclear & quantum physics

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.

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E.5 Fusion & stars practice questions

Question 1Multiple choice · 1 mark

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 stagefinal state
Ared giantwhite dwarf
Bred supergiantwhite dwarf
Cred supergiantneutron star or black hole
Dwhite dwarfblack 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?

  1. A 6.7 × 10−9 W m−2
  2. B 1.3 × 10−8 W m−2
  3. C 8.4 × 10−8 W m−2
  4. 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
PartAnswer and marking pointsMarks
(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
PartAnswer and marking pointsMarks
(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
PartAnswer and marking pointsMarks
(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
PartAnswer and marking pointsMarks
(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☉ ✓
2

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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