Option: Astrophysics – A-level Physics exam questions with mark scheme

1Core3 marks

An astronomical refracting telescope in normal adjustment has an objective lens of focal length 1.20 m and an eyepiece lens of focal length 0.040 m.

(a) Calculate the angular magnification of the telescope. [1]

(b) State the distance between the two lenses. [1]

(c) The Moon subtends an angle of 9.0 × 10$^{-3}$ rad at the unaided eye. Calculate the angle subtended by the image of the Moon seen through the telescope. [1]

(a) $M = \frac{f_o}{f_e} = \frac{1.20}{0.040}$ = 30(1)
(b) $f_o + f_e$ = 1.24 m(1)
(c) $30 \times 9.0\times10^{-3}$ = 0.27 rad (ecf from (a))(1)
Answer: (a) 30 (b) 1.24 m (c) 0.27 rad
Examiner tip: Angular magnification has no unit, and normal adjustment means the lens separation is the sum of the focal lengths.

2Standard4 marks

A radio telescope has a dish of diameter 76 m and observes radiation of wavelength 0.21 m.

(a) Calculate the minimum angular resolution of the telescope. [1]

(b) Two radio sources have an angular separation of 0.10°. Determine whether the telescope can resolve them. [2]

(c) Charge-coupled devices (CCDs) are used to detect light in optical telescopes. State one advantage of a CCD over the human eye. [1]

(a) $\theta \approx \frac{\lambda}{D} = \frac{0.21}{76}$ = 2.8 × 10$^{-3}$ rad (allow 2.76 × 10$^{-3}$ rad)(1)
(b) $0.10^\circ = 1.7\times10^{-3}$ rad (or 2.8 × 10$^{-3}$ rad = 0.16°)(1)
(b) separation is smaller than the minimum resolvable angle, so the sources cannot be resolved(1)
(c) any one: higher quantum efficiency (about 80% compared with about 1% for the eye) allow detects a greater fraction of incident photons / better resolution (smaller pixels than receptor spacing in the eye) / can take long exposures (collects light over time) / detects a wider range of wavelengths / gives a digital record that can be stored, processed or shared(1)
Answer: (a) 2.8 × 10$^{-3}$ rad (b) 0.10° = 1.7 × 10$^{-3}$ rad, smaller than 2.8 × 10$^{-3}$ rad, so not resolved (c) e.g. higher quantum efficiency
Examiner tip: Convert degrees to radians before comparing — the Rayleigh criterion gives θ in radians.

3Standard4 marks

A star has an apparent magnitude of 3.5 and is 25 pc from the Earth.

(a) Calculate the absolute magnitude of the star. [2]

(b) A second star has the same absolute magnitude and is 250 pc from the Earth. Calculate its apparent magnitude and state whether it can be seen with the naked eye. [2]

(a) $m - M = 5\log\frac{d}{10}$, so $M = 3.5 - 5\log\frac{25}{10}$(1)
(a) $M$ = 1.5 (1.51)(1)
(b) $m = 1.51 + 5\log\frac{250}{10}$ = 8.5 allow 3.5 + 5 = 8.5 (distance 10 times greater so m larger by 5)(1)
(b) not visible: faintest naked-eye stars have magnitude about 6, and 8.5 is fainter (larger number)(1)
Answer: (a) 1.5 (b) 8.5, not visible to the naked eye
Examiner tip: A larger magnitude means a fainter star — don't say 8.5 is 'brighter'.

4Hard5 marks

The spectrum of a star has a peak intensity at a wavelength of 350 nm. The star's luminosity is 40 times that of the Sun. The Sun's surface temperature is 5800 K. Assume that both the star and the Sun behave as black bodies.

(a) Calculate the surface temperature of the star. [1]

(b) Calculate the ratio $\frac{\text{radius of star}}{\text{radius of Sun}}$. [3]

(a) $T = \frac{2.9\times10^{-3}}{350\times10^{-9}}$ = 8.3 × 10$^3$ K accept 8286 K / 8.29 × 10$^3$ K(1)
(b) uses $P = \sigma A T^4$ with $A = 4\pi r^2$, so $L \propto r^2T^4$(1)
(b) $\frac{r_s^2}{r_\odot^2} = 40 \times \left(\frac{5800}{8286}\right)^4$ = 9.6 ecf from (a)(1)
(b) ratio = 3.1 (3.10) allow 3.09 (from T = 8300 K); ecf from (a)(1)
(c) A (ecf from (a): class A is about 7500–11 000 K)(1)
Answer: (a) 8.3 × 10$^3$ K (b) 3.1 (c) A
Examiner tip: In ratio questions, cancel σ and 4π first — it avoids huge numbers and calculator errors.

5Hard4 marks

The hydrogen-alpha line has a wavelength of 656.3 nm when measured in a laboratory. In the spectrum of a distant galaxy the same line is observed at 682.5 nm.

(a) Calculate the recessional speed of the galaxy. [2]

(b) Calculate the distance to the galaxy in Mpc. Take the Hubble constant to be 65 km s$^{-1}$ Mpc$^{-1}$. [2]

(a) $z = \frac{\Delta\lambda}{\lambda} = \frac{682.5 - 656.3}{656.3}$ = 0.0399(1)
(a) $v = zc$ = 1.2 × 10$^7$ m s$^{-1}$ (1.20 × 10$^4$ km s$^{-1}$) allow 1.19–1.20 × 10$^7$; ignore sign(1)
(b) $d = \frac{v}{H_0}$ with $v$ in km s$^{-1}$: $\frac{1.20\times10^4}{65}$ (or equivalent working in SI units, converting back to Mpc)(1)
$d$ = 1.8 × 10$^2$ Mpc (184 Mpc; allow 180–185)(1)
Answer: (a) 1.2 × 10$^7$ m s$^{-1}$ (b) 1.8 × 10$^2$ Mpc (184 Mpc)
Examiner tip: Keep units consistent with H₀: put v in km s⁻¹ to get d directly in Mpc.

6Hard6 marks

Describe the life cycle of a star with a mass similar to that of the Sun, from its formation to its final state. Your answer should explain what happens in the star at each stage. [6]

a cloud of dust and gas (nebula) collapses under gravity, forming a protostar; loss of gravitational PE heats it(1)
when the core is hot and dense enough, hydrogen nuclei fuse to helium(1)
main sequence: stable because outward radiation (gas) pressure balances inward gravitational force; star stays here for most of its life(1)
hydrogen in the core runs out; core contracts and heats; hydrogen fusion continues in a shell around the core(1)
outer layers expand and cool: red giant(1)
core becomes hot enough for helium to fuse into carbon (and oxygen)(1)
outer layers are ejected as a planetary nebula(1)
the hot, dense core remains as a white dwarf; no fusion, it cools slowly (electron degeneracy pressure is extra detail, not required)(1)
on the HR diagram: forms and settles on the main sequence, moves up and to the right to the giant region, then down and to the left to the white dwarf region(1)

Level	Marks	What it looks like
Level 3	5–6	A full, logically sequenced account from formation through main sequence, red giant, planetary nebula and white dwarf, giving correct reasons for each stage (what is fusing, balance of pressure and gravity, core contraction, loss of outer layers).
Level 2	3–4	Most stages described in the correct order with some explanation, or all stages named with limited explanation.
Level 1	1–2	Some stages named (e.g. main sequence, red giant, white dwarf) with little or no explanation; order may be partly wrong.
Level 0	0	No relevant content.

Answer: nebula → protostar → main sequence → red giant → planetary nebula + white dwarf
Examiner tip: For the top level, explain why each stage happens (balance of pressure and gravity, what is fusing) — a list of stage names alone stays in Level 1.

7Stretch7 marks

(a) Using a Hubble constant of 65 km s$^{-1}$ Mpc$^{-1}$, estimate the age of the Universe in years. 1 Mpc = 3.08 × 10$^{22}$ m; 1 year = 3.15 × 10$^7$ s. [3]

(b) Observations of type Ia supernovae suggest that the expansion of the Universe is accelerating. Explain how these observations lead to this conclusion. [2]

(c) A massive star collapses to form a black hole of mass 10 solar masses. Calculate the Schwarzschild radius of the black hole. Mass of Sun = 1.99 × 10$^{30}$ kg. [2]

(a) $H_0 = \frac{65\times10^3}{3.08\times10^{22}}$ = 2.1 × 10$^{-18}$ s$^{-1}$(1)
(a) age $\approx \frac{1}{H_0}$ = 4.7 × 10$^{17}$ s(1)
(a) $\frac{4.74\times10^{17}}{3.15\times10^7}$ = 1.5 × 10$^{10}$ years(1)
(b) type Ia supernovae have a known (peak) absolute magnitude/luminosity, so act as standard candles, giving their distance(1)
(b) distant ones were dimmer than expected, so further away than predicted from their redshift (for constant expansion), so expansion is accelerating allow reference to dark energy as the suggested cause(1)
(c) $R_s = \frac{2GM}{c^2} = \frac{2\times6.67\times10^{-11}\times10\times1.99\times10^{30}}{(3.00\times10^8)^2}$(1)
(c) $R_s$ = 3.0 × 10$^4$ m (29.5 km) allow 2.9–3.0 × 10$^4$ m(1)
Answer: (a) 1.5 × 10$^{10}$ years (b) type Ia supernovae are standard candles; distant ones are dimmer, so further away, than expected for their redshift (c) 3.0 × 10$^4$ m
Examiner tip: To get 1/H₀ in seconds, both distance units must be metres: convert km to m and Mpc to m.

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