Astronomy · CIE A2 Physics 9702 · 2025

10(b)(i) · Radius of the Sun (3 marks)

Given: Period of rotation \(T = 2.07 \times 10^{6}\,\text{s}\). The observed surface speed (from Doppler) is about \(v \approx 2.10 \times 10^{3}\,\text{m s}^{-1}\).

Using circular motion: \( v = \frac{2\pi R}{T} \quad \Rightarrow \quad R = \frac{vT}{2\pi} \) Substituting: \( R = \frac{(2.10 \times 10^{3})(2.07 \times 10^{6})}{2\pi} \) \( R \approx 6.93 \times 10^{8}\,\text{m} \)

10(b)(ii) · Wavelength shift (2 marks)

At position Z (moving away from us due to rotation), the observed wavelength is **greater** than the emitted wavelength. Reason: Doppler effect shifts the wavelength to the **red** side of the spectrum. Examiner phrasing: “Observed λ is longer than emitted λ (redshift).”

10(b)(iii) · Surface temperature of the Sun (2 marks)

Use luminosity \(L = 3.8 \times 10^{26}\,\text{W}\). Stefan–Boltzmann law: \( L = 4 \pi R^2 \sigma T^4 \) Rearranging: \( T = \left(\frac{L}{4 \pi R^2 \sigma}\right)^{1/4} \) Substituting: \( T = \left(\frac{3.8 \times 10^{26}}{4\pi (6.93 \times 10^{8})^2 (5.67 \times 10^{-8})}\right)^{1/4} \) \( T \approx 5.8 \times 10^{3}\,\text{K} \)

10(a) · Statement (2 Marks)

Hubble’s law: the recession speed of galaxies is proportional to their distance from the observer (i.e. speed ∝ distance). Accept phrasing: “v ∝ d” or “v = H₀d”.

10(b)(i) · Recession & observed wavelength (2 Marks)

Observational statement: the galaxy is receding from the Earth. The observed wavelength is redshifted: observed \( \lambda \) is greater than the emitted \( \lambda \) (observed > emitted).

10(b)(ii) · Calculate \(v\) from spectral lines (2 Marks)

Given (from paper): \( \lambda_{\text{obs}} = 4.91\ \text{nm},\ \lambda_{\text{em}} = 4.62\ \text{nm}.\)

Use fractional shift: \( \frac{\Delta\lambda}{\lambda}=\frac{\lambda_{\text{obs}}-\lambda_{\text{em}}}{\lambda_{\text{em}}}=\frac{4.91-4.62}{4.62}\approx 0.0628. \) For small \(z\), \( v = \dfrac{\Delta\lambda}{\lambda}\,c \): \( v \approx 0.0628\times(3.00\times10^{8}\,\text{m s}^{-1}) \approx 1.9\times10^{7}\ \text{m s}^{-1}. \) Boxed answer: \( \displaystyle v \approx 1.9\times10^{7}\ \text{m s}^{-1} \)

10(b)(iii) · Wavelength → temperature comment (B1 + B1)

Wien’s law: \( \lambda_{\max}\propto 1/T \). If the observed wavelength is larger than expected (i.e. redder), the temperature inferred from that peak would be lower. Short statement: “Observed λ too large ⇒ inferred T too low.”

10(c) · Hubble distance (C1, A1)

Rearrange Hubble’s law: \( d = \dfrac{v}{H_0} \). Use \( H_0 = 2.3\times10^{-18}\ \mathrm{s}^{-1} \) (paper value).

\( d = \frac{1.9\times10^{7}}{2.3\times10^{-18}} \approx 8.3\times10^{24}\ \text{m}. \) Boxed answer: \( \displaystyle d \approx 8.3\times10^{24}\ \text{m} \)

10(a) · Statement (2 Marks)

Speed ∝ distance ~ Hubble’s law: galaxies recede from the observer and recession speed increases with distance. (Accept: "Hubble’s law" / "cosmological expansion" wording.)

10(b)(i) · Recession & observed wavelength (2 Marks)

Observational statement: the galaxy is receding from the Earth; the observed wavelength is redshifted (observed > emitted) due to Doppler / cosmological redshift.

10(b)(ii) · Calculate v from spectral lines (2 Marks)

Given wavelengths from the paper: observed \( \lambda_{\text{obs}} = 4.91\,\text{nm} \), emitted \( \lambda_{\text{em}} = 4.62\,\text{nm} \).

Use \( \dfrac{\Delta\lambda}{\lambda} = \dfrac{v}{c} \Rightarrow v = \dfrac{\lambda_{\text{obs}} - \lambda_{\text{em}}}{\lambda_{\text{em}}}\,c \).

\( v = \frac{4.91 - 4.62}{4.62}\times(3.00\times10^{8}\,\text{m s}^{-1}) \approx 1.9\times10^{7}\,\text{m s}^{-1}. \)

10(b)(iii) · Wavelength / temperature comment (2 Marks)

By Wien’s law \( \lambda_{\max}\propto 1/T \). If the observed \( \lambda_{\max} \) is larger than expected, the temperature inferred would be lower. (Short explanation: observed wavelength too high ⇒ inferred T too low.)

10(c) · Hubble distance (2 Mark)

Use Hubble’s law \( v = H_0 d \). Use the given \( H_0 = 2.3\times10^{-18}\,\text{s}^{-1}\). Rearranging:

\( d = \frac{v}{H_0} = \frac{1.9\times10^{7}}{2.3\times10^{-18}} \approx 8.3\times10^{24}\,\text{m}. \)

Box final answers:

• \( v \approx 1.9\times10^{7}\,\text{m s}^{-1} \)
• \( d \approx 8.3\times10^{24}\,\text{m} \)

11(a)(i) · Name of process (1 mark)

Nuclear Fusion ~ the combining of light nuclei (deuterium) to form heavier nuclei (helium-4) with release of energy.NB: Carefully analyse what's given and understand the difference; Fission is the splitting of a large nucleus (like uranium-235) into two smaller nuclei, plus neutrons and energy

11(a)(ii) · Energy released per helium nucleus (3 marks)

Mass defect data: • Deuterium: Δm = 0.002388 u • Helium-4: Δm = 0.030377 u

For 3 deuterons: total = \(3 \times 0.002388 = 0.007164\,u\). Mass defect difference: \(\Delta\ m = 0.030377 - 0.007164 = 0.023213,u\)

Convert to energy: \( E = \Delta m \, c^2 = (0.023213)(931.5\,\text{MeV}) \approx 21.6\,\text{MeV} \) In joules: \( E = 21.6 \times 1.6\times10^{-13} \approx 3.47\times10^{-12}\,\text{J} \)

11(b)(i) · Luminosity of the star (3 marks)

Mass of one He-4 nucleus: \(4.00\,u \approx 6.64 \times 10^{-27}\,\text{kg}\). Mass production rate: \(7.34\times10^{16}\,\text{kg s}^{-1}\).

Number of He-4 nuclei produced per second: \( N = \frac{7.34\times10^{16}}{6.64\times10^{-27}} \approx 1.11\times10^{43} \) Energy per He-4 = \(3.47\times10^{-12}\,\text{J}\). Total power (luminosity): \( L = N \times E = (1.11\times10^{43})(3.47\times10^{-12}) \approx 3.85\times10^{31}\,\text{W} \)

11(b)(ii) · Surface temperature of the star (2 marks)

Stefan–Boltzmann law: \( L = 4\pi R^2 \sigma T^4 \) Rearranging: \( T = \left(\frac{L}{4\pi R^2 \sigma}\right)^{1/4} \)

Substituting: \( T = \left(\frac{3.85\times10^{31}}{4\pi(6.96\times10^8)^2(5.67\times10^{-8})}\right)^{1/4} \) \( T \approx 1.2\times10^{7}\,\text{K} \)