Astronomy and cosmology

12(a) total power of radiation emitted (by the star) B1 12(b) L C1 F = 4πd2 3.83×1026 = 4×π×1.51×10112 = 1340 W m −2 A1 © UCLES 2022 Page 16 of 17 12(c) E A1 m= c2 3.83×1026 = 3.00×1082 = 4.26×109 kg 12(d) L = 4πσr2T 4 B1 3.83×1026 = 4×π×5.67×10 −8 ×6.96×1082×T4 leading to T = 5770 K 12(e) 1 C1 λ ∝ (max) T 5.00×10 −7 9940 = λ 5770 λ =2.90 × 10 −7m A1 © UCLES 2022 Page 17 of 17

10(a) wavelength of maximum intensity is inversely proportional to (thermodynamic) temperature B1 10(b)(i)  = 0.50 m for A and 0.65 m for B C1 MAX T = 5800  (0.50 / 0.65) A1 = 4500 K 10(b)(ii) (star B has) greater peak / average wavelength B1 (star B looks) redder B1 10(c)(i) apparent wavelength is greater B1 or wavelength is greater than known value (due to) movement of star away (from observer) B1 10(c)(ii) by examining the (lines in the) spectrum (of light from the star) B1 and comparing with known spectrum B1 © UCLES 2022 Page 16 of 16

9(a)(i) speed is (directly) proportional to distance M1 where speed is speed of recession of galaxy (from observer) and distance is distance of galaxy away from observer A1 9(a)(ii) wavelengths (of spectral lines) are greater (than their known values) B1 redshift shows stars (in distant galaxies) moving away from Earth B1 9(b) (all) parts of Universe moving away from each other B1 more distant objects are moving away faster B1 matter must have been close together / very dense in the past B1 © UCLES 2022 Page 15 of 16

10(a) wavelength of maximum intensity is inversely proportional to (thermodynamic) temperature B1 10(b)(i)  = 0.50 m for A and 0.65 m for B C1 MAX T = 5800  (0.50 / 0.65) A1 = 4500 K 10(b)(ii) (star B has) greater peak / average wavelength B1 (star B looks) redder B1 10(c)(i) apparent wavelength is greater B1 or wavelength is greater than known value (due to) movement of star away (from observer) B1 10(c)(ii) by examining the (lines in the) spectrum (of light from the star) B1 and comparing with known spectrum B1 © UCLES 2022 Page 16 of 16

9(a) total power of radiation emitted (by the star) B1 9(b)(i) F = L / (4d 2) C1 = 9.86  1027 / [4  (8.14  1016)2] A1 = 1.18  10–7 W m–2 9(b)(ii) L = 4 r 2T4 C1 9.86  1027 = 4    5.67  10–8  r 2  98304 radius = 1.22  109 m A1 9(c) wavelength of peak intensity determined (from spectrum of star) B1 wavelength of peak intensity from object of known temperature determined B1 Wien’s displacement law used B1 or wavelength of peak intensity inversely proportional to temperature © UCLES 2022 Page 14 of 15

9(a)(i) • energy of photon has a corresponding frequency B3 • change in electron energy level emits a single photon • photon energy = difference in energy levels • discrete frequencies must have come from discrete energy gaps • discrete energy changes imply discrete energy levels Any three points, 1 mark each 9(a)(ii) transition (to – 3.400 eV) from X corresponds to 658 nm line C1 E – E = hc /  C1 1 2 E – (– 3.400) = (6.63  10–34  3.00  108) / (658  10–9  1.60  10–19) A1 1 and so E = –1.51 eV (full substitution and answer needed) 1 9(b)(i) redshift B1 9(b)(ii) moving away (from observer) B1 9(b)(iii)  /  = v / c C1 e.g. for 658 nm line:  = 686 – 658 ( = 28 nm) (other lines may be used) 28 / 658 = v / (3.00  108) (other lines may be used) C1 v = 1.3  107 m s–1 A1 9(c) v = H d C1 0 H = (1.3  107) / (5.7  1024) A1 0 = 2.3  10–18 s–1 © UCLES 2022 Page 15 of 16

9(a) total power of radiation emitted (by the star) B1 9(b)(i) F = L / (4d 2) C1 = 9.86  1027 / [4  (8.14  1016)2] A1 = 1.18  10–7 W m–2 9(b)(ii) L = 4 r 2T4 C1 9.86  1027 = 4    5.67  10–8  r 2  98304 radius = 1.22  109 m A1 9(c) wavelength of peak intensity determined (from spectrum of star) B1 wavelength of peak intensity from object of known temperature determined B1 Wien’s displacement law used B1 or wavelength of peak intensity inversely proportional to temperature © UCLES 2022 Page 14 of 15

10(a) brighter star could be closer (to Earth) B1 brighter star could have a greater luminosity (in the visible wavelengths) B1 10(b) object with known luminosity B1 10(c)(i) 660.9−656.3 v B1  leading to 2.1 106 m s–1 656.3 3.0108 10(c)(ii) v = H d C1 o d = 2.1  106 / 2.3  10–18 A1 = 9.1  1023 m 10(c)(iii) wavelength has increased / light is redshifted B1 star within galaxy is moving away / receding (from Earth) B1 Universe is expanding B1 © UCLES 2023 Page 18 of 18

10(a) speed is (directly) proportional to distance M1 speed is speed of recession of galaxy from an observer, and distance is the distance of the galaxy from the observer A1 10(b) F = L / (4d2) C1 = (3.8  1031) / [4  (1.8  1024)2] A1 = 9.3  10–19 W m–2 10(c)(i) galaxy is moving away (from the Earth) B1 wavelength (of light from the galaxy) increased by the Doppler effect / due to redshift B1 10(c)(ii)  /  = v / c C1 v = [(492 – 486)  3.00  108] / 486 (v = 3.7  106 m s–1) H = v / d C1 0 = (3.7  106) / (1.8  1024) A1 = 2.1  10–18 s–1 © UCLES 2023 Page 16 of 16

9(a) difference between mass of nucleus and (total) mass of nucleons M1 when infinitely separated A1 9(b)(i) neutron B1 9(b)(ii) E = m c2 C1 m = (0.030377 – 0.002388 – 0.009105)u C1 ( = 0.018884u) energy release = (0.030377 – 0.002388 – 0.009105)  1.66  10–27  (3.00  108)2 = 2.8  10–12 J A1 9(c)(i) number of atoms per unit time = (1.4  1028) / (2.8  10–12) C1 ( = 5.0  1039 s–1) mass of one atom = 4  1.66  10–27 or (4  10–3) / (6.02  1023) C1 ( = 6.64  10–27 kg) mass per unit time = 6.64  10–27  5.0  1039 A1 = 3.3  1013 kg s–1 9(c)(ii) L = 4σr2T4 C1 1.4  1028 = 4  5.67  10–8  (2.3  109)2  T4 T = 7800 K A1 © UCLES 2023 Page 14 of 15

10(a) speed is (directly) proportional to distance M1 speed is speed of recession of galaxy from an observer, and distance is the distance of the galaxy from the observer A1 10(b) F = L / (4d2) C1 = (3.8  1031) / [4  (1.8  1024)2] A1 = 9.3  10–19 W m–2 10(c)(i) galaxy is moving away (from the Earth) B1 wavelength (of light from the galaxy) increased by the Doppler effect / due to redshift B1 10(c)(ii)  /  = v / c C1 v = [(492 – 486)  3.00  108] / 486 (v = 3.7  106 m s–1) H = v / d C1 0 = (3.7  106) / (1.8  1024) A1 = 2.1  10–18 s–1 © UCLES 2023 Page 16 of 16

10(a) temperature inversely proportional to wavelength M1 temperature is thermodynamic temperature of surface, and wavelength is the wavelength at which maximum emission rate A1 occurs 10(b)(i) (astronomical) object of known luminosity B1 10(b)(ii) star / galaxy is moving away from the student B1 10(b)(iii) one tick placed in correct column in each row: B1 wavelength: too high surface temperature: too low B1 distance: unchanged B1 radius: too high B1 © UCLES 2023 Page 15 of 15

10(a)(i) L = 4σr2T4 C1 3.85  1026 = 4  5.67  10–8  r2  57804 r = 6.96  108 m A1 © Cambridge University Press & Assessment 2024 Page 13 of 14 10(a)(ii) F = L / 4d2 C1 = (3.85  1026) / (4  (1.50  1011)2) = 1.36  103 W m–2 A1 10(a)(iii) line of same shape showing peak intensity at greater wavelength B1 line of same shape showing lower peak intensity B1 10(b)(i) 5 lines in same pattern shifted to longer wavelengths B1 10(b)(ii)  /  = v / c C1  = (21400 / 300000)  656 = 46.8 nm wavelength = 656 + 46.8 A1 = 703 nm 10(b)(iii) (peak) wavelength too high so temperature too low B1 © Cambridge University Press & Assessment 2024 Page 14 of 14

10(a)(i) total power B1 power radiated (by the star) B1 10(a)(ii) standard candle has known luminosity B1 radiant flux intensity measured by observer B1 (distance calculated using) F = L / 4d2 B1 10(b)(i) luminosity = 4 r2T4 C1 = 4  5.67  10–8  (6.96  108)2 × 57804 = 3.85 × 1026 W A1 10(b)(ii)  T = constant C1 MAX temperature = (5780  501) / 624 A1 = 4640 K © Cambridge University Press & Assessment 2024 Page 15 of 15

8(a)(i) movement of star causes change in (observed) frequency B1 or movement of star causes redshift observed frequency is lower (than emitted frequency) B1 8(a)(ii) all three lines shown to left of corresponding printed lines B1 distance between drawn line and corresponding printed line approximately the same for all three lines B1 8(b)(i) E = hf and  = c / f C1 E = (6.63  10–34  3.00  108) / (488  10–9) A1 = 4.08  10–19 J 8(b)(ii) photon energy = (4.08  10–19) / (1.60  10–19) C1 = 2.55 eV energy level = –3.40 + 2.55 A1 = –0.85 eV 8(b)(iii)  =   (v / c) C1 = (488  6.2  106) / (3.00  108) ( = 10 nm) observed wavelength = 488 +  = 488 + 10 A1 = 498 nm © Cambridge University Press & Assessment 2024 Page 13 of 16 8(c) v = H d C1 0 d = (6.2  106) / (2.3  10–18) A1 = 2.7  1024 m © Cambridge University Press & Assessment 2024 Page 14 of 16

10(a)(i) total power B1 power radiated (by the star) B1 10(a)(ii) standard candle has known luminosity B1 radiant flux intensity measured by observer B1 (distance calculated using) F = L / 4d2 B1 10(b)(i) luminosity = 4 r2T4 C1 = 4  5.67  10–8  (6.96  108)2 × 57804 = 3.85 × 1026 W A1 10(b)(ii)  T = constant C1 MAX temperature = (5780  501) / 624 A1 = 4640 K © Cambridge University Press & Assessment 2024 Page 15 of 15

10(a) • redshift is the increase in observed wavelength / decrease in observed frequency (caused by Doppler effect) B3 • radiation from distant galaxies is observed to be redshifted • redshift provides evidence that galaxies are moving apart • galaxies moving apart means Universe must be expanding Any three points, 1 mark each 10(b)(i) F = L / 4d2 C1 d = √(1.90  1036 / [4  8.42  10–16]) A1 = 1.34  1025 m 10(b)(ii)  /  = v / c C1 (726 – 658) / 658 = v / (3.00  108) v = 3.1  107 m s–1 A1 10(c)(i) line with positive gradient passing through the origin B1 straight line with positive gradient B1 10(c)(ii) Hubble constant B1 © Cambridge University Press & Assessment 2024 Page 15 of 15

2(a) force per unit mass B1 2(b)(i) g = GM / x2 C1 = (6.67  10–11  1.99  1030) / (1.47  1011)2 A1 = 6.14  10–3 N kg– 1 2(b)(ii) E = – GMm / x C1 P = – (6.67  10–11  1.99  1030  2.63) / (1.47  1011) = – 2.37  109 J A1 2(c)(i) F = L / 4x2 C1 (g = GM / x2 and so) x2 = GM / g M1 and x2 = L / 4F elimination of x and subsequent algebra shown leading to g = 4GMF / L A1 2(c)(ii) correct read-off of pair of values of g and F and full substitution of values of g, G, M and F into equation C1 e.g. L = (4  6.67  10–11  1.99  1030  1.83  103) / (8.0  10–3) L = 3.8  1026 W A1 2(c)(iii) L = 4 r2T4 C1 3.8  1026 = (4  5.67  10–8  57804)  r2 r = 6.9  108 m A1 © Cambridge University Press & Assessment 2024 Page 6 of 14

10(a) • redshift is the increase in observed wavelength / decrease in observed frequency (caused by Doppler effect) B3 • radiation from distant galaxies is observed to be redshifted • redshift provides evidence that galaxies are moving apart • galaxies moving apart means Universe must be expanding Any three points, 1 mark each 10(b)(i) F = L / 4d2 C1 d = √(1.90  1036 / [4  8.42  10–16]) A1 = 1.34  1025 m 10(b)(ii)  /  = v / c C1 (726 – 658) / 658 = v / (3.00  108) v = 3.1  107 m s–1 A1 10(c)(i) line with positive gradient passing through the origin B1 straight line with positive gradient B1 10(c)(ii) Hubble constant B1 © Cambridge University Press & Assessment 2024 Page 15 of 15

10(a)(i) total power of radiation emitted (by the star) B1 10(a)(ii) standard candle has known luminosity B1 measure the radiant flux intensity B1 use F = L / (4d2) to calculate d B1 10(b)(i) v = 2R / T C1 v = 3.00  108  (656.2877 – 656.2831) / 656.2831 C1 R = [3.00  108  (656.2877 – 656.2831) / 656.2831]  (2.07  106) / 2 = 6.93  108m A1 10(b)(ii) Z is moving towards Earth M1 so observed wavelength is less than the emitted wavelength A1 10(b)(iii) L = 4 r2T4 C1 3.8  1026 = 4  5.67  10–8  (6.93  108)2  T4 T = 5800K A1 © Cambridge University Press & Assessment 2025 Page 14 of 14

10(a) speed is (directly) proportional to distance M1 speed is speed of recession of galaxy from an observer, and distance is the distance of the galaxy from the observer A1 10(b)(i) galaxy is receding from the Earth B1 observed wavelength is redshifted from emitted wavelength B1 10(b)(ii)  /  = v / c C1 (4.91 – 4.62) / 4.62 = v / (3.00 × 108) v = 1.9 × 107 m s–1 A1 10(b)(iii) wavelength (of maximum intensity) is inversely proportional to temperature B1 observed wavelength too high, so determined temperature too low B1 10(c) v = H d C1 0 d = (1.9 × 107) / (2.3 × 10–18) A1 = 8.3 × 1024 m © Cambridge University Press & Assessment 2025 Page 19 of 19

10(a) speed is (directly) proportional to distance M1 speed is speed of recession of galaxy from an observer, and distance is the distance of the galaxy from the observer A1 10(b)(i) galaxy is receding from the Earth B1 observed wavelength is redshifted from emitted wavelength B1 10(b)(ii)  /  = v / c C1 (4.91 – 4.62) / 4.62 = v / (3.00 × 108) v = 1.9 × 107 m s–1 A1 10(b)(iii) wavelength (of maximum intensity) is inversely proportional to temperature B1 observed wavelength too high, so determined temperature too low B1 10(c) v = H d C1 0 d = (1.9 × 107) / (2.3 × 10–18) A1 = 8.3 × 1024 m © Cambridge University Press & Assessment 2025 Page 19 of 19

9(a) temperature inversely proportional to wavelength M1 temperature is thermodynamic temperature of surface of star and wavelength is the wavelength at which maximum A1 emission rate from star occurs 9(b) Any three points from: B3 • (surface) temperature of star X = 7000 K or star X has a higher temperature than the Sun • star X has a higher luminosity than the Sun • luminosity of star X = 2.7  1027 W • radius of star X = 1.3  109 m 9(c) light (from star X) is redshifted B1 wavelength of peak emission rate would be greater (using observed data) B1 © Cambridge University Press & Assessment 2025 Page 16 of 17

9(a) difference between mass of nucleus and mass of (constituent) nucleons M1 when nucleons are separated to infinity A1 9(b) m = (2  2.013553) – (4.001505) (u) C1 ( = 0.025601 u) E = c2m C1 energy from one He-4 nucleus= 0.025601  1.66  10–27  (3.00  108)2 C1 (= 3.82  10–12 J) energy to form 1.00 mol= 3.82  10–12  6.02  1023 A1 = 2.30  1012 J 9(c)(i) L = 1.09  1011  (3.00  108)2 C1 = 9.81  1027 W A1 9(c)(ii) L = 4 r2T4 C1 9.81  1027 = 4  5.67  10–8  (1.19  109)2  T4 T = 9930 K A1 9(d) standard candles have known luminosity B1 radiant flux intensity (from star) measured (on the Earth) B1 distance found from F = L / (4d2) B1 © Cambridge University Press & Assessment 2025 Page 16 of 17

9(a) temperature inversely proportional to wavelength M1 temperature is thermodynamic temperature of surface of star and wavelength is the wavelength at which maximum A1 emission rate from star occurs 9(b) Any three points from: B3 • (surface) temperature of star X = 7000 K or star X has a higher temperature than the Sun • star X has a higher luminosity than the Sun • luminosity of star X = 2.7  1027 W • radius of star X = 1.3  109 m 9(c) light (from star X) is redshifted B1 wavelength of peak emission rate would be greater (using observed data) B1 © Cambridge University Press & Assessment 2025 Page 16 of 17