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Active Physics Full Solutions to Textbook Exercises 71 Photoelectric Effect 64 Starlight: Messengers from the Stars Checkpoint 1 (p. 104) (a) angle (b) angle (c) distance 3. (c) F (d) F (a) Star R It is 2.512 times brighter. (b) Star S Since the stars are almost the same distance from us, we can deduce that S has a larger absolute magnitude from the fact that it has a larger apparent magnitude. Note that 1 pc is defined as the distance between the Sun and a star whose parallax is 1″. Do not confuse the parallax angle with the parsec. 2. 3. Mars is farther from the Earth when the angular diameter is smaller. (a) Yes (b) Yes (c) No. 1 AU ≈ 1.50 × 1011 m. p.1 apparent magnitude when they look equally bright in the sky. Checkpoint 1. | 4. Difference in magnitude = 14.5 The increase in brightness is 100(14.5/5) = 631 000 times 5. ∵(1360)∙(12) = I∙(1.52) ∴ I = 604 W m−2 4. θ = 6779/55.8 × 106 = 0.000 121 rad 5. See the table below. In this case, we do not need to convert the distances to metres if both of them have the same unit. distance star parallax Sirius 0.379″ Altair 0.195″ Dumbbell Nebula 6. 0.00238″ (in pc) (in AU) 2.64 pc 5.44 × 105 AU 5.13 pc 420 pc (in ly) 8.60 ly 1.06 × 106 AU 16.7 ly 8.66 × 107 AU 1370 ly Checkpoint 3 (p. 117) 1. A blackbody is ‘black’ because it does not reflect any EM radiation. 2. θ/2 = 1/10 ⟹ θ = 0.2″ The parallax between the Sun and the star is θ/2 but not θ. Then we have θ/2 (in arc seconds) = 1/(10 pc) by definition. C 3. (a) T (b) T (c) T (d) F (c) < (b) < (a) < (d) 4. Checkpoint 2 (p. 109) 1. (a) yes; no (b) yes; yes (c) no; no The difference between apparent magnitude and absolute magnitude is that we have taken account of how far the star is from us when considering the apparent magnitude. Obviously, the farther the star, the dimmer it appears in the sky. 2. (a) (b) F P may be farther away and have a smaller absolute magnitude than Q such that the stars still have the same apparent magnitude. T By definition, the stars have the same Note that a class F star has a higher surface temperature than a class G star which in turn has a higher surface temperature than a class K star. Checkpoint 4 (p. 120) 1. B Recall that 2. (a) . T In daily use, luminosity refers to how bright something is. But in this context, luminosity Active Physics Full Solutions to Textbook Exercises 71 Photoelectric Effect refers to how much EM radiation a star radiates and not only visible light. 3. (b) F Recall that L = 4πR2∙σT4. Luminosity also depends on the radius of the star. (c) F L = 4πR2∙σT4 = 4π(2.8 × 109)2∙σ(15 000)4 = 2.828 × 1029 ≈ 2.83 × 1029 W (a) (b) (c) 2. (a) T The Sun is a class G star which must have a higher temperature than a class K star. 2. 3. (a) S (b) P and Q Δλ = 589.543 – 588.997 = 0.546 nm T From the H–R diagram on p. 123, we can see that a class B main-sequence star is larger than a class G main-sequence star. ∴vr = (9.270 × 10−4)∙(3 × 108) = 2.781 × 105 m s−1 We can also deduce this from L ∝ R2∙T4. For a class B main-sequence star, its luminosity is 100 times that of a class G main-sequence star but the temperature is only doubled. Hence a class B main-sequence is larger. You can determine that the galaxy is moving away from the shift in wavelength. Longer observed wavelength ⇒ red shift ⇒ the body is moving away. ≈ 2.78 × 105 m s−1 The galaxy is moving away from the Earth. 4. F Luminosity depends on the temperature as well as the size. (a) The radial velocity is (1.2 × 105) cos (180°−150°) = 1.039 × 105 ≈ 1.04 × 105 m s−1 (b) The radial velocity is (1.2 × 106)∙cos (180°−20°) = −1.128 × 106 ≈ −1.13 × 106 m s−1 P: red giant Compare with Fig. 4.24 on p. 121. Star Y shows a larger Doppler shift because its radial velocity has a larger magnitude. 5. (a) F A star shows a Doppler shift when the star is moving with a radial velocity relative to the Earth. (b) T In reverse, the star is moving towards the Earth if there is a blue shift. (c) T Recall that M, G, A, A When looking up their spectral classes, you need to consider the surface temperatures only. (i) S<Q<R<P (ii) P<Q<R=S (iii) S < Q < R < P Note that L ∝ R2∙T4 ⇒ R2 ∝ L / T4. P is more than 100 times larger in luminosity than R but has a temperature of about 1/3 of that of R. From the above relation, we can see that P has a larger radius. 3. No Δλ/λ = 0.546/588.997 = 9.270 × 10−4 S: white dwarf (c) (c) R does not show red shift or blue shift because its radial velocity is zero, i.e. it does not move along the line joining the star and the Earth. Q and R: main-sequence star (b) Yes (no difference) R2 = 3.85 × 1026 / (4π(5.67 × 10−8)(6000)4) = 4.169 × 1017 6. (a) P (b) 7. S vr = (−8.3 × 10−5)∙(3 × 108) = − 2.49 × 104 m s−1 Hence the expansion speed is 2.49 × 104 m s−1. Checkpoint 6 (p. 131) Checkpoint 7 (p. 138) 1. 1. Yes (no difference) . At P, the gas expands (moves) with a velocity perpendicular to the line of sight, i.e. the radial velocity is zero. Therefore, the fractional change in wavelength is also zero. ∴R = 6.457 × 108 ≈ 6.46 × 108 m (a) p.2 The light speed observed is independent of the relative motion of the observer and the emitter. It depends on the medium where light travels. Checkpoint 5 (p. 124) 1. (b) | (a) R and S Active Physics Full Solutions to Textbook Exercises (b) P (c) Q The radial velocity is positive when the star is receding from us, negative when approaching us and zero when moving in a direction to the line of sight. 71 Photoelectric Effect p.3 invisible is pulling the celestial bodies and hence affecting their velocities. This is an evidence for the existence of dark matter. Exercise Exercise 4.1 (p. 110) 1. 2. | D The answer can be found by expressing the distances in parsecs(pc): ≈ 5.02 × 1011 m ≈ 1.88 × 1033 kg 3. A 200 ly 61 pc B 50 pc 50 pc C 400 000 AU 1.9 pc D 0.015″ 67 pc 2. A The distance in parsecs is 1/0.05″ = 20 pc. The distance in light years is 20 × 3.26 = 65.2 ly. 3. A See the table below. The larger the value of θ, the larger the apparent diameter. Checkpoint 8 (p. 142) 1. (a) (b) 2. F Dark matter is ‘dark’ such that it does not emit or reflect any radiation. F In fact, dark matter is the main composition of the universe and not only found in the centres of galaxy. (c) T (d) F Dark matter is detected as it exerts gravitational force on visible matter. (a) D F By Hubble’s law, v = H∙d. That means, the receding speed depends on the distance from us. T (c) F Since the galaxies are receding from us, their radial velocities are positive. Their spectra should be red-shifted instead of blue-shifted. 3. (a) (b) The disk is revolving around its centre. The positive radial velocity suggests that the side of the disk is receding from us while the negative velocity suggests the reverse. The velocity curve suggests that the velocity does not decrease as predicted. This can be explained by assuming that something θ ≈ D/d A 1.41 × 105 ly 2.54 × 106 ly 0.0555 rad B 3470 km 384 000 km 0.00903 rad C 1.39 × 109 m 1.50 × 1011 ly 0.00927 rad D 150 ly 18 000 ly 0.000833 rad 4. C The stellar parallax is (4 × 0.05″)/2 = 0.1″. Therefore, the distance is 1/0.1 = 10 pc. 5. C The ratio can be found by 6. (a) The distance is 7000/3.26 = 2147 pc. The parallax is 1/2147 ≈ 0.000466″. (b) In fact, some galaxies close to us, e.g. the Andromeda Galaxy, are approaching us. (b) d 7. No. Since the parallax of the star is smaller than the uncertainty of the measurement, the distance to the star cannot be measured accurately. The max. apparent diameter is θ = D/d = 3470/363 000 ≈ 9.56 × 10−3 rad The min. apparent diameter is θ = D/d = 3470/406 000 ≈ 8.55 × 10−3 rad 8. (a) The apparent diameter is (b) The actual size is D = dθ = (2.81× 106)(0.02036) ≈ 5.72 × 104 ly. Active Physics Full Solutions to Textbook Exercises 9. 71 Photoelectric Effect (a) The distance to the nearest star is (b) The largest and smallest apparent magnitudes of the star are 3.4 and 2.1, respectively. (b) The percentage uncertainty is 0.001/0.042 × 100% ≈ 2.38%. (c) The difference in magnitude is 3.4 − 2.1 = 1.3. (c) Let x be the angular measurement with a maximum uncertainty of 20%. | p.4 Therefore, the brightness of the star changes by 2.5121.3 ≈ 3.31 times. Exercise 4.2 (p. 125) 1. The maximum distance that can be measured is Recall the mnemonic ‘Oh Be A Fine Girl Kiss Me’. d = 1/p = 1/0.005″ = 200 pc. 10. (a) X appears brighter because its apparent magnitude is smaller. 2. Thus, X is 2.5122.5 ≈ 10.0 times brighter than Y. 12. (b) Both X and Y are at the same distance from the Earth but X appears brighter. Therefore, X emits more light per unit time than Y does. (a) P and Q appear equally bright as they have the same apparent magnitude. 3. C Since the peaks are the same, the stars should have the same temperature. As X has a higher peak for intensity, we can deduce that the star is brighter due to a larger size (radius) 4. B From the H–R diagram, we can see that white dwarfs have a lower luminosity than main-sequence stars. In general, the higher the class of the main-sequence stars, the higher the luminosity. Therefore, a class B main-sequence star should have the highest luminosity. (b) Q emits more light per unit time as it has a smaller absolute magnitude. (c) Q is farther away from the Earth because Q emits more light per unit time but it appears to be as bright as P. (a) The difference in magnitude of Mars is 1.83 −(−2.91) = 4.74. 5. C Refer to the H–R diagram on p. 121. Therefore, the apparent brightness of Mars changes by 2.5124.74 ≈ 78.7 times. 6. (a) Yes. A hotter star has a blackbody radiation curve peaked at a shorter wavelength. A blue star emits more light around the blue range (shorter wavelength). In contrast, a red star emits more light around the red range (longer wavelength). Hence we can deduce that blue stars are hotter. (b) No. Planets are not blackbodies. (a) A>G>M (b) The distance between the Earth is the longest when they are on the opposite side of the Sun, and the shortest when they are on the same side as shown (not to scale). Mars Earth Mars Sun Earth Sun 7. The intensity is inversely proportional to the square of the distance. As a result, the apparent magnitude can change so greatly. 13. C Since the stars are equally bright in the sky, the distance ratio can be found by The difference in magnitude between X and Y is 4.9 − 2.4 =2.5. 11. D The temperature drops from class O to class M. (a) The period is about 60 h. Note the time lapse between the two troughs. The temperature drops from class O to class M. Recall the mnemonic ‘Oh Be A Fine Girl Kiss Me’. Active Physics Full Solutions to Textbook Exercises 71 Photoelectric Effect (b) 13. 8. 9. (a) Vega Its spectrum peaks at a shorter wavelength. (b) Class A Sun is a class G star. As Vega is hotter than a class G star, it is likely to be a class A star. (a) They should have the same (similar) surface temperature because they are from the same class. (b) Q should be more luminous while P is less luminous. (b) Since L ∝ R2T4, we have (a) X: white dwarf Y: main-sequence star (b) Since L ∝ R2T4, we have | The ratio is 4 : 25 = 0.16 : 1. (c) Since L ∝ R2T4, we have (d) Since L ∝ R2T4, we have (a) (i) P: 10 000 K Q: 3000 K (ii) P: class A Q: class M The Sun is a class G main-sequence star. Q is more luminous because a giant is luminous than a class G main-sequence star. For P, it is also a main-sequence star but has a lower class (M as compared with O). Therefore, it should be less luminous. 10. (a) surface temperature / K (b) 14. (c) See the figure above for the positions. luminosity: (3) < (2) < (1) temperature: (2) < (3) < (1) radius: (3) < (2) < (1) 11. (a) (b) (iii) P: main-sequence star Q: red giant By L = 4πR2∙σT4, the luminosity is 4π(3.13 × 1011)2∙(5.67 × 10−8)(3000)4 ≈ 5.65 × 1030 W. (b) Star Q is bigger. P and Q have the same luminosity but Q is colder (lower temperature). From L = 4πR2∙σT4, we can deduce that Q has a larger radius and hence it is bigger. (c) Since L ∝ R2T4, we have It should be a supergiant. It has a relatively low temperature and a large radius as compared with the Sun. (c) The distance is (3.13 × 1011)/(1.50 × 1011) ≈ 2.09 AU. The Earth would be engulfed by the star. 12. (a) It is a main-sequence star. The ratio is 0.09 : 1. p.5 Active Physics Full Solutions to Textbook Exercises 71 Photoelectric Effect Applying , the radius is 8 (1.010 × 10 )/(2π) × (1.89 × 104) = 3.038 × 1011 ≈ 3.04 × 1011 m. A Since the body is receding, it should be a red shift and hence Δλ > 0. Options B and D are incorrect. By (b) , we have 2. A It has the largest magnitude of radial velocity. 3. D Dark matter does not emit or reflect any EM waves; it exerts gravitational force on matters and that is how we can ‘see’ it. 4. A Option B is incorrect. The galaxies in the Local Group are close to us and do not obey Hubble’s law. Applying 9. (a) , we have Since the spectral line is red-shifted (Δλ > 0), the galaxy is receding from the Earth. (a) (b) Applying , the highest speed is 10. (a) C The periods of A, B and C are TA = 0.25 y, TB = 0.5 y and TC = 1 y, respectively. (b) A The speeds of A, B and C are vA = 2 × 105 m s−1, vB = 105 m s−1 and vC = 5 × 104 m s−1, respectively. (c) 8. (a) Since and the values of Tv are the same for A, B and C, we can deduce that the stars have the same orbital radius. Applying (ii) TP : TQ = 1 : 2 Since , we have 11. 12. (i) 100 km s−1 (ii) 125 km s−1 (b) The existence of dark matter, which exerts gravitational force on other matter but does not emit or reflect any EM radiation, is the view held by many astronomers as the cause of the deviation between the theoretical and observed curves. (a) The recession velocity v of a distant galaxy is proportional to its distance d from us, i.e. v = H∙d. (b) The motion of the galaxies is greatly affected by the gravity of other galaxies in the Local Group. (a) The distance to the star is The angular distance in radians is (a) vP : vQ = 2 : 1 Therefore, the ratio is 4 : 1. Applying D = dθ, the distance is d = D/θ = 0.1006/(5.963 × 10−7) ≈ 1.69 × 105 ly. 7. (i) (iii) Since and the values of Tv are the same for P and Q, we have rP : rQ = 1 : 1. (b) 6. The mass is (c) Options C and D are incorrect. The recession velocity obeys Hubble’s law, i.e. v = H∙d. 5. p.6 The period in seconds is 3.2 × 365.25 × 24 × 60 × 60 = 1.010 × 108 s. Exercise 4.3 (p. 143) 1. | d = 1/p = 1/0.545″ = 1.8349 ≈ 1.83 pc. (b) Applying , we have , the orbital speed is The shift in wavelength is −0.234 nm. Active Physics Full Solutions to Textbook Exercises The star is approaching as the observed wavelength becomes shorter (blue shift). (c) 71 Photoelectric Effect 15. (a) See the diagram below. | p.7 The radius in metres is (25 000)(9.461 × 1015) = 2.365 × 1020 m. The period is 2πr/v = 2π(2.365 × 1020)/(220 × 103) = 6.755 × 1015 s ≈ 2.14 × 108 y. (b) The total mass is 1011M☉ = 1.99 × 1041 kg. (c) The orbital speed is (d) No. Astronomers believe that there are some ‘invisible’ mass, known as dark matter, that extend much farther than the visible edge of the galaxy. The speed of the star is (d) The transverse displacement of the star is D = vtt = (90 × 103)(365.25 × 24 × 60 × 60) = 2.840 × 1012 m. Applying D = dθ, the change in angular position is Chapter Exercise 13. Chapter Exercise: Multiple-choice Questions (p. 149) (a) 1. A Statement (2) is incorrect. An object looks bigger when it is closer to the observer. That mean, the parallax becomes larger. Statement (3) is incorrect. Whether an object is faint or not depends on its luminosity. (b) (c) The two radial velocity curves are out of phase because the two stars revolve around their centres of mass in opposite directions. (a) No. In this case, the radial velocity of the small star is always zero. No Doppler shift can be observed on the Earth. Therefore, the mentioned physical quantities cannot be determined by spectroscopic analysis. (b) D The larger the value of (M – m), the smaller the distance. (M: absolute magnitude and m: apparent magnitude) 3. A Statement (2) is incorrect. The luminosity of a star depends on both the temperature and the radius. 4. C Statement (1) is correct. We can analyse the chemical composition by observing the absorption lines on the spectrum. 5. B Recall L = 4πR2∙σT4. 6. C The small star has a higher orbital speed. By , we know that r ∝ v. The smaller star has a larger orbital radius and hence a higher orbital speed. 14. 2. All of them can be measured. Statement (2) is incorrect. By L = 4πR2∙σT4, Q should have a larger radius if it is colder than S. For the orbital radius, it can be found by dθ where d = 1/p and θ is given. For the orbital period, it can be measured by recording the time taken for the small star to orbit once around the massive star. For the orbital speed, it can be determined from the orbital period and the orbital radius using . 7. B Option D is not the correct answer because it demonstrates a blue shift. For the remaining options, the radial velocities should be vA = 35 km s−1, vB = 58 cos 30° ≈ 50 km s−1 and vC = 0. Since the star in B moves Active Physics Full Solutions to Textbook Exercises 71 Photoelectric Effect away with the greatest radial velocity, it should have the largest red shift. 8. 9. 10. 11. B The fractional Doppler shift (Δλ/λ) should be the same for the spectral lines coming from the same star. p.8 is more luminous, given that the stars should have the same surface temperature. Chapter Exercise: Structured Questions (p. 151) 17. (a) The distance is 1/0.007 41″ = 135.0 pc. (1M) In light years, it should be 135.0 × 3.26 = 439.9 ≈ 440 ly. (1A) B Star Y is receding from the Earth when it is at position P. The corresponding spectrum should be red-shifted. In contrast, the star is approaching the Earth when it is at position R. The corresponding spectrum should be blue-shifted. (b) The angular size in radians is The actual size is D = dθ = 439.9 × 0.036 65 ≈ 16.1 ly (1M+1A) C Statement (2) is correct. The orbital speed is equal to the maximum radial velocity. The orbital radius can be calculated by (2.5 × 104 × 500 × 24 × 60 × 60)/(2π) = 1.719 × 1011 m. (c) L > L☉, R > R☉ and T > T☉. (3A) Refer to the H–R diagram on p. 123. (d) Since L ∝ R2T4, we have D For a star of surface temperature around 3000 K, the blackbody radiation curve peaks in the infrared range (close to the red end of the visible spectrum). In contrast, if the surface temperature is around 20 000 K, the curve peaks in the ultraviolet region. (1M+1A) 18. (a) By Wien’s displacement law (p. 114), the peak λmax is given by λmax = b/T where b = 2.90 × 10−3 m K is a constant and T is the surface temperature in kelvins. 12. B An absorption spectrum is produced when a spectrum of light passes through a gas and the absorption lines can be regarded as a ‘fingerprint’ of the gas. 13. B Options A and C are incorrect. Since the stars have the same luminosity, Q must be farther away so that it looks fainter. By , the intensity is inversely proportional to the distance squared. Therefore, Q’s distance should be 5 times that of P such that its brightness is 1/52 = 1/25 that of P. 14. | A Note that the line is blue shifted, i.e. Δλ < 0. Options C and D MUST be incorrect. The fractional shift is −50/410 = −0.1220. Therefore, the observed wavelength of the red line should be 656 × (1−0.1220) ≈ 576 nm. 15. C The two curves peak at the same wavelength and this implies that the surface temperatures are the same. 16. B Since L ∝ R2T4, X should be larger in size so that it W: main-sequence star; X: main-sequence star; Y: red giant; Z: white dwarf (4A) [1A for each star’s class and position on the diagram] (b) Y should be the farthest. (1A) It has the highest luminosity but appears the dimmest as seen from the Earth. (1A) Z should be the closest. (1A) It has the lowest luminosity but appears the brightest as seen from the Earth. (1A) (c) Since L ∝ R2T4, we have (1M+1A) Active Physics Full Solutions to Textbook Exercises 19. 71 Photoelectric Effect (a) The radius and the surface temperature (2A) (b) The distance of Arcturus from us is | p.9 36.7 × (9.461 × 1015) = 3.472 × 1017 m By , the intensity should be (1M+1A) (c) (i) When d = 10 pc, (1A) At t = 0 or 2 years, the spectrum is red-shifted. This implies that the star is moving away from the Earth. (1A) At t = 0.5 and 1.5 years, no Doppler shift has been observed. This implies that the star is moving in a direction perpendicular to the line of sight from the Earth. (1A) At t = 1 year, the spectrum is blue-shifted. This implies that the star is moving towards the Earth at that time. (1A) (1A) (ii) The distance of Arcturus in pc is 36.7/3.26 = 11.26 pc (1M) If m = −0.04, (1A) 20. (a) (b) From the graph, the greater the mass, the smaller the radius and hence the volume. (1A) Since density is mass over volume, we can deduce that white dwarfs of larger mass should have a higher density. (1A) Since L ∝ R2 T 4 , Since L ∝ R2 T 4 , (b) Applying , we have (1A) we have (c) The radial velocity of the star is as shown. (1M+1A) (c) we have (3A) [1A for axes + 1A for curve + 1A for labels] (d) (1M+1A) (d) Applying strength is The radius of the orbit is (60 000)(2 × 365.25 × 24 × 60 × 60)/2π = 6.027 × 1011 m. Applying , the gravitational field , the mass is (1M+1A) (e) (1A) The weight is mg = (70)(3.67 × 106) ≈ 2.58 × 108 N. (1A) 21. (a) Yes, it can be a black hole. (1A) Its mass is 3.25 × 1031/1.99 × 1030 ≈ 16M☉. (1A) 22. (a) The orbital motion of the star is as shown. The rotational speeds of the lumps of ice can be measured by observing the Doppler shift in the spectra observed from the lumps at both sides of the rings. (1A) By , the radial velocities of the lumps can be found, and hence the rotational speeds. (1A) (b) The rotational speeds measured would be differed by a factor of cos 20°. (1A) Active Physics Full Solutions to Textbook Exercises 71 Photoelectric Effect The measured speeds are lower than the actual speeds as all the measurements of velocities would be differed by this factor. (1A) (c) (b) Different sets of dark lines correspond to different elements. Hence the existence of a specific set of dark lines reveals the existence of an element in the star. (1A) (i) The star is receding and there is a red-shift for all minima. (2A) (ii) Δλ = 120.9 – 119.5 = 1.4 nm (1M) Applying (1A) (d) Applying p.10 (ii) The speed decreases with the distance. (1A) Suppose the centripetal force on the lumps is provided by the gravitational force from Saturn. | , , (1M) v = (1.4/119.5)∙(3 × 108) (1M) v ≈ 3.51 × 106 m s−1 (1A) (1M) Applying 25. , the mass is (a) It measures how bright the star is as seen from the Earth. (2A) (b) (i) dY = 3.8 pc ⇒ 1/dY2 = 0.0693 pc−2. dZ = 4.6 pc ⇒ 1/dZ2 = 0.0473 pc−2. (1A) 23. (a) The recession velocity is directly proportional to the distance. (1A) (b) The relation in (a) suggests that the universe is expanding. (1A) This is because the farther away a galaxy, the faster it is moving away from us. (1A) (c) 65 km s−1 Mpc−1 (1A) (ii) Since v = H∙d, the slope of the graph is actually the Hubble constant. (d) (i) (ii) The universe is expanding. (1A) All distant galaxies, including the galaxy shown, are receding from us and thus all the measured spectra are red-shifted. (1A) The difference in the degree of Doppler shifts at the two sides of the galaxies suggests that the galaxy is rotating. (1A) Neglecting the recession velocity of the galaxy, the stars on the right are receding from us while those on the left are approaching us. (1A) Hence the galaxy is rotating in a anticlockwise direction when viewed from above. (1A) (2A) [1A for data points, 1A for the line] (iii) k = slope of the graph (1A) Hence k = 7 × 10−3 W m−2 pc2. (c) (i) log I = −2 log d −2.2 (2A) (ii) m = −2.5 (−2 log d −2.2) + a ⇒ m = 5 log d + b where b = a + 5.5 (2A) (d) m – M = 5 log (d/10) (1A) 3.0 – M = 5 log (2.1/10) ⇒ M ≈ 6.39 (1A) 26. (a) (i) Since L ∝R2T4, we have (1A) (iii) It is the existence of dark matter in the galaxy that causes the orbital speeds of the stars increase with their distances from the galactic centre. (1A) 24. (a) (i) The dark lines are formed when certain wavelengths of the visible spectrum are absorbed by the gases on the surface of the star and then re-radiated in all directions. (2A) (1A) (ii) From the above, (1M+1A) Active Physics Full Solutions to Textbook Exercises (b) (i) By 71 Photoelectric Effect , the calculated L would be larger if the measured I remains unchanged while d becomes larger. (1A) Since L ∝R2, the radius would also become larger. (1A) (ii) (c) The parallax is too small. (1A) The brightness of the sun is The distance of Betelgeuse in AU is 200 × 206 265 = 4.126 × 107 AU (1A) The brightness of Betelgeuse is (1A) From the above calculation, it is not true that Betelgeuse is as bright as the Sun. (1A) 27. (a) (i) Substitute L = 4πR2∙σT4 into , the intensity is (1A) Rearrange I = P/A, the power is (1A) (ii) Assuming that the planet is a blackbody, (1A) Simplify the above and we have (1A) (b) (i) The temperature is (1M+1A) (ii) Water can be in the liquid form (1A) because the temperature is between the ice point and the steam point. (1A) It may be favourable. (iii) A class K star is cooler than our Sun. (1A) This suggests that the surface temperature can be lower as the power received from the star is lower. (1A) | p.11