1a. 5.0 x 10^{-4}

1b. 12 x 10^{3-4} = 12 x 10^{-1} = 1.2

1c. 1/3 x 1/10^{3} = 0.333 x 10^{-3} = 3.33 x 10^{-4}

1d. 3^{4} x (10^{3})^{4} = 81 x 10^{12} = 8.1 x 10^{13}

1e. 3^{-2} x (10^{3})^{-2} = 1/9 x 10^{-6} = 1.111 x 10^{-7}

2. False. Green has a smaller wavelength and hence larger frequency than red light (recall the spectrum : red, orange, yellow, green, blue, indigo, violet, going from longer to shorter wavelengths).

3. d. Use wave equation : speed = wavelength x frequency. Speed
is 3 x 10^{8} m/s (note, use meters/sec), frequency is
6 x 10^{5} Hz (dont forget that kHz is kilohertz so 600 kHz is
600 x 1000 Hz = 6 x 10^{5} Hz). Hence wavelength = speed/frequency
= 3 x 10^{8} / 6 x 10^{5} = 0.5 x 10^{3} =
5 x 10^{2} meters.

4. c. They are both electromagnetic waves, but of different wavelength.

5. False. Ultraviolet light is blocked by the Earth's atmosphere, so UV telescopes must be in orbit above the earth's atmosphere.

6. (a) Use the Planck relation : E = h.f so f=E/h where h is Planck's
constant. f = 4.09 x 10^{-19}/6.62 x 10^{-34}=
6.17 x 10^{14} Hz. (b) Use wave equation : c = lambda x frequency
so lambda = c/frequency = 3.0 x 10^{8}/6.17 x 10^{14}=
4.85 x 10^{-7} m = 485 nm (nanometers). (c) This is in the middle
of the visible part of the spectrum, so the emission line appears green.

7. False. Hotter objects tend to emit bluer light (Wein's Law), so orange stars are cooler than yellow stars.

8. (a) 35 + 273 = 308 K (273 is the offset which converts degrees
centigrade to degrees Kelvin). (b) Wein's Law is : lambda(peak) in
nanometers = 2.9 x 10^{6}/ T (in degrees K). So for T=308 we have
lambda(peak)=9.41 x 10^{3}nanometers.
(c) This is in the Infra-red part of the spectrum (remember, red visible
light has wavelength of about 700 nm). (d) Stefan-Boltzmann relation
says how the power in thermal radiation = area x sigma x T^{4}.
For a person, the area is 1 and T is 308 K. Taking sigma, the constant,
from the numbers listed at the beginning of the questions, we have the
power radiated = 1 x 5.67 x 10^{-8} x (308)^{4}.
Plug this in your calculator and you find the power is about 500 Watts.
So you are a half kilowatt radiator !

9. d. From Stefan-Boltzmann law : energy radiated is proportional
to T^{4} for a given area. So, due to temperature alone, star A
radiates 2^{4}=16 times more than star B for the same area. But star
B has 4 times the radius, so it has 4^{2}=16 times more area than
star A. So while star B is fainter by x16 due to temperature, it is brighter
by x16 due to its larger size. So the net result is that star B has the
same brightness as star A.

10. Similarly to question 9 : if the filaments have the same area, their
brightness depends only on T^{4}. Thus the hotter filament
radiates 3^{4}=81 times more energy than the cooler one.

11. d. For neutral atoms, the number of electrons exactly matches the number of protons, since each carry the same (but opposite) charge.

12. b.

13. True. Remember, the Balmer series (optical) start from level 2, while the Paschen series (infrared) start from level 3. The Lyman series are in the ultraviolet, and start from level 1.

14. e. Remember the other two of Kirchoff's Laws : a thick hot gas emits a continuous spectrum; and a thin gas in front of a hotter thick gas produces absorption lines.

15. a. The shift is to the red, so the velocity is away from us. The Doppler formula is v/c = change in wavelength / original wavelength = (700 - 500)/500 = 200/500 = 0.4

16. Balmer absorption lines occur when electrons jump from level 2 to higher levels in hydrogen atoms. In very hot stars (O type) all the hydrogen is IONIZED and so there is no neutral hydrogen to form the absorption lines. Conversely, in cool stars (eg K type) although the hydrogen is not ionized, all the electrons are sitting in the ground state (level 1), and so again there are no Balmer absorption lines. Only around 10,000 K (ie A type stars) is hydrogen basically neutral but there are many atoms with their electrons bumped up into the 2nd level.

17. False. The colors seen in paper result from dyes which absorb some of the colors of the "white" light falling on them from either the sun or artificial lights. The colored light is not itself created by the paper.

18. Many gaseous nebulae contain dilute hydrogen gas. When electrons jump between levels 3 and 2 in the hydrogen atoms, then the red Balmer alpha emission line is produced. It is stronger than the other Balmer lines, and so the overall color of the nebula is red.

19. False. Because the orbit of the earth is approximately circular around the sun, we are neither moving towards nor away from the sun. Hence we see NO doppler redshift or blueshift --- our motion is TRANSVERSE compared to the direction to the sun.

20. Surface area of a sphere is 4 pi R^{2}. Here, R = 6000 km
= 6 x 10^{6} m. So the surface area of the earth is
4 pi x 36 x 10^{12} m^{2} = 4.52 x 10^{14} m^{2}.

21. Gamma-rays, X-rays, Ultraviolet, infra-red, (only radio and optical pass through the earth's atmosphere).

22. False. The atomic nucleus fills a VERY small part of the atom. Its
diameter is 10^{-5} that of the atom. (Of course, almost all
the MASS of the atom is in the nucleus, but the question asks about the
space within atoms).

23. True. In harsh astronomical environments, atoms can lose electrons in a number of ways --- eg by collisions with other atoms; by being struck by a photon with enough energy (these two processes are termed, not surprisingly, collisional ionization and photoionization).

24. 1800 arcsec. Remember, 1 degree contains 60 arcmins, and one arcmin contains 60 arcsec. So one degree contains 3600 arcsec, so the moon, which is half a degree across, spans 1800 arcsec in the sky.

25. A star is a ball of hot dense gas, and so by Kirchoff's first law, it produces a continuous (thermal, or blackbody) spectrum. However, it has a cooler atmosphere of low density gas, and so as the light passes through this atmosphere, colors are absorbed by the atoms in the gas and so we see an absorption line spectrum.

26. Hydrogen (about 75%) and Helium (about 23 %)

27. a. The photosphere of the sun has a much lower density than the air on earth.

28. b. Surprisingly, although the temperature drops from the base of the photosphere to the top, it begins to increase again in the chromosphere and goes on increasing to a couple of million degrees in the corona.

29. True. They result from the impact of the solar wind as it is channelled along the earth's magnetic field lines which dive into the atmosphere near the north and south magnetic poles. The aurora is the air glowing from the collisions between fast protons and the air molecules.

30. d. Solar storms are fundamentally different from storms on earth, which do indeed have the rotation properties described. On the sun, the predominant forces are vertical, since they are related to magnetic fields which come from below and emerge up into the chromosphere.

31. E = m.c^{2}, so for 1 kg we have 1 x
(3 x 10^{8})^{2}
= 9 x 10^{16} Joules (note we need c in m/s and m in kg). If we convert
1 kg of hydrogen into helium, the amount of mass converted into energy is
0.7% (ie the reaction is 0.7% efficient). Hence the energy released is
0.007 x 9 x 10^{16} = 6.3 x 10^{14} Joules. This is still
a huge amount of energy, equivalent to a 100 MegaWatt power station
running for 10 weeks.

32. b.

33. False. Remember, brightness obeys the "inverse square law" so that pushing the star to 5 times the distance makes it appear 25 times fainter (ie brightness becomes 1/25)

34. True. Remember, there is a brightness ratio of a factor of 100 for
each difference of 5 magnitudes. The galaxy is 15 magnitudes different
from Vega, which is 5 + 5 + 5 in magnitudes so we have 100 x 100 x
100 = 10^{6} = million times in brightness. Since magnitudes
increase for fainter objects, the galaxy (15 mag) appears FAINTER (not
brighter) than Vega (0 mag)

35. False. In the temperature sequence O,B,A,F,G,K,M, the K stars come after the G stars, hence they are cooler, not hotter. The "2" refers just to the subcategory along the sequence, but wont alter this answer.

36. Spectral type M. This is at the coolest end of the sequence, where the temperatures are sufficiently low to allow the survival of molecules.

37. True. The lower pressures in the extended atmospheres of giant stars mean that atoms collide less frequently, hence their energy levels can "settle down" and hence the absorption lines are narrower.

38. L = b x 4pi d^{2} so we need distance, d. From parallax: d = 1/p = 1/0.02 = 50pc = 50 x 3.08 x 10^{16} = 1.54 x 10^{18} meters. Hence L = 3 x 10^{-9} x 4 x pi x (1.54 x 10^{18})^{2}
= 8.94 x 10^{28} Watt = (8.94 x 10^{28}) / (3.8 x 10^{26}) = 235 L_{sun}. Check you understand the algebra and powers of 10. Note that to get to L_{sun} we divided the star power in Watts by the Sun's power in Watts.

39. False. Remember, parallax angle (in arcsec) is 1/(distance in parsecs). Hence the distance is 5pc for a parallax of 0.2 arcsec.