ASTR 3220 - Homework 5
Due Thursday Feb. 26

Total of 15 points. Show your work if a calculation is required. Give units for your answer if applicable. Do your own work.

1. H II Regions (3 points) -
(a) Define the term `H II region'. (... in two sentences or less).
ANS: An H II region is a region of hot ionized gas, comprised mainly of (singly) ionized hydrogen in which the hydrogen atoms have lost their electrons. H II regions are often found surrounding hot O or B stars, which are capable of providing sufficient ultraviolet radiation to ionize the hydrogen atoms.
(b) Why would you NOT expect to see a 21 cm emission line from an electron spin-flip transition when observing a HII region?
ANS: The spin-flip transition can only occur in neutral (un-ionized) hydrogen, where the electron an proton are bound together. In this closely bound configuration, the spin of the electron and the spin of the proton interact magnetically (... like two tiny bar magnets). Once the atom becomes ionized (as in H II regions), the proton and electron are no longer bound together and their spins no longer interact.

2.* Energy per Unit Area (Surface Flux) (3 points) -
The total energy per second emitted by a star, divided by the surface area of the star, is called the `surface flux'.
(a) Starting with the Stefan-Boltzmann Law, obtain a simple expression for the surface flux of a star with radius R and temperature T. Assume the star radiates like a spherical blackbody.
ANS: Surface flux is just the stellar luminosity (L) divided by the surface area of the star ( = 4*pi*R^2).
By the Stefan-Boltzmann Law, L = 4*pi*R^(2)*sigma*T^(4).
So, L/Area = [4*pi*R^(2)*sigma*T^(4)] / [4*pi*R^(2)] =
sigma*T^(4). So, Surface Flux (F) = sigma*T^(4) .
(b) What is the ratio of the energy emitted in one second from one square centimeter of the surface of an O-star (T = 29,000 K) to that emitted in one second from one square centimeter of the sun (T = 5800 K)?
ANS: The energy emitted in one second from one square centimeter is just the surface flux F. By part (a), F = sigma*T^(4).
So, F(O-star)/F(sun) = [sigma*T(O-star)^4] / [sigma*T(sun)^4] =
[(29,000)^4] / [(5800)^4] = 5^4 = 625.
(c) Generally, the flux at any distance D from a star is just the total energy emitted per second by the star divided by the surface area of a sphere of radius D. What is the ratio of the flux at distance D to the surface flux of a star whose radius is R?
ANS: The ratio of the two fluxes is just the ratio of (luminosity/area) for the two spheres, one with radius R and the other with radius D. But the luminosity L must be the SAME for the two spheres - that is, the same amount of total energy passes through both spheres.
So, Flux(at radius D)/Flux(at radius R) = [L / (4*pi*D^2)] / [L / (4*pi*R^2)] =
(R/D)^2. (... note, L and the constants cancel out.)

3. Stars are not Perfect Blackbodies (3 points) -
(a) Define the term `blackbody'(... in two sentences or less).
ANS: A blackbody is an ideal body that absorbs all radiation that falls upon it. In steady-state conditions (that is, at constant temperature) the blackbody emits as much radiation as it absorbs.
(b) Explain why the sun's energy output as a function of wavelength cannot be exactly reproduced using a single-temperature blackbody curve (see Fig. 4.2).
ANS: The radiant energy from the sun is produced at different depths in the photosphere, ranging in temperature from about 4500 K up to about 8000 K. Thus, the sun cannot be described accurately as a single-temperature body. Also, the suns spectrum contains absorption lines, which cannot be reproduced with a blackbody curve.
(c) By Wien's Law, a blackbody of T = 5800 K emits most of its radiation at a wavelength of 5000 Angstroms, which is in the green part of the visible band. In that case, why doesn't the sun's light look green?
ANS: Even though the radiation peaks at green wavelengths, the sun's light is an admixture of light from a broad range of wavelengths (or 'colors'). If you mix predominantly green light with some light of other colors, then you dont see green light. Also, the short wavelength (blue and blue-green) light from the sun is scattered out of the line-of-sight by dust in the Earth's atmosphere, making the sun appear redder than if the light were unobstructed.

4. Bolometric Correction and Bolometric Magnitude (3 points) -
(a) Why can the bolometric correction (BC) of a star never be a positive number?
ANS: BC = M(bol) - M(V) . If BC were positive, then M(bol) would be larger than M(V). Since M(bol) is a magnitude, this would imply that the star would be FAINTER at all wavelengths combined than it is at visual wavelengths. This is impossible, since the bolometric magnitude includes the light from visual wavelengths (as well as non-visible radiation such as infrared and ultraviolet).
(b) Which type of star would have a more significant bolometric correction, a star like the sun or a hot O-type star (T = 30,000 K)?
ANS: By Wien's Law, the O-star emits most of its light in the ultraviolet while the sun (T = 5800 K) emits most of its light in the visible. Thus, the BC would be more significant for the O-star.
. (c)* If a star has a bolometric magnitude Mbol(*) = +3.75, what is the ratio of the star's true luminosity to that of the sun (... that is, L(*)/L(sun) = ? ). For the sun, use Mbol(sun) = +4.75.
ANS: Using the equation on page 62 of Aller,
log [L(*)/L(sun)] = 0.4 (4.75 - 3.75) = 0.4 (1) = 0.4.
Thus, L(*)/L(sun) = 10^(0.4) = 2.512.

5.* Stellar Spectra and Rotation (3 points) -
Suppose that a star which is NOT rotating emits an emission line at a sharply-defined wavelength of 6000 Angstroms. Sketch what this emission line would look like if the star were rotating like a solid body, with a surface rotation speed of 300 km/sec. Assume you are observing the star in its equatorial plane. Note: `Solid body rotation' means the star is rotating rigidly, like a record on a record player. The rotation speed of a particle on the other with radius R. star's axis of rotation is then zero, and the speed increases at a constant rate as you move away from the axis, reaching a value of 300 km/sec at the surface of the star.
ANS: The emission line will be broadened by Doppler effects. Gas on the part of the stellar surface that is coming toward the observer will contribute blueshifted emission lines and gas on the part of the surface that is moving away will contribute redshifted emission lines. The maximum blueshift/redshift can be found by the Doppler formula, noting that 300 km/s = 0.001 c. This gives:
(lambda - 6000)/6000 = 0.001 , so lambda = 6006 Angstroms (maximum redshift). Similarly, the maximum blueshift will be (6000 - 6) = 5994 Angstroms. Gas moving at intermediate radial speeds between 0 and 300 km/s will produce blueshifts/redshifts of less than 6 Angstroms. So, the resulting emission line will be broadened into a broad emission line spanning a wavelength range 5994 - 6006 Angstroms.

* Problems marked with an asterisk require a numerical calculation.