SCCC 116
Homework 1 - Due 26 January 2006
Review and Discussion: 17-11, 18-13, 18-14
Problem: 17-9, 18-10
Prof. Christina Lacey

R&D 17-11

The Hertzsprung-Russell (H-R) diagram is a plot of star's absolute magnitude against spectral type. Each star to be plotted must have its spectral type determined. The apparent magnitude must be observed and the distance determined by some method such as parallax so that the absolute magnitude can be determined. Absolute magnitude is on the vertical axis, with the brightest stars at the top of the scale. Spectral types (or temperature) are plotted on the horizontal axis with O type stars (hottest) stars on the left and M type stars (coolest) on the right.

R&D 18-13

21 cm radio emission is emitted by cold, neutral hydrogen gas anywhere in the Galaxy. This wavelength passes through interstellar dust clouds without being scattered, so the entire Galaxy can be observed. The temperature and density of the entire Galaxy has been measured with 21 cm radio radiation.

R&D 18-14

21 cm radio emission is emitted by cold, neutral (not ionized) hydrogen gas anywhere in the Galaxy. But 21 cm radio radiation also passes through interstellar dust clouds without being scattered, so the entire Galaxy can be observed. Because it is also an emission line, it can be used to measure the velocities of the clouds that emit it, allowing astronomers to study the dynamics (motions) of the Galaxy.

P 17-9

What is the absolute magnitude, M, of a star with apparent magnitude, m=4.0, and distance D=100 pc?

$m-M=5\log{\frac{D}{10~pc}}$

$M=m-5\log{\frac{D}{10~pc}}=4.0-5\log{\frac{100~pc}{10~pc}}= 4.0-5\log{10}=4.0-5(1)=-1.0$

P 18-10

To ionize the electron in a hydrogen atom you need a photon with a wavelength, $\lambda<9.12 \times 10^{-8}~m= 91.2 nm$. Thus, the maximum wavelength (minimum energy) needed to ionize an electron is $\lambda < 9.12 \times 10^{-8}~m$. Use Wien's Law to calculate the temperature of a star with a peak energy to ionize hydrogen.

Wien's Law: $\lambda_{peak}=\frac{0.0029m}{T(K)}$

$T= \frac{0.0029m}{\lambda_{peak}}=\frac{0.0029m}{9.12 \times 10^{-8}~m}=3.2\times 10^{4}~K=32,000~K$

A temperature of at least 32,000 K is required to ionize hydrogen. Thus a star needs to have a temperature of at least 32,000 K in order to ionize hydrogen, so an O or B type star is needed.