## AST 324: HOMEWORK 1 ANSWERS

1. Because as wavelength gets shorter frequency gets higher, the highest frequency of any band corresponds to the shortest wavelength. On your spectrum chart, we refer to the right-hand edge of each band, but we are given wavelengths and need to convert to frequency with the equation (lambda)(nu) = c, or (nu) = c/(lambda). We can then use the frequency (nu) to get energy for a photon E = h(nu). Below is a chart of the answers:

BANDSHORTEST WAVELENGTHHIGHEST FREQUENCY ENERGY PER PHOTON
radio0.1 cm3 x 1011 Hz 1.98 x 10-15 ergs
infrared10-4 cm3 x 1014 Hz 1.98 x 10-12 ergs
visible3 x 10-5 cm1015 Hz 6.6 x 10-12 ergs
ultraviolet10-6 cm3 x 1016 Hz 1.98 x 10-10 ergs
X-rays10-9 cm3 x 1019 Hz 1.98 x 10-7 ergs

Remember that Hz are sec-1 and E/s units are (gm cm2 sec-1) x (sec-1) = gm cm2 sec-2 = ergs, an energy unit (it is about the amount of energy a fly needs to do a push-up).

2. We would not want to make a telescope's reflecting surface out of a blackbody because a blackbody does not reflect. Therefore, it would not work at all. No light from a star would be focused at a point where one could measure it. The temperature of a blackbody with certain (lambda)max values can be found from the Wien's law: (lambda)maxT = 0.3 cm K. We rewrite as (lambda)max = (0.3 cm K/T), and again make a chart:

BANDLONGEST WAVELENGTHTEMPERATURE
infrared0.1 cm3 K
visible10-4 cm 3,000 K
ultraviolet3 x 10-5 cm 10,000 K
X-rays10-6 cm300,000 K
(gamma)-rays10-9 cm3 x 108 K

3. We have two wavelengths: one from a distant object, and one from something local, presumably not moving relative to us. From this we can calculate the redshift

z = ((lambda)obs - (lambda)em)/(lambda)em = (657.36 nm - 656.7 nm)/(656.7 nm) = 0.00101.

Note that the units (nm) cancel, and z is a dimensionless number. Note also that this number is positive, so the velocity is 'positive,' meaning that this galaxy is moving away from us. Now we can get the velocity, rewriting z = v/c as:

v = zc = 0.00101 x 3 x 105 km/s = 301.5 km/s.

We can use the simple form for relating v to z because z is very small. Now that we have a velocity, we could use Hubble's law to find the distance.

v = Hr, so r = v/H = 301.5 km/s/(30 km/s/Mly) ~~ 1 Mly.

(I didn't write all the digits, because we can't be that accurate here anyway.) However a few hundred km/s is typical for the random motion of galaxies independent of the Hubble expansion, so the most we can say about this galaxy is that it isn't very far away. It would be in the local group, in fact.

4. If a galaxy recedes at 60% = 0.6 times the speed of light, we need to use the relativistic Doppler formula. To see this, let us calculate it both ways:

 non-relativistic 1 + z = 1 + v/c = 1 + 0.6c/c = 1 + 0.6 = 1.6, relativistic 1 + z = [(1 + v/c)/(1 - v/c)]1/2 = [(1 + 0.6)/(1 - 0.6)]1/2 = [1.6/0.4]1/2 = 41/2 = 2.

Since these are so different, we can see that when the velocity is this high the approximate formula (nonrelativistic) doesn't work and we need to use the real formula. We now have 1 + z, and so we use the formula 1 + z = (lambda)obs/(lambda)em, and we know that this is the factor desired. Also, we can find z = 2 - 1 = 1.