Observational Astronomy
1911.231: HW #5

1. Energy per photon is proportional to frequency and inversely proportional to wavelength.
The blue 430 nanometer has higher energy than the red l = 656 nanometer photon.

2. A bigger telescope better as measured by aperture (diameter of primary optical element) is better because it has
(1) better resolution (ignoring atmospheric effect)
(2) greater light gather power (most important for point sources)

One can argue that a larger focal length is better because it will result in higher magnification, but you will have a smaller field of view (i.e., plate scale).

3. We find angular resolution with the forumula:
q = 2.5 X 10⁵ l / D  in arcseconds

We substitute D = 0.4 m, l = 510 nm = 510 X 10^-9 m = 5.1 X 10^-7 m

q = 2.5 X 10⁵ X 5.1 X 10^-7m / (0.4m)
q = 0.33 arcseconds

Alternatively: 1.6 X 10^-6 radians

4. We know q = 1.0 arcsecond and l = 10 micrometers = 1 X 10^-5 meters

q = 2.5 X 10⁵ l / D
1 = 2.5 X 10⁵  X 1 X 10^-5 m / D
D = 2.5 m
2.5 meter aperture!

5. We have l = 121.6 nm = 1.216 X 10^-7 m

E = hc / l where h = 6.626 X 10^-34 J*s and c = 3.0 X 10⁸ m/s
Thus, hc = 6.626 X 10^-34 X 3.0 X 10⁸ J*m = 2.0 X 10^-25 J*m
E = 2.0 X 10^-25 J*m / (1.216 X 10^-7 m) = 1.64 X 10^-18 J
also E = 1.64 X 10^-18 J / 1.6 X 10^-19 = 10.2 eV

6. We have l_max = 1000 nm

l_max T = 2.9 X 10⁶ nm*K

Solving for T:
T = 2.9 X 10⁶ nm*K / l_max = 2.9 X 10⁶ nm*K / 1000nm = 2900 K

7. We have T = 3000K?

l_max T = 2.9 X 10⁶ nm*K

Solving for l_max:
l_max = 2.9 X 10⁶ nm*K / T = 2.9 X 10⁶ nm*K / 3000K = 967 nm

8. Telescope A: D = 0.1 m = 100 mm, f = 600 mm
Telescope B: D = 1.0 m, f = 9.0 m

(a) Telescope B has a larger focal length and focal ratio so Telescope A will have larger field of view

(b) Telescope A: f_obj = 600mm, mag = f_obj / f_eye = 600mm / 30mm = 20
Telescope B: f_obj = 9.0m = 9000mm, mag = f_obj / f_eye = 9000mm / 30mm = 300

(c) Would you want to use this eyepiece in Telescope B?
300 times magnification would have a narrow field of view
Most likely the angular resolution will be limited by environmental effects so that the image would be blurry at this magnification. Not very useful.

9. An electron dropped from level 4 to level 1 would emit a photon of energy 12.8 eV
E = 12.8 eV = 12.8 * 1.6 X 10^-19 J = 2.05 X 10^-18 J

E = hc / l = 2.0 X 10^-25 J*m / l

l = 2.0 X 10^-25 J*m / 2.05 X 10^-18 J = 9.76 X 10^-8 m
l = 97.6 nanometers Ultraviolet

10. NO, a 3 eV photon cannot absorbed because the energy must exactly match the energy difference between the ground state and another allowed energy level for this atom’s electron.

11. This energy is called ionization energy. From the figure it is 13.6 eV.
E = 13.6 eV = 13.6 * 1.6 X 10^-19 J = 2.18 X 10^-18 J

E = hc / l = 2.0 X 10^-25 J*m / l

l = 2.0 X 10^-25 J*m / 2.18 X 10^-18 J = 9.17 X 10^-8 m
l = 91.7 nanometers Ultraviolet

12.

Cloud 1: Sodium

Cloud 2: Hydrogen

Cloud 1 exhibits a redshift so it must be moving away.
Cloud 2 shows not Doppler shift so there is no discernable motion.