Download ASTRONOMY 120

Astronomy 120
HOMEWORK - Chapter 4
Spectroscopy
Use a calculator whenever necessary.
For full credit, always show your work and explain how you got your answer in full, complete
sentences on a separate sheet of paper.
Be careful about units!
Please CIRCLE or put a box around your final answer if it is numerical.
If you wish, you may discuss the questions with friends, but please turn in your own hand-written
solutions, with questions answered in your own way.
1. Chaisson Review and Discussion 4.1
What is spectroscopy? Explain how astronomers might use spectroscopy to determine the
composition and temperature of a star. (3 points)
When light from a star is passed through a prism or similar device, a spectrum is produced.
Spectroscopy is the observation and study of these spectra. For example, by studying the
patterns of dark absorption lines in the spectrum of a star, a variety of information can be
determined, including composition and temperature.
2. Chaisson Review and Discussion 4.2
Describe the basic components of a simple spectroscope. (3 points)
The basic structure of a spectroscope consists of a slit, a prism, and an eyepiece or screen.
The slit turns the light into a narrow beam. The prism (or similar device) spreads the beam of
light out into its various wavelengths or colors. The eyepiece or screen then allows the
spectrum to be observed and photographed.
3. Chaisson Review and Discussion 4.3
What is a continuous spectrum? An absorption spectrum? (3 points)
A continuous spectrum is emitted by a “blackbody.” It consists of light of all wavelengths,
but the amount of light emitted at each wavelength varies and depends on the temperature
of the blackbody. An absorption spectrum is created when the light from a blackbody
shines through a cooler gas. It looks like a continuous spectrum, but specific wavelengths
of light are missing, creating gaps called spectral lines. These dark vertical bands can be
quite narrow or very broad, and are connected to elements found in the object emitting the
spectrum.
4. Use Kirchhoff’s rules to predict what would happen if the emission-line spectrum from a hot,
transparent gas passed through another transparent gas that was hotter and had a different
chemical composition. (3 points)
You would see both emission spectra superimposed. The emission lines from the original hot
transparent gas would only be absorbed by the same elements at a cooler temperature – so
you see both.
5. Arrange the following kinds of electromagnetic radiation in order from the least to the most
energetic: x-rays, radio, ultraviolet, infrared. (2 points)
Radio, infrared, ultraviolet, and x-rays
6. Arrange the following kinds of electromagnetic radiation in order from longest to shortest
wavelengths: gamma rays, radio, ultraviolet, infrared. (2 points)
Radio, infrared, ultraviolet, and gamma rays
7. Chaisson Review and Discussion 4.4
Why are gamma rays generally harmful to life-forms, but radio waves generally harmless? (3
points)
Because of their short wavelengths and higher frequencies, gamma rays carry much more energy
in every photon than radio waves. As they pass through cells, gamma rays can break up important
molecules, such as DNA, and also ionize atoms to cause further cellular damage.
8. Chaisson Review and Discussion 4.9
What does it mean to say that a physical quantity is quantized? (2 points)
When a physical quantity is quantized it means that the quantity can take on only specific
discrete values rather than having an infinite number of possible values.
9. Chaisson Review and Discussion 4.10
What is the normal condition for atoms? What is an excited atom? What are orbitals? (4
points)
Normally the number of electrons in an atom equals the number of protons in the nucleus,
and the electrons are in their lowest energy level, the “ground state.” When an atom is
excited, an electron absorbs energy from an outside source and moves to a higher energy
orbital. The precisely-defined energy states or energy levels are referred to as orbitals.
They are the regions surrounding the nucleus that can be occupied by electrons.
10. Chaisson Review and Discussion 4.11
Why do excited atoms absorb and reemit radiation at characteristic frequencies? (4 points)
In order for a photon to be absorbed by an electron, it must have an energy that is
precisely equal to the energy difference between the energy level occupied by the electron
and a higher energy level. When the electron absorbs the photon, it moves immediately to
the higher energy level. Very quickly thereafter the electron moves back down to a lower
energy level by emitting a photon. The emitted photon must have an amount of energy
equal to the energy difference between the two levels. These discrete amounts of photon
energy translate into particular frequencies of radiation.
11. Chaisson Review and Discussion 4.16
How do molecules produce spectral lines unrelated to the movement of electrons between
energy levels? (3 points)
Molecules can rotate and they can vibrate. These two motions have quantized energy levels
just like the motions of electrons between orbitals. Changes in the rotational or vibrational
state of a molecule will therefore absorb certain amounts of photon energy from a source,
and produce specific absorption lines unique to each molecule.
12. Chaisson Review and Discussion 4.18
How can the Doppler Effect cause broadening of a spectral line? (4 points)
If an atom is stationary compared to an observer, it will produce spectral lines at the
wavelengths expected from that particular element. However, if there is relative motion
between the source and observer, the Doppler Effect will alter those wavelengths. An atom
approaching the observer will produce a line that is observed to be shifted to shorter
wavelengths, and an atom moving away will produce an observed wavelength that is
longer. In any hot gas there are atoms moving randomly in all directions; the hotter the gas
the faster they move. The net result is a broadening of the spectral line, since some atoms
will produce shifts to longer wavelength, others shifts to shorter wavelength, and some no
shift at all. Mass motions of the gas and stellar rotation will also produce broadening of
the line in much the same way.
13. Chaisson Review and Discussion 4.20
List three properties of a star that can be determined from observations of its spectrum. (3
points)
Many stellar properties can be deduced from the spectrum: Radial velocity of the star,
elemental abundances, temperature, rate of rotation, presence of turbulence, strength of
magnetic field, and atmospheric pressure.
14. Chaisson Problem 4.8
A distant galaxy is receding from Earth with a radial velocity of 3000 km/s. At what
wavelength would its Lyα line be received by a detector above Earth’s atmosphere? (4 points)
The Lyα line has a wavelength of 121.6 nm. Using the Doppler Effect formula:


obs  121.6 1 
3000 
  122.8 nm
300000 
15. The conservation of energy states that an object’s total energy must remain constant. If an
object’s potential energy decreases, its kinetic energy must increase. This is summed up by
the equation, Etot  K. E. P. E. where Etot is the total energy and K . E . is the kinetic energy
and P. E . is the potential energy. The Earth has a certain amount of gravitational potential
energy due to the Sun's gravity. If it suddenly started falling to the Sun, with what speed
would it crash into the Sun's surface? You solve this by realizing that all the potential energy
would be converted to kinetic energy just before impact.
K. E. =
1
 1 1
mplanet v 2 and P.E. = GMsunmplanet R - r 
2
the above equations are correct if:
r is the earth-sun distance in meters
R is the radius of the sun in meters
2
G is equal to 6.673  10 11 N  m
kg 2
M sun and m planet are in kilograms (kg)
v is velocity in meters per second ( m s )
Set K. E .  P. E . and solve for v . This would be the velocity of the Earth at impact.
Solution:
(1 point)
1
1 1
2


2mplanetv = GMsunmplanet R - r 
solve for :
 1 1
2
2 = GMsunmplanetR - r  x m


planet
 1 1


2 = 2GMsunR - r 
(2 points)
=
 1 1
2GMsunR - r 


Nm2
1
1

30 kg 
x
2
x
10

2
8
11
kg
7 x 10 m 1.5 x 10 m
(2 points)
=
(1 point)
 = 6.16 x 105 m/s or 616 km/sec!
2 x 6.67 x 10-11
16. Einstein found a relationship between the energy of a photon and the frequency:
E ph
= h  f
where E ph is the energy of the photon in Joules
h is a constant h  6.63  1034 J  s
f is the frequency in Hertz ( cycles sec )
a) What is the energy of a photon ( E ph ) of red visible light?
(2 points)
Eph = 6.63 x 10-34J s x 4.29 x 1014 Hz = 2.84 x 10-19 J
b) What is the energy of a photon ( E ph ) of X-ray light?
(2 points)
Eph = 6.63 x 10-34J s x 3 x 1018 Hz = 2 x 10-15 J
c) How much more energetic is an X-ray photon than a visibly red
photon? (Express it as a percentage)
(2 points)
EX-ray
2 x 10-15J
=
x 100 = 700,000% !!!
Ered
2.84 x 10-19J