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Understanding Atomic Energy Levels and Stellar Spectra - Prof. Paul Camp, Assignments of Astronomy

How electrons in atoms can only occupy certain orbits, leading to specific energies and wavelengths of absorbed or emitted light. The concept of spectral lines and their relation to atomic transitions, as well as the connection between a star's temperature and the wavelength of its peak emission (wien's law). The document also touches upon the differences in size and luminosity between various stars, such as capella and regulus.

Typology: Assignments

Pre 2010

Uploaded on 08/04/2009

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Homework Solutions
Chapter 12
4. In atoms, electrons may only occupy certain orbits, not any possible
orbit. This means that there are only certain allowed energies that the
electrons may absorb or emit, corresponding to the amount of energy
gained or lost when an electron moves from one allowed orbit to
another. By absorbing light of the right energy, an electron moves to a
higher energy orbit. When it reemits that light, the electron moves
back down to a lower energy orbit. Since the energy of light is
determined by its wavelength, the existence of a discrete set of orbits
means that only certain energies may be absorbed or emitted which
means that only certain wavelengths of light may be absorbed or
emitted. The Sun’s spectrum shows a lot of absorption lines, each of
which corresponds to the specific wavelength of light that is absorbed
by a specific atom or molecule.
5. When electrons in the helium atom move between the lowest
energy orbit and higher orbits, the energy changes correspond to
spectral lines that are not in the visible part of the spectrum. To
produce visible absorption lines, helium must first be excited to the
second orbit by some other process and then it may absorb and emit
light in the visible part of the spectrum. It reaches the second orbit by
energy absorbed from collisions with other atoms, but the amount of
energy available in a collision depends on the temperature of the gas.
In the photosphere, the gas is not hot enough to lift a significant
number of helium atoms into the second higher orbit, so there is no
significant absorption in the visible band.
7. If we model the Sun as a black body radiator, its continuous
spectrum must peak (that is, be brightest) at a specific wavelength
(see the picture on pg. 259 of your book). This wavelength is related
to the temperature by Wien’s law. So you measure the intensity of the
light at each wavelength, graph it to find the peak of the curve,
determine what wavelength corresponds to that peak intensity, and
apply Wien’s law to calculate the temperature. This is an estimate
since the Sun is not truly a black body, but it actually comes pretty
close so it is a pretty good estimate.
12. We can see the top of the convection flow in the form of
granulation in the photosphere. We saw a movie of this in class. Hot
material convecting upward reaches the photosphere, cools and sinks
back down. At the highest point of the convection flow, it produces
bumps in the photosphere which can be seen fairly easily.
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Homework Solutions Chapter 12

  1. In atoms, electrons may only occupy certain orbits, not any possible orbit. This means that there are only certain allowed energies that the electrons may absorb or emit, corresponding to the amount of energy gained or lost when an electron moves from one allowed orbit to another. By absorbing light of the right energy, an electron moves to a higher energy orbit. When it reemits that light, the electron moves back down to a lower energy orbit. Since the energy of light is determined by its wavelength, the existence of a discrete set of orbits means that only certain energies may be absorbed or emitted which means that only certain wavelengths of light may be absorbed or emitted. The Sun’s spectrum shows a lot of absorption lines, each of which corresponds to the specific wavelength of light that is absorbed by a specific atom or molecule.
  2. When electrons in the helium atom move between the lowest energy orbit and higher orbits, the energy changes correspond to spectral lines that are not in the visible part of the spectrum. To produce visible absorption lines, helium must first be excited to the second orbit by some other process and then it may absorb and emit light in the visible part of the spectrum. It reaches the second orbit by energy absorbed from collisions with other atoms, but the amount of energy available in a collision depends on the temperature of the gas. In the photosphere, the gas is not hot enough to lift a significant number of helium atoms into the second higher orbit, so there is no significant absorption in the visible band.
  3. If we model the Sun as a black body radiator, its continuous spectrum must peak (that is, be brightest) at a specific wavelength (see the picture on pg. 259 of your book). This wavelength is related to the temperature by Wien’s law. So you measure the intensity of the light at each wavelength, graph it to find the peak of the curve, determine what wavelength corresponds to that peak intensity, and apply Wien’s law to calculate the temperature. This is an estimate since the Sun is not truly a black body, but it actually comes pretty close so it is a pretty good estimate.
  4. We can see the top of the convection flow in the form of granulation in the photosphere. We saw a movie of this in class. Hot material convecting upward reaches the photosphere, cools and sinks back down. At the highest point of the convection flow, it produces bumps in the photosphere which can be seen fairly easily.

Chapter 13

  1. Refer to figure 13.15 for this problem. Remember that size increases along a diagonal from lower left to upper right. Luminosity class I is supergiants, furthest up in the right hand corner, so the MI star is the largest. Being furthest up (the direction of increasing luminosity), it is also the most luminous.
  2. (a) Capella is further toward the upper right corner and so is larger. Physically, it has roughly the same temperature but is brighter so it must be emitting light over a larger surface area. (b) Capella is further toward the upper right corner and so is larger. Physically, they have the same luminosity even though Regulus is hotter. In order to compensate for its lower temperature and still have the same brightness, Capella must be emitting light from a larger surface. (c) Vega is slightly further up the main sequence and so is somewhat more luminous, but there is not a lot of difference between them. (d) You have to look up Pollux online since it is not on the chart. Google it and you will find Pollux is a K0 star with luminosity class IIIb. This puts it on the same gray line (on figure 13.15) as Capella at class K which is roughly on the same horizontal line as Regulus. Since Pollux’s spectral class is much further toward the cool end, it would appear redder.
  3. These would be stars toward the lower left hand corner of the HR diagram. In order to be very hot while at the same time being very dim, they must be emitting light from a very small surface area. These are tiny stars and that is why they are called white dwarfs.
  4. For one thing, they are brighter. Being the same surface temperature means that every square meter of photosphere for each star is emitting the same amount of light. For giants and supergiants to be brighter, then, requires them to have more square meters of surface which means they are larger stars.