Physics 121, Problem Set #7 Solutions
Review Questions (Chapter 6)
4. Explain why ionized calcium can form absorption lines but ionized
hydrogen cannot.
Solution:
Unionized calcium contains 20 electrons while unionized hydrogen
contains one electron. If you ionize calcium and remove one of the
electrons, one of the remaining electrons will fill the outer most shell
and be capable absorbing photons, thereby producing absorption
spectra. Once ionized hydrogen, on the other hand, has no electrons.
Therefore, there are no electrons to make transitions and energy
cannot be absorbed by ionized hydrogen.
5.Describe two ways an atom can become excited.
Solution:
Atoms can become excited from the absorption of a photon or from
collisions. If a photon has an energy that is equal to the energy
between any two of the energy levels in an atom and the electron of
that atom is in the lower of these two levels, the atom can absorb the
energy of that photon and the electron will move to the upper energy
level. In a collision, some of the energy associated with the speed of
the two particles (i.e. the kinetic energy) can be absorbed by an
electron in a lower energy level and cause the electron to move to an
excited level.
10. Why are Balmer lines strong in the spectra of medium-temperature
stars and weak in the spectra if hot and cool stars?
Solution:
The visible Balmer lines of the hydrogen spectrum occur as a result of
transitions from the first excited level (n=2) to higher energy levels.
In cool stars, nearly all of the hydrogen is in the ground state so that
only transitions from n=1 to higher levels are observed. In hot stars
much of the hydrogen is ionized by the collisions that occur between
the atoms. Since the atoms now contain no electrons at all, hydrogen
can not produce an absorption spectrum. In the medium temperature
stars, the temperature is high enough that collisions keep a large
number of the hydrogen atoms excited to the first excited level (n=2).
In these stars there is a sufficient number of hydrogen atoms in the
first excited level to readily absorb the visible light photons associated
with the Balmer spectrum of hydrogen.
Problems (Chapter 6)
1. Human body temperature is about 310 K (98.6 Fahrenheit). At what
wavelength do humans radiate the most energy? What kind of
radiation do we admit?
For this problem we will use the black body radiation equation given on
page 98. And to determine the type of radiation we will use figure 52 on page 70.
3000000 3000000
λ max =
=
T
310
λ max = 9680nm
This falls in the Infrared region.
5. If one star has a temperature of 6000 K and another star has a
temperature of 7000 K, how much more energy per second will the
hotter star radiate from each square meter of it’s surface?
The energy given off per square meter per second is given off by the
Stefan-Boltzmann law found on page 98 (by the numbers 6-1).
E = σ ×T
EA
σ × T4
A
=
EB
( )
7000
6000
4
σ × T4
B
=
 T A
 
 TB
4
4
= 1.9
Therefore the hotter star is giving of 1.9 times more radiation per
square meter.
7. Determine the temperature of the following stars based on their
spectra. (use Figures
6-7, and 6-9).
a) medium-strength Balmer lines, Strong helium lines
b) medium-strength Balmer lines, weak ionized-calcium lines
c) Strong TiO bands
d) very weak Balmer lines, strong ionized-calcium lines
It is easer to use figure 6-7 in conjunction with table 6-1 to find the
values
a) 25,000 K (B class star), b) 8,000 K (A class star), c) 3,000 K (M
class star), d) 4,500 K (K class star)