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Spring 2005
Dr. Mike Fanelli
Solutions to Assigned Problems
Chapter 18
PROBLEM 18-1:
Given the density of gas within the “Local Bubble”, a region of space surrounding
the solar system, you are asked to determine the total mass within a volume
equal to the size of Earth.
ANSWER: You are given a density (the number of hydrogen atoms per volume)
and asked to determine the total amount of mass within that volume. The volume
is the volume of Earth. Given the definition of density,
Density = total mass  volume,
rearrange this expression to get the total mass.
Remember that the volume of a sphere = 4/3    R3, and the radius of Earth is
6400 kilometers.
Multiplying each side of the expression by volume gives:
Total mass = density  volume
= # of atoms  mass of 1 atom  volume of Earth
= 103 atoms per meter3  1.7  10-27 kilogram per atom
 volume of Earth
-24
= 1.7  10 kilogram per meter3  4/3    R 3
= 1.7  10-24 kilogram per meter3  13.51  (6.4  106 meters)3
= 1.7  10-24 kilogram per meter3  1.1  1021 meters3
= 0.0019 kilograms or 1.9 grams (!)  A very small mass.
PROBLEM 18-10:
To ionize interstellar hydrogen, a photon must have a wavelength smaller than
91.2 nanometers (9.12  10-8 m). Assuming a star had its peak wavelength at
this value, what is the surface temperature of this star ?
ANSWER: Use Wien’s law from chapter 3, to relate peak wavelength to the
peak temperature. Note that you will need to convert the wavelength to
centimeters in order to use the expression in More Precisely 3-2. The result will
then be expressed in units of degrees Kelvin.
Wien’s law states: (max) = 0.29  T, where (max) is the peak (maximum)
wavelength of emitted radiation, measured in centimeters, and T is the star’s
surface temperature, expressed in K. Taking the peak wavelength as defined in
the problem:
9.12  10-8 m
= 0.29  T,
converting to centimeters, 9.12  10-6 cm = 0.29  T,
rearranging,
Temp = 0.29  9.12  10-6 cm = 31,800 K
A star of this temperature is an O star, O7-O9 spectral class.
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