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Astronomy 114
Due:
Problem Set # 3
26 Feb 2007
Name: SOLUTIONS
1 A healthy person has a body temperature of 98.6◦ F. I mentioned in class that
the peak radiation is the infrared. Use the properties of blackbody radiation to
work out the precise wavelength of the peak radiation.
According to Wien’s Law:
λpeak =
2.9 × 106
nm.
T (K)
However, as we discussed in class, there are three different temperature scale:
Kelvin temperature scale TK ; Fahrenheit temperature scale TF and Centigrade temperature scale TC . The temperature in Wien’s law is Kelvin which
defines zero degrees to be absolute zero, the temperature at which there are
no vibrations.
The relations between these different temperature scales are:
TK = TC + 273.15
9
TF =
TC + 32
5
5
TC =
(TF − 32)
9
(1)
So to apply Wien’s law, we convert degrees Fahrenheit to degrees Kelvin
TK =
5
(TF − 32) + 273.15 K = 310.15K
9
Substituting this into Wien’s Law gives
λpeak = 9.35 × 103 nm.
Checking the electromagnetic spectrum from your textbook, this wavelength
belongs to the infrared.
2 The two stars have identical diameters. One has a temperature of 5800 K; the
other has a temperature of 2900 K. What are the colors of these stars? Which
had the larger luminosity? How much more luminous is it?
This problem uses both Wien’s Law and the Stefan–Boltzmann Law.
For TK =5800K:
λpeak = 500 nm.
Again comparing with the electromagnetic spectrum chart, this visible wavelength is green (of course, from discussion in class, we know that we perceive
this as white).
For TK =2900K:
λpeak = 1000 nm,
This star’s peak energy wavelength is in the near infrared (near meaning just
beyond visible).
The luminosity for a star is L = 4πR2 σT 4 . Since these two stars have same
diameters and therefore the same radii, the relative luminosity is proportional
to T 4 . This implies that the one with higher temperature (5800K), has larger
luminosity. Using our usual technique the ratio of luminosities:
L5800
5800K
=
L2900
2900K
4
= 24 = 16.
3 What is the temperature of a star with a wavelength of maximum light of 290
nm (recall: nm is the abbreviation for nanometers)?
More Wien’s Law:
λpeak =
so
T (K) =
2.9 × 106
nm
T (K)
2.9 × 106
K = 104 K.
λpeak ( nm)
4 Freedman & Kaufmann, pg. 120, Problem # 26
According to Wien’s Law
λpeak =
2.9 × 106
2.9 × 106
nm =
nm = 2.9 nm.
T (K)
106
Checking electromagnetic spectrum chart, this wavelength lies in the X-ray.
Those of you that keep up with astronomy news in the press will note that
much of the recent evidence for black holes as come from the Chandra X-Ray
Observatory.
5 Can an atom in its ground (lowest-energy) state emit a photon? Absorb a
photon? Explain.
A photon is emitted by the atom as the electron falls from a higher energy
level down to a lower energy level. A photon is absorbed by the atom if the
energy of the photon matches the energy between levels and causes electron
to jump from a low energy level to a high energy level.
For an atom in its ground state, which is the lowest energy level, it is impossible for electron to fall down lower energy and therefore can not emit a
photon. However, it can get to a higher energy level absorbing a photon.
6 Explain why absorption lines of an element have the same wavelengths as emission lines of an element.
Most of this explanation appears in the explanation for the previous problem.
The energy of an electron bound to an atom depends on its energy level.
When an electron decends an energy level, the energy of the photon emitted
depends the energy difference between the levels. Similar, as I said above,
an electron in a lower level can move to a higher level by absorbing a photon
whose energy is the difference between the two levels.
Mathematically, we can write this verbal explanation as follows: Assume
n=1 and n=2, two energy level, E2 > E1 , The energy of absorption line is
∆Ea =|E1 − E2 |, The energy of emission line is ∆Ee =|E2 − E1 |, thus they
have same energy, and the wavelength can be determined by:
∆E = hν = hc/λ.
So the absorption line and emission line have the same wavelength.