Download Solutions to Homework# 2 Chapter 3 : Review and Discussion

Solutions to Homework# 2
Chapter 3 : Review and Discussion
3. The longer the wavelength, the lower the frequency; the shorter the wavelength, the higher the
frequency. Wavelength and frequency are inversely related.
9. A star contains many charged particles that are moving. This motion creates waves in the electric
fields of the charged particles and these waves propagate or move outward and away from the star.
Traveling at the speed of light, a few of these waves will finally reach a person’s eye, which also
contains charged particles. The waves make the charged particles move, and this motion is sensed by
nerves and transmitted to the brain as an image of the star.
13. A blackbody is an idealized object that absorbs all radiation falling on it. It also re-emits all this
radiation. The radiation emitted occurs at all wavelengths but peaks at a wavelength that depends on
the temperature of the blackbody. The hotter the temperature, the shorter the wavelength of the peak
radiation.
17. The Doppler Effect is the observed change in the wavelength (or frequency) of a wave due to the
motion of the emitter, observer, or both, towards or away from each other. If the motion is towards
each other, the observed wavelength appears shorter than the wave emitted. If the motion is away from
each other, the observed wavelength appears longer than the wave emitted.
Chapter 3 : Problems
2. The relationship between frequency, wavelength, and wave velocity is λf = v.
The frequency is 100 MHz or 108/sec and v = 3×108 m/sec. ∴ λ = 3 m.
6. Wien's law states that the peak wavelength is inversely proportional to the temperature.
Comparing 200 to 650 nm gives a factor of 3.25; therefore, the object with a peak wavelength of 200
nm must be 3.25 times hotter than the object that peaks at 650 nm.
9. λ max = 2900 nm/T(1000°K) = 2900 nm or 2.9 μm, which is in the infrared.
11. The Sun’s temperature is 5778°K. Using this in Stefan’s Law gives,
E = σT4 = 5.67×10-8 × (5778)4 = 6.32×107 J.
Its radius of 6.96×108 m will give a surface area of
A = 4π R2 = (6.96×108)2 = 6.09 × 1018 m2.
Multiplying these two results gives the total solar luminosity of 3.85×1026 W.
15. The true wavelength must be exactly 3.00000 m. Using the Doppler formula
3.00036
v
= 1+
3.00000
300, 000
v = 36 km/s
The spacecraft is moving a distance equal to one circumference in this speed. The period of the orbit
can therefore be determined. Using velocity = distance / time, time = distance / velocity. 2π ×
(100,000 km) / 36 km/s = 17,450 s. To use Kepler’s third law, change units to years and A.U. There
are 3.15×107 seconds in a year, so this period is 0.000553 yr. The orbital radius is
100,000 / 150,000,000 = 0.000667 A.U. Kepler’s third law gives:
0.0006673
0.0005532 =
M
M = 0.000969 solar masses or M = 1.93 × 1027 kg or 323 Earth masses or about 1 Jupiter mass.
Chapter 4 : Review and Discussion
2. A simple spectroscope is made up of a slit, a prism, and an eyepiece or screen. The slit defines a
narrow beam of light. The prism spreads the light out into its various wavelengths or colors. The
eyepiece or screen allows the spectrum to be observed.
8. The hydrogen atom has one proton in its nucleus and one electron moving around it. The electron is
found in one of many possible energy levels or orbitals.
13. According to Kirchhoff’s first law, a luminous solid liquid or dense gas will emit light of all
wavelengths and produce a continuous spectrum.
18. When an atom produces a spectral line, the wavelength observed depends on the motion of the
atom. The Doppler Effect tells us that an atom moving towards us will produce a line that is observed
to be shifted to shorter wavelengths; an atom moving away will produce an observed wavelength that is
longer. In any hot gas there are atoms moving in all directions; the hotter the gas the faster they move.
The net result is a broadening of the spectral line. Mass motions of the gas and stellar rotation will also
produce broadening of the line in much the same way.
Chapter 4 : Problems
3) 2 eV red photon: = 3.2×10-19 J; = 4.8×1014 Hz; = 620 nm.
0.1 eV infrared photon: 1.6×10-20 J; = 2.4×1013 Hz; = 12,400 nm or 12.4 µm.
5000 eV photon: 1240/5000 = 0.25 nm. See More Precisely 4-1.
7a) Carefully note that the transition is from the tenth to the ninth excited state, so the value of n
changes from 11 to 10. Using the formula in the More Precisely 4.2, the energy in the eleventh level is
E11 = 13.6(1-1/112) eV, E11 = 13.488 eV. Likewise, E10 = 13.464 eV.
∴ ∆E = 0.024 eV = 1.6×10-19 × 0.024 = 3.84 ×10-21 J.
Using E = hf, the frequency of this wave can be calculated. 3.84 ×10-21 J = 6.63×10-34 J s × f.
∴ f = 5.79 ×1012 Hz.
Converting this frequency to wavelength, using c = fλ gives 3×108 m/s = 5.79 ×1012 Hz × λ.
λ = 5.18×10-5 m = 52.0 µm, which is in the infrared.
b) Because the energy of the two levels are so similar, the calculation needs to be carried out to more
decimal places. E101 = 13.59867 eV, E100 = 13.59864 eV. ∴ ∆Ε = 3× 10-5 eV = 4.8×10-24 J.
∴ 4.8×10-24 J = 6.63×10-34 J s × f & f = 7.24×109 Hz. This is 7240 MHz which is in the radio range. Its
wavelength is 3×108 m/s = 7.24×109 Hz × λ. ∴ λ = 0.041 m or 4.1 cm.
c) E1001 = 13.599986427 eV & E1000 = 13.599986400 eV, ∴ ∆E = 2.7×10-8 eV = 4.32×10-27 J.
∴ 4.32×10-27 J = 6.63×10-34 J s × f. ∴ f = 6.52×106 Hz. This is 6.52 MHz, which is in the short wave
radio range. Its wavelength is 3×108 m/s = 6.52×106 Hz × λ. ∴ λ = 46.0 m.
11) The H-α line has a wavelength of 656.3 nm. Using the Doppler formula from Chapter 3 gives
656/656.3 = 1 + v / (3×108 m/s). v = -5.94×105 m/s or -137 km/s.
13) The Doppler equation given in Chapter 3 can also be expressed as
change in wavelength
velocity
=
true wavelength
speed of light
change in wavelength
12 km/s
=
656.3 nm
300,000 km/s
change in wavelength = 0.026 nm
total width = 0.052 nm or 0.00008 of the original wavelength
15) From Problem 8, the thermal broadening is 12 km/s. Using this as the Sun’s equatorial velocity,
and a distance of one circumference traveled in one rotation period gives:
2 × π × 700, 000 km
12 km/s =
P
P = 370,000 s = 4.2 days
Rotation = 0.24 revolutions per day