AY 2 Home Work 4 SOLUTION


There is a correction to the units listed on the homework:
E=h n where h=6.626 x 10-34 Joule s
in other words, the units of h are J s (Joule sec)
Also, please note that in the formula: v/c = (lshifted- l0) /l0
l0 is the same as llab
and lshifted is the same as lobserved.

  1. Vega is a bright nearby star. Let's say that the wavelength at which Vega emits most of its light is 3052 Angstroms.
    a) What is the surface temperature of Vega?

    lmax T = 2.9 x 10-3m K (Wien's Law)
    Please note the "K" is a unit of temperature. Unlike the units Celcius and Fahrenheit, the unit Kelvin does not use a superscript of the circle symbol to denote "degrees".
    T = 2.9 x 10-3m K /lmax
    convert units of wavelength:
    (3052 Angstroms)(10-10m/1 Angstrom)=3.052 x 10-7m
    T = 2.9 x 10-3m K / 3.052 x 10-7 m (units of meters cancel out)
    T = 9502 K

    The sun has a surface temperature of 5860 K.
    b) What is the wavelength at which the sun emits most of its light?

    lmax = 2.9 x 10-3m K / T
    lmax = 2.9 x 10-3m K / 5860 K (units of K cancel out)
    lmax = 4.945 x 10-7m = 4948 Angstroms

    c)What can you say about the color of the sun compared to the color of Vega?

    The Sun is cooler than Vega. The Sun's peak wavelength is longer than Vega's peak wavelength, so its color appears redder than Vega (i.e. Vega appears bluer than the Sun).

  2. Rigel and Betelgeuse are two bright stars in the constellation Orion. (FYI, you can see a picture of the constellation on p. 40 of the text, or look up at night towards the West before 8:30 pm.) Betelgeuse is red, and Rigel is blue.
    a)Which star is hotter?
    The bluer star is hotter, so Rigel is hotter than Betelgeuse.

    If two stars were located at the same distance, and if the blue star appeared fainter than the red star.
    b)Which star, the blue star or the red star, would have a smaller radius?
    The color of the star tells you about the temperature of the star. The blue star has a higher temperature than the red star.
    Both stars are at the same distance. Apparent Brightness = Luminosity / 4 pD2 , (where D is the distance to the star)
    Luminosity depends both on the radius of a star and on its temperature.
    L=4 p s R2 T4, (where p and s are just constants, R is radius and T is Temperature)
    Substituting the above Eq'n for Luminosity into the Eq'n for Apparent brightness, one can see that if the two stars are located at the same distance, and if the blue star appears fainter (i.e. less bright) than the red star, then the blue star must have a smaller radius than the red star.

  3. Using data from Table F.1 and F.2 in the Appendix section of your text "The Cosmic Perspective," we see that a Centauri is a binary star i.e. it consists of two stars orbiting around each other. Inspecting the second star of the pair a Centauri (spectral type K0 V), it has a luminosity that is just over half the luminosity of the sun
    (L a Centauri =0.53 Lsun). Using additional data from the table
    which star will appear brighter to us on Earth: the Sun or a Centauri (spectral type K0 V)?
    The Sun appears brighter to us on Earth because it is more luminous and because it is so much closer to us.

    How much brighter will that star appear?
    Apparent Brightness = Luminosity / 4 pD2
    This question is basically asking you to take into account that the two stars are located at different distances. So take the ratio of the apparent brightness of the sun and the apparent brightness of a Centauri (spectral type K0 V).
    So first write two equ'ns- one for the brightness of each star.
    Bsun = Lsun / 4 pDsun2
    Ba Cen = La Cen / 4 pDa Cen2
    Then plug in La Cen = 0.53 Lsun and divide Bsun by Ba Cen
    Bsun / Ba Cen = Da Cen2/ 0.53 Dsun2 ( the 4 p and Lsun all cancel out)
    Bsun / Ba Cen = (4.4 ltyr)2/ 0.53 (0.000016 ltyr)2
    Bsun / Ba Cen = 19.36 / (0.53)(2.56 10-10)
    The Sun appears 1.42 1011 times brighter than a Centauri (spectral type K0 V).

  4. In the hydrogen atom, a transition of the electron from one energy level to another energy level has a rest wavelength of 4340 Angstroms. Suppose you see this line in the spectrum of a star with a wavelength of 4342 Angstroms.

    a) Is the star moving away or moving towards the Earth?

    The observed wavelength of light is longer than the rest wavelength (observed in a lab on Earth), so the star must be moving away from Earth.

    b)What is the velocity of that star?

    v/c = D l/ l0
    where v is the radial velocity of the object (the ambulance or a star) moving away or towards us
    where l0 = lrest
    and where D l is the doppler shift in wavelength lshifted- lrest
    c=300,000 km/s
    v/c = (4342 Ang - 4340 Ang) / 4340 Ang = 2 / 4340 = 0.00046
    v=(0.00046)(3 105km/s)=
    v=138 km/s

  5. Imagine a hydrogen atom, with an electron in an "excited" state. The electron absorbs light of wavelength 4340 Angstroms and is further excited.
    a)What is the energy of this wavelength of light? in units of Joules and in units of eV?

    l= c/n
    E=h n
    E=hc/l
    convert Angstroms to units of meters:
    (4340 Angstrom)(10-10m / 1 Angstrom) = 4.34 x 10-7 m
    plug into energy eq'n: E=hc/l=(6.626 x 10-34 kg m2/ s2)(3 x 108m/s) / 4.34 x 10-7 m
    E=4.58 10-19 J
    convert to Joule to units of eV: E=(4.58 10-19 J )(1 eV / 1.6 x 10-19 Joule) = 2.86 eV

    b) Energy is gained by the electron when it absorbs this light.

    c)How much energy is lost or gained by the electron?
    exact same answer as 5 a. i.e. the energy gained by the electron is the energy of the light it absorbs.

    d)If the electron originally has an energy of 10.2 eV, what is its energy in eV after absorbing the photon of light at 4340 Angstroms?
    10.2 eV + 2.86 eV = 13.06 eV