Homework on Light; newest version with solutions added.

This is the final up-to-date version of this assignment with solutions added. This assignment was due on Friday (March 6).

Ch 26. Problems: 3 (1st part only), 5, 6, 7
[hint for 5: radio waves travel at the speed of light=3 x 10^8 m/s. time is distance/speed.
For 7, someone in the class got a wavelength to atom size ratio of about 10,000. Does that seem about right? Feel free to post your result and comment.]

3. time =distance/speed of light = 500 sec

5. time =distance/speed of light = 1.4 x 10^8 sec (a much longer time than 500 sec)

6. f=c/lamda = 5 x 10^14 sec-1 . It is important to get the exponent right!

7. lamda =c/f = 0.5 x 10^6 meters = 500 nm. In these problems (6 and 7) it is important to get the exponents right and to be mindful of any nm to meter conversions. Atom size is typically 0.1 to about 0.3 nm. So that is about 1000 times (or a little more) smaller than the wavelength of visible light.
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Ch 30: Exercises: 3, 5, 15, 48
Problem "1" (there is only one problem, it has no number)

3. blue light emitted corresponds to a greater change in energy of the electron in the atom.

5. If you double with wavelength, the frequency decrease by a factor of 2 (f=c/lamda), so the energy of a photon would be half as much. (Not twice.)

15. infrared is lowest frequncy range, UV is highest frequency and therefore has highest photon energies. Visible is in between IR (lowest photon energies) and UV (highest phton energies).

48. see attached jpg image. After you do this you can sort of see the solution visually from the picture!

Problem: One key thing for this problem is to realize that wavelength is inversely related to frequency, so if the frequency for a transition is higher, the wavelength must be shorter!

Balmer Series Problem: [Because this problem was posted late, it will count as extra credit and, because it is long, triple credit. I suggest you do it because it is relevant to things we will emphasize regarding both the quantum nature of the atom and spectroscopy.]

In the emission and absorption spectra of a hydrogen atom gas, sharp lines are seen at the following wavelenghts: 656 nm, 486 nm, 434 nm and 410 nm.

a) Determine the light frequency (in Hz) corresponding to each of these wave-lengths.

b) Sketch a spectroscopy set-up for an absorption spectrum, and then graph an absorption spectrum of intensity vs frequency. (This means intensity as a function of frequency, with intensity on the vertical axis and frequency on the horizontal axis. It is highly preferable to start the frequency axis at zero frequency; that way it is easier to see patterns of frequencies.) For your sketch of the spectroscopy set-up, you can limit it to a source, a source aperture, a prism, a moving aperture and a detector. (The book shows mirrors and that is more correct, but for your sketches you can skip the mirrors for simplicity.)
Remember that for an absorption spectrum there is a cold gas of atoms in the beam. Actually, there should be two graphs: one of the intensity vs frequency when there is no H-atom gas in the beam, which is called a reference spectrum; the other the actual absorption spectrum that you get when the H-atom gas in in the beam. (For the reference spectrum assume a broad, featureless spectrum spanning the entire visible range as well as some IR and UV.)

c) Sketch a spectroscopy set-up for an emission spectrum, and then graph an emission spectrum of intensity vs frequency for the same four line sequence. For an emission spectrum the source is a hot gas of H-atoms.

d) Using h= 4.14 x 10^-15 eV-sec, calculate the energy of a photon at each of these four photon frequencies in eV.