Q2
The part of the hydrogen atom spectrum which occurs in the visible region arises from electrons in excited levels falling to the n = 2 level. The quantum mechanical expression for the frequencies in this case, corresponding to equation (3) of the text for the Lyman series, is: (4)
The energy of an emitted photon for a jump from level n to level 2 is: (5)
Equation (5) predicts that a plot of the photon energies versus (1/n2) should be a straight line. Furthermore, it predicts that the intercept of this line with the energy axis, corresponding to the value of 1/n2 = 0, i.e., n = ¥, should equal (¼)K where: (6)
The point of this problem is to test these theoretical predictions against the experimental results.
Experimentally we measure the wavelength of the emitted light by means of a diffraction grating. A grating for the diffraction of visible light may be made by marking a glass plate with parallel, equally spaced lines. There are about 10,000 lines per cm. The spacings between the lines in the grating d is thus about 1 ´10-4 cm which is the order of magnitude of the wavelength of visible light. The diffraction equation is: (7)
as previously discussed in Problem I-1. We measure the angle q for different orders n = 1, 2, 3, ... of the diffracted light beam. Since d is known, l may be calculated. The experimentally measured values for the first four lines in the Balmer series are given below.
Balmer Series
l(Å)
|
n
|
6563
|
3
|
4861
|
4
|
4341
|
5
|
4102
|
6
|
The value of the principal quantum number n which appears in equation (5) is given for each value of l. (This n is totally unrelated to the n of equation (7) for the experimental determination of l.) Calculate the energy of each photon from the value of its wavelength.
Plot the photon energies versus the appropriate value of 1/n2. Let the 1/n2 axis run from 0 to 0.25 and the energy axis run from 0 to 3.6 ev. Include as a point on your graph e = 0 for 1/n2 = 0.25, i.e., when n = 2, the excited level and the level to which the electron falls coincide.
- Do the points fall on a straight line as predicted?
- Determine the value of K by extending the line to intercept the energy axis. This intercept should equal K/4. Read off this value from your graph.
- Compare the experimental value for K with that predicted theoretically by equation (5). Use e = 4.803 ´ 10-10 esu, express m and h in cgs units and the value of K will be in ergs (1 erg = 6.2420 ´ 1011 ev). Recall that K is the ionization potential for the hydrogen atom. An electron falling from the n = ¥ level to the n = 2 level will fall only (¼)K in energy as is evident from the energy level diagram shown in Fig. 3-2.