|
Problem On page 219: Calculate the photon energies corresponding to the first emission line in the fourth spectral series (n1 = 4) for atomic hydrogen.
Subject The Spectrum of Atomic Hydrogen, Fine: pages 179-182
Solution The spectrum of atomic hydrogen is given by the Rydberg equation at the top of page 180. The equation gives us the wavelength and frequency of the emission lines as:
|
n
c
|
= RH |
é
ê
ë |
|
1
n12
|
- |
1
n22
|
ù
ú
û |
, |
|
or equivalent:
| n = RH ·c |
é
ê
ë |
|
1
n12
|
- |
1
n22
|
ù
ú
û |
, |
|
where c = 3.00 ×108 m·s-1 is the speed of light, RH = 109677.58 cm-1 º 10967758. m-1 is the Rydberg constant, and n1, n2 are the two integers which specify which lines we are talking about. The number n1 gives the series, n2 is the number of the line within this series.
In this case, n1 = 4, which equates to the Brackett line (see table 5.4 on page 180). As the caption of the Rydberg equation (top of page 180) tells us, n2 > n1. Then, the first line has to be n2 = 5. Instead of a frequency we are asked for an energy. For this, we can express the photonenergy in terms of the frequency of the line using the Planck relationship on page 171:
Filling everything in into the Rydberg equation then gives us:
|
|
|
|
|
| E=h ·RH ·c |
é
ê
ë |
|
1
n12
|
- |
1
n22
|
ù
ú
û |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
