# Solutions 11

- Page ID
- 47388

### Q11.1

Calculate the wavelength of the highest energy Balmer emission line of hydrogen.

*The equation for the emission wavelength of hydrogen is: *

*\[\frac{1}{\lambda} = R_H(\frac{1}{n_f}^2 - \frac{1}{n_i}^2)\]*

*where \(R_H = 109737.32 cm^{-1}\), \(n_f = 2\) and, \(n_i = infinity\). *

*\[\frac{1}{\lambda} = 109737.32 cm^{-1}(.25 - 0)\]*

*\[\lambda = 3.65x10^{-5} cm = 365 nm\]*

### Q11.2

Calculate the wavelength of an ultraviolet transition in the *Lyman *emission series of Hydrogen gas from the n = 2 level and to the n= 1 level.

*\[\frac{1}{\lambda} = R_H(\frac{1}{n_f}^2 - \frac{1}{n_i}^2)\]*

*\[\frac{1}{\lambda} = 109737.32 cm^{-1}(1 - .25)\]*

*\[\lambda = 1.22x10^{-5} cm = 122 nm\]*

### Q11.3

Calculate the energy of a photon **absorbed **for an electron to jump from a n=3 and n=4 state of atomic hydrogen? Calculate the energy of a photon emitted for an electron to jump from a n=4 and n=3 state of atomic hydrogen?

*a) \[\frac{1}{\lambda} = R_H(\frac{1}{n_f}^2 - \frac{1}{n_i}^2)\]*

*\[\frac{1}{\lambda} = 109737.32cm^{-1}(\frac{1}{9} - \frac{1}{16})\]*

*\[\frac{1}{\lambda} = 5334.45cm^{-1}\]*

*\[\lambda = 1.875x10^{-6}m\]*

*\[E = \frac{hc}{\lambda} = 1.06x10^{-19} or .66 eV\]*

*b) Same answer as a.*

### Q11.4

Calculate the wavenumber of the wavelength of the light emitted from the \(n=8\) to \(n=6\) transition.

*\[\nu = \frac{1}{\lambda} = R_H(\frac{1}{n_f}^2 - \frac{1}{n_i}^2)\]*

*\[\nu = \frac{1}{\lambda} = 109737.32cm^{-1}(\frac{1}{36} - \frac{1}{64}) = 1333.86 cm^{-1}\]*

### Q11.5

Explain what the expected emission spectrum would be for atomic hydrogen if continuum mechanics (i.e., classical mechanics) were applicable for describing it.

*The emission spectrum of atomic hydrogen without quantization would be a continuous broadband spectrum. *