Atomic Spectra and Bohr Model (5)
Calculate DE for e- in H “falling” from
n = 2 to n = 1 (higher to lower energy) .
so, E of emitted light = (3/4)R = 2.47 x 1015 Hz
and l = c/n = 121.6 nm (in ULTRAVIOLET region)
DE = Efinal - Einitial = -C[(1/12) - (1/2)2] = -(3/4)C
C has been found from experiment. It is now called R,
the Rydberg constant. R = 1312 kJ/mol or 3.29 x 1015 Hz
This is exactly in agreement with experiment!
- (-ve sign for DE indicates emission (+ve for absorption)
- since energy (wavelength, frequency) of light can only be +ve it is best to consider such calculations as DE = Eupper - Elower