Hydrogen’s sole electron can absorb varying amounts of energy. Imagine quadrillions of exited atoms. (How does one imagine quadrillions!?) In some atoms, for example, the excited electron may have only reached the n = 3 level. Upon returning to the n = 2 level, these atoms will emit light with a wavelength corresponding to the red colour. In the so-called Balmer series, most of the emissions are visible. These include the n = 6 to 2(violet), n = 5 to 2(blue), n = 4 to 2(blue-green), and as mentioned, the n = 3 to 2(red).

But can’t the electron return to its ground state? Yes, but the emissions form such transitions fall in the invisible ultraviolet spectrum. These are the so-called Lyman series.

For any series, the wavelength of the radiation emitted can be calculated by using two formulas:

(1)First to calculate the energy emitted, DE, from the electron:

DE = -R( 1/n_f² – 1/n_i²)

where R = rydberg’s constant = 2.178 X 10^-18 J

n_f = final energy level

n_i = initial energy level

(2) to obtain the wavelength associated with the energy:

l = hc/ -DE

where h = Planck’s constant = 6.626 X 10^-34 Js

c = speed of light = 2.9979 X 10⁸ m/s

To convert to angstroms, A, multiply by A/10^-10m

Balmer		E(J)	l (A)
nf	ni
2	6	-4.8400E-19	4.103E+03
2	5	-4.5738E-19	4.343E+03
2	4	-4.0838E-19	4.864E+03
2	3	-3.0250E-19	6.567E+03
Lyman
nf	ni
1	5	-2.0909E-18	9.500E+02
1	4	-2.0419E-18	9.728E+02
1	3	-1.9360E-18	1.026E+03
1	2	-1.6335E-18	1.216E+03