# Hydrogen emission spectrum diagram

## Homework Statement

Hydrogen atom with ionisation energy 13.6 eV is found to have an emission spectrum with lines at 1.89 eV, 10.20 eV and 12.09 eV. Draw a labelled diagram to show the energy levels of hydrogen atom by showing the transitions of electrons causing the emission of the lines stated. Label also the quantum number of the corresponding energy levels in your diagram.

## Homework Equations

E = hc/λ ---> λ = hc/E

and

1/λ = RH [1 - 1/n^2]

## The Attempt at a Solution

alright, now i understand how to draw an energy diagram, im just having trouble with this question:

thus far, i think ive correctly gotten to what the 'λ'are for this diagram:

using equation: E = hc/λ ---> λ = hc/E

1) 13.6 --> 1.89 = 1.8736x10^-18 J
λ = (6.626x10^-34)(3x10^8) divided by 1.8736x10^-18
λ = 106.095nm (Lyman Series)

2) 13.6 --> 10.2 = 5.44x10^-19
λ = (6.626x10^-34)(3x10^8) divided by 5.44x10^-19
λ = 365.4 nm (Balmer Series)

3) 13.6 ---> 12.09 = 2.416x10^-19
λ = (6.626x10^-34)(3x10^8) divided by 2.416x10^-19
λ = 822.76 nm (paschen series)

alright, assuming those are correct, im now having trouble finding the quantum numbers 'n'.... i made an attempt:

using : 1/λ = RH [1 - 1/n^2]

Lyman
1/(106.095x10^-9) = 1.097x10^7 [1 - 1/n^2]
n = 1.89

and i did that for the other 2:

Balmer: n=43.59
Paschen: n =56.25

buuut, now i dont really get what next.... or if the n is correct... as i know, the 'n' in an energy diagram, has gotta be whole number starting 2,3,4,5..... so im not sure what now... i mean, i get how to draw the diagram, i just dont know what the quantum numbers are... which the question is asking me to find.

^ah, forgot about that, the first one it's = R [1/1^2 - 1/n^2] which is 1/1 = 1 x 1/n^2 and what not, forgot to change the formula back when i was posting...

but im not sure if that's even correct... i mean, specially for the other 2, what will the n prime be? i don't know what the n is.... o:

ehild
Homework Helper
n and n' are the principal quantum numbers of those states between which the electron transition occurs, accompanied by photon emission. The energy of the photon is equal to the energy difference between the states. As energy is given, better to use the energy formula

hv=13.6(1/n'2-1/n2)

The photon energies can be converted to wavelength with the formula λ = hc/E, and you will see that the highest energy corresponds to a photon in the Lyman series, in the UV range of the spectrum: It means electron transitions to the n'=1 level from the n=2, 3, 4 ... ones. Find out n with the Rydberg formula. The lowest energy photon is in the visible range and belongs to the Balmer series, where the electron jumps on the n'=2 level from the n=3, 4 .... levels. Determine n.

ehild