Hydrogen emission spectrum diagram

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Homework Help Overview

The discussion revolves around the emission spectrum of a hydrogen atom, specifically focusing on the energy levels and transitions associated with given photon energies of 1.89 eV, 10.20 eV, and 12.09 eV. Participants are tasked with drawing a diagram to illustrate these energy levels and the corresponding quantum numbers.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of wavelengths from energy values using the formula E = hc/λ. There are attempts to identify the principal quantum numbers associated with the energy levels, but uncertainty arises regarding the correct values and the application of the Rydberg formula. Some participants express confusion about the quantum numbers and the proper setup for the transitions.

Discussion Status

There is an ongoing exploration of the relationships between energy, wavelength, and quantum numbers. Some participants have provided guidance on using the correct formulas and have pointed out potential errors in the initial attempts. Multiple interpretations of the quantum numbers are being considered, and the discussion remains open without a clear consensus.

Contextual Notes

Participants are navigating the constraints of homework rules, which may limit the information they can use or share. There is a focus on ensuring that the quantum numbers are whole numbers and correctly correspond to the energy levels in the hydrogen atom.

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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, I am just having trouble with this question:

thus far, i think I've 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, I am 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 don't really get what next... or if the n is correct... as i know, the 'n' in an energy diagram, has got to be whole number starting 2,3,4,5... so I am not sure what now... i mean, i get how to draw the diagram, i just don't know what the quantum numbers are... which the question is asking me to find.
 
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^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 I am 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:
 
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
 

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