Using e = hc/λ to find wavelength of light emitted

[miniradman]

Homework Statement

Hello there, a question that I have been given is to "find the wavelenght of light emitted from an electron jumping from the 2nd to the 6th orbital (or vise versa if that matters). The atom is hydrogen"

Homework Equations

E= \frac{hc}{λ}

The Attempt at a Solution

I've already worked out this question using the Rydberg Formula where I got 410 nm (which I know is correct because I crossed checked it using wikipedia and my textbook). However, this is part of a quatumn mechanics assignment, and I was suppose to use the forumla E= \frac{hc}{λ}

But I have no idea about how to apply the forumla to get the wavelenght. But if I made λ the subject, it would read to be

\frac{hc}{E} = λ

I think what I need to find out is the value of E (well, obviously that's what I need to find). But how do I relate the different energy levels with the orbials in which the electron travels?

Surely its not just 6 -2 = 4

thanks - miniradman

 

[Steely Dan]

The Rydberg formula is based on that formula for the energy of a photon; giving an answer that way ought to be sufficient. The way it is derived is to say that the energy levels are given by E_n = -\frac{13.6\ \text{eV}}{n^2} and then to subtract E_{n_i} from E_{n_f}.

 

[turin]

The Rydberg formula is based on that formula for the energy of a photon;

No, it is not. It is based on Rydberg's direct empirical observation that wavenumbers come in series. It was basically a matter of (discrete) curve fitting (and almost two decades before Einstein's relation).

@miniradman,

Your instructor probably wants you to use the quantum-mechanical fact that the energy levels in the Hydrogen atom are discrete, the Einstein relation that you give, and the conservation of energy (where the electron's energy is lost as a photon). So:

E_i = (-13.6 eV)/n_i²

E_f = (-13.6 eV)/n_f²

E_γ = (h⋅c)/λ

E_i = E_f + E_γ

(Of course, you need the hc to be in eV⋅m, not J⋅m.)