# Balmer Wavelength

1. Jun 7, 2014

### Meekay

I have been having trouble with this Balmer wavelength problem and was hoping I can get a little guidance.

The question: Find the balmer wavelength (n=3 --> n=2) emitted from a hydrogen-like Fe atom (z=26)

The answer is supposed to be 0.971 nm

My attempt:

1/lambda = R(1/n'^2 - 1/n2)
Since this is a hydrogen-like atom, it should be straight forward

I get lambda = 36/5*R which comes out to be 654 nm

And if I use one of Bohrs equations: E = -Z^2(R/n^2) and E3 = E2+ (hc/lambda) I get a value of 1200 nm

2. Jun 7, 2014

### Meekay

I figured it out, my book did not give me this equation for hydrogen-like atoms:

1/lambda = R*Z^2(1/n'^2 - 1/n^2) this gives me 0.97 nm

Thanks wikipedia!

Balmer Series for $Li^{2+}$ Oct 12, 2016