# Homework Help: Wavelength - Yellow line in the spectrum of a sodium lamp

1. Aug 24, 2011

### mstud

1. The problem statement, all variables and given/known data
The outermost electron in a sodium atom can be in these energy states: (1)-0.82 aJ, (2) -0,48 aJ, (3) -0.31 aJ $1 aJ= 10^{-18} J$.

When we study the spectrum of a sodium lamp, we see a yellow line.
Use the above given energy states to determine the wavelength of this yellow light. Visual light: 400 nm to 750 nm

2. Relevant equations

$E_{photon}=hf$ where h is planck's constant: $h=6.63 \cdot 10^{-34}$

$c=f\lambda$

3. The attempt at a solution

I can only find one ultraviolet and one infrared wavelength from the given energy states.

frequency: $f=\frac {E_f}{h} = \frac {-.31 \cdot 10^{-18}J -(-.82 \cdot 10^{-18} J))}{6.63 \cdot 10^{-34} Js} \approx 7.69 \cdot 10^{14}$

Wavelength: $\lambda = \frac{c}{f}=\frac {3 \cdot 10^8 m/s}{7.69 \cdot 10^{14}}=3.90 \cdot 10^{-7}= 390 nm$

This is in the UV-part of the spectrum. Similar problem arises with the other energy transiteration.

However the answer key says 585 nm.

2. Aug 24, 2011

### kuruman

You must have done something wrong. One of the differences does give 585 nm. Check your work. Also, you can use E = hc/λ to relate directly the energy to the wavelength.

3. Aug 25, 2011

### mstud

That's true, yesterday I only thought of two possible movements between the energy states, but there were three. And the third one was the one I should use. Thank you for taking time to help...