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

#### mstud

1. Homework Statement
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. Homework 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.

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#### kuruman

Homework Helper
Gold Member
Similar problem arises with the other energy transiteration.

However the answer key says 585 nm.

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.

#### 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...

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