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**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 [itex]1 aJ= 10^{-18} J[/itex].

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**

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

[itex]c=f\lambda[/itex]

**3. The Attempt at a Solution**

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

frequency: [itex]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}[/itex]

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

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

However the answer key says 585 nm.

Please help!