Atomic physics: sodium D-line transition missing 2 pi

In summary, to find the wavelength of the sodium transition 3p^{1},^{2}P -> 3s^{1},^{2}S, you can use the equation \frac{1}{\lambda} = -R[\frac{1}{(n_{i} - d_{i}(n,l))^{2}} - \frac{1}{(n_{f} - d_{f}(n,l))^{2}}], which is proportional to \hbar and the energy difference \Delta E between the initial and final states. It is important to note that the correct form of the equation is \lambda = \frac{hc}{\Delta E}, not \frac{\hbar c}{\Delta
  • #1
El Hombre Invisible
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Homework Statement



Find the wavelength of the sodium transition [tex]3p^{1},^{2}P -> 3s^{1},^{2}S[/tex]

Homework Equations



[tex]E_{n,l} = -\frac{hcR}{(n - d(n,l))^{2}}[/tex]

d(3,s) = 1.374
d(3,p) = 0.884

[tex]\lambda = \frac{\hbar c}{\Delta E}[/tex]

The Attempt at a Solution



Plugging the numbers in ain't even close. However, I've found another equation that works:

[tex]\frac{1}{\lambda} = -R[\frac{1}{(n_{i} - d_{i}(n,l))^{2}} - \frac{1}{(n_{f} - d_{f}(n,l))^{2}}][/tex]

but since E is proportional to [tex]\hbar[/tex] and the RHS on the top equation just to h... where did the [tex]2\pi[/tex] go..?
 
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  • #2
'Sokay. Just realized [tex]\lambda = \frac{hc}{\Delta E}[/tex] not [tex]\frac{\hbar c}{\Delta E}[/tex]. Knew it would be something stupid.
 

Related to Atomic physics: sodium D-line transition missing 2 pi

1. What is the significance of the sodium D-line transition in atomic physics?

The sodium D-line transition refers to the specific wavelengths of light emitted by excited sodium atoms when they return to their ground state. This phenomenon is important because it allows scientists to study the energy levels and transitions of atoms, which provides insight into the fundamental properties of matter.

2. Why is the sodium D-line transition sometimes described as "missing 2 pi"?

The D-line transition of sodium atoms is actually a doublet, meaning there are two distinct wavelengths of light emitted. However, one of these wavelengths is slightly shifted due to the relativistic effects of the atom's electrons. This small shift is equivalent to "missing 2 pi" in the wavelength measurement.

3. How is the sodium D-line transition used in practical applications?

The distinct wavelength of the sodium D-line transition is used in a variety of applications, such as in street lamps, where the orange-yellow glow is produced by excited sodium atoms. It is also used in spectroscopy to identify and measure the composition of materials.

4. What factors affect the wavelength of the sodium D-line transition?

The wavelength of the D-line transition is affected by the energy levels of the sodium atom, which in turn are influenced by factors such as temperature, pressure, and the presence of other elements. The relativistic effects mentioned earlier also play a role in the exact wavelength measurement.

5. Can the sodium D-line transition occur in other elements?

Yes, similar transitions can occur in other elements, but they will have different wavelengths depending on the specific energy levels of the atom. For example, potassium has a D-line transition in the red part of the spectrum, while lithium has a D-line transition in the violet part of the spectrum.

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