Atomic physics: sodium D-line transition missing 2 pi

El Hombre Invisible
Messages
691
Reaction score
0

Homework Statement



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

Homework Equations



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

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

\lambda = \frac{\hbar c}{\Delta E}

The Attempt at a Solution



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

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

but since E is proportional to \hbar and the RHS on the top equation just to h... where did the 2\pi go..?
 
Last edited:
Physics news on Phys.org
'Sokay. Just realized \lambda = \frac{hc}{\Delta E} not \frac{\hbar c}{\Delta E}. Knew it would be something stupid.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top