Spectral lines in the emission spectrum for an electron at n= 3

Click For Summary
An electron transitioning from n=3 can emit spectral lines by moving directly to n=1 or indirectly through n=2. This results in one or two lines based on the transition path. However, in a broader context, three lines are typically observed in the emission spectrum due to multiple atoms undergoing these transitions. The emission spectrum reflects both direct and indirect transitions from n=3. The discussion confirms the understanding of spectral line generation in a simplified hydrogen model.
Pushoam
Messages
961
Reaction score
53
Homework Statement
Find the spectral lines of in the emission spectrum for an electron excited to 3rd Orbit.
Relevant Equations
One line belonging to n=3 to n =1or two lines belonging to n=3 to n =2 and then n=2 to =1.
Since there is only one excited electron, it could come from n=3 to n =1directly or n=3 to n =2 and then n=2 to =1.

Hence, there could be one or two lines depending upon the path taken by electron.
Is this right?
 
Physics news on Phys.org
Yes. This assumes you're considering a simplified model of hydrogen, say, where effects due to electron spin are ignored. In particular, the assumption here is that there is only one energy value corresponding to a particular value of n.

Instead of saying there could be one or two lines depending upon the path taken by the electron, I think most people would say there are three lines that can be produced corresponding to the initial state n = 3. This is because emission spectra are usually generated by repeated transitions from many atoms. Some of these transitions are directly from ##3 \rightarrow 1## and some are the indirect ##3 \rightarrow 2 \rightarrow 1##. So, all three lines would occur in the spectrum. But you are thinking about it correctly.
 
Thank you.
 
At first, I derived that: $$\nabla \frac 1{\mu}=-\frac 1{{\mu}^3}\left((1-\beta^2)+\frac{\dot{\vec\beta}\cdot\vec R}c\right)\vec R$$ (dot means differentiation with respect to ##t'##). I assume this result is true because it gives valid result for magnetic field. To find electric field one should also derive partial derivative of ##\vec A## with respect to ##t##. I've used chain rule, substituted ##\vec A## and used derivative of product formula. $$\frac {\partial \vec A}{\partial t}=\frac...