Li atom and its valence e bound energy.

[prehisto]

Homework Statement

Find bond energy of valence electron in principal state in Li atom 2S. If first line of the sharp series is 0.813microm and short wave boundary is 0.349 microm.

Homework Equations

i think I have to use $\tilde{v}$=R[( 1/(x-(Δ)) )^2-( 1/(n-(Δ)) )^2]
Δ-quantum defect,which mostly is Δ(l) and constants.
x- i don't now how this number is determined.

The Attempt at a Solution

So I have
1)atom of Li ; where l=0 and s=2.
2) line where "sharp" stands for s( here l too is l=0)
3) and some kind of boundary, i do not now what means "short wave boundary"

I suppose that short wave boundary has something to do with x in the boundary formula.
Could someone help me to make sense of all this,where should I start?

 

[mark726]

Hello,

Thank you for your post. It seems like you have a good understanding of the basic concepts involved in this problem. To find the bond energy of a valence electron in the principal state of a Li atom in the 2S state, you will need to use the Rydberg formula, which is:

1/λ = R[(1/(n1)^2) - (1/(n2)^2)]

Where λ is the wavelength of the light emitted or absorbed, R is the Rydberg constant, and n1 and n2 are the initial and final energy levels, respectively.

In this case, n1 = 2 and n2 = 1 (since the first line of the sharp series corresponds to the transition from n=2 to n=1). The short wave boundary is most likely referring to the shortest wavelength in the series, which would correspond to the transition from n=∞ to n=1. This wavelength can be calculated using the Rydberg formula with n1 = ∞ and n2 = 1.

Once you have both wavelengths, you can use the formula for the bond energy, which is:

E = hc/λ

Where h is Planck's constant and c is the speed of light.

I hope this helps you get started on solving the problem. Let me know if you have any further questions or need clarification on any of the concepts.