- #1
hitmeoff
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Homework Statement
The ground state of lithium (Z=3) has two electrons in the 1s level and one in the 2s. Consider an excited state in which the outermost electron has been raised to the 3p level. Since the 3p wave functions are not very penetrating, you can estimate the energy of this electron by assuming it is completely outside both the other electrons.
a) In this approximation, what is the potential energy function felt by the outermost electron?
b) IN the same approximation write the formula for the energy of the outer electron if its principle quantum number is n.
c) Estimate the energy of the 3p electron in this way and compare to the observed value of -1.556 eV
d) repeat for the case that the outer electron is in the 3d level, whose observed energy is -1.513 eV
e) Explain why the agreement is better for the 3d level than the 3p. Why is the observed energy for 3p lower than for the 3d?
Homework Equations
U(r) = -Zeff(r)[tex]\frac{ke^{2}}{r}[/tex]
Zeff [tex]\approx[/tex] Z [r inside all other electrons]
Zeff [tex]\approx[/tex] 1 [r outside all other electrons]
Most probable radius: rmp [tex]\approx[/tex] [tex]\frac{n^{2}a_{B}}{Z_{eff}}[/tex]
The Attempt at a Solution
For a, I take the first equation U(r) = -Zeff(r)[tex]\frac{ke^{2}}{r}[/tex] and since the 3p electron is so far outside of the other 2 electrons Zeff [tex]\approx[/tex] 1
For b, I substituted r for rmp, so U(r) becomes [tex]\approx[/tex] -[tex]\frac{ke^{2}}{n^{2}a_{B}}[/tex] which got me -3.02 eV
But I am completely stumped on the rest. HOw do I actually calculate the difference in energies from 3p and 3d? I know 3p should have a lower energy than 3d because it feels more of the nucleus but how do I actually work this out given my equations (or any other you may know)?
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