Ground state of Potassium and Hydrogen

[shyguy79]

Homework Statement

For potassium in ground state configuration (Z=19) how would you expect the energy of the least tightly bound electron to compare with the energy of the electron in hydrogen excited to a state of the same principal quantum number n. Explain your answer.

Homework Equations

None

The Attempt at a Solution

Ok, so in it's ground state in standard notation Potassium 19 is:

K 1s$^{2}$ 2s$^{2}$ 2p$^{6}$ 3s$^{2}$ 3p$^{6}$ 4s$^{1}$

I'm guessing that the principal quantum number that it's asking for is when n=4 in the hydrogen? If so where do I go from here?

Any help is gratefully received

 

[shyguy79]

Is it perhaps that when the quantum number n=4 in the hydrogen atom the eV value ≈ -0.85eV (-13.6ev/4^2)? Because the angular momentum quantum number l = 0 then it has less energy and is closer to the nucleus?

I'm kinda clutching at straws

 

[einstein1921]

1s 2 2s 2 2p 6 3s 2 3p 6 electron will screen the nucleus of K, it seems that the 4s electron is attracted by 1 charge(19-18),but it doesn't screen so good,so the 4s electron will see more than 1 charge,for example 2,this situation will increase the attraction force to the 4s electron resulting in the decreasing of the energy.In contrast to hydrogen atom ,the nucleus is just 1 charge.so the energy of 4s electron will lower than that of the 4s electron of hydrogen.

 

[Rooted]

Hmmm, interesting question. I have a feeling the energy equation for hydrogenic atoms are related to the eigenvalue n like this;

E= -(e²/8πε₀a₀) Z²/n²I'll try and find a reference... hang on a sec

Edit - does this page help at all? http://electron6.phys.utk.edu/phys250/modules/module 3/Multi-electron atoms.htm