Ground state of Potassium and Hydrogen

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SUMMARY

The discussion focuses on comparing the energy of the least tightly bound electron in potassium (K) in its ground state configuration (Z=19) with that of an electron in hydrogen (H) excited to the same principal quantum number n=4. The ground state configuration of potassium is K 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹. The energy of the 4s electron in potassium is lower than that of the 4s electron in hydrogen due to increased nuclear charge and electron shielding effects, resulting in a stronger attraction to the nucleus in potassium.

PREREQUISITES
  • Understanding of atomic structure and electron configurations
  • Familiarity with quantum numbers and their significance
  • Knowledge of energy levels in hydrogenic atoms
  • Basic principles of electron shielding and effective nuclear charge
NEXT STEPS
  • Study the concept of effective nuclear charge in multi-electron atoms
  • Learn about the energy level equations for hydrogenic atoms, specifically E = -(e²/8πε₀a₀) Z²/n²
  • Research electron shielding effects and their impact on atomic energy levels
  • Explore the differences in electron configurations between hydrogen and multi-electron atoms
USEFUL FOR

Students studying atomic physics, educators teaching quantum mechanics, and anyone interested in the comparative analysis of atomic energy levels in hydrogen and multi-electron systems like potassium.

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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



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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
 
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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
 
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.
 

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