The relative penetration of the inner core electrons by the outer e- in Na can b

Question

The relative penetration of the inner core electrons by the outer e- in Na can be described by the calculation Zeff from E = -[Z^2eff(13.6)eV]/n^2 and comparing with E=-13.6eV/n^2 for no penetration. (a) Find the energies of the outer electron in the 3s,3p, and 3d states from Figure 7-22. (hint:a accurate method is to use -5.14eV for the ground state as given and find the energy of the 3pa nd 3d states from the photon energies of the indicated transitions.) (b) Find Zeff for the 3p and 3d states. (c) Is the approximation -13.6eV/n^2 good for any of these states?

Explanation / Answer

From the figure a) E3s = -5.14 eV we will find energy at 3p state E3p -E3s = hc/?-------(1) E3p - (-5.14 eV) = (1239.8  eV nm)/(589.59) nm E3p = -3.03 eV using energy at 3p state we will find th eenergy at 3d state E3d -E3p = hc/?-------(1) E3d - (-3.03 eV) = (1239.8  eV nm)/(818.33) nm E3d = -1.51 eV b)For 3p state: E = -[Z^2eff(13.6)eV]/n^2 -3.03 eV = -[Z^2 eff(13.6)eV]/(3)^2 Z eff = 1.416 For 3d state: E = -[Z^2eff(13.6)eV]/n^2 -1.51eV = -[Z^2 eff(13.6)eV]/(3)^2 Z eff = 0.99 c)The approximation is good for 3d state since -13.6eV/n^2 = -13.6eV/(3)^2                   = -1.51 eV E3d -E3p = hc/?-------(1) E3d - (-3.03 eV) = (1239.8  eV nm)/(818.33) nm E3d = -1.51 eV b)For 3p state: E = -[Z^2eff(13.6)eV]/n^2 -3.03 eV = -[Z^2 eff(13.6)eV]/(3)^2 Z eff = 1.416 For 3d state: E = -[Z^2eff(13.6)eV]/n^2 -1.51eV = -[Z^2 eff(13.6)eV]/(3)^2 Z eff = 0.99 c)The approximation is good for 3d state since -13.6eV/n^2 = -13.6eV/(3)^2                   = -1.51 eV E = -[Z^2eff(13.6)eV]/n^2 -1.51eV = -[Z^2 eff(13.6)eV]/(3)^2 Z eff = 0.99 c)The approximation is good for 3d state since -13.6eV/n^2 = -13.6eV/(3)^2                   = -1.51 eV

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.