What is the de Broglie wavelength of an electron? What determines the value of the de Broglie wavelength for an electron?

Solution 16E Step 1 Effective nuclear charge is that portion of charge that a given electron in an atom experiences.It is the net electrical field experienced by the electron Z eff- equation 1 where, Z is the atomic number of the element and is the screening factor depending on the electrons present Step 2 a)Case 1 Zeff experienced by the outermost electron in both Si and Cl is calculated by assuming that core electrons contribute 1.00 and valence electrons contribute 0.00 to the screening constant The atomic number and electronic configurations of Si is as given below Si- Z= 14 Electronic configuration is 1s 2s 2p 3s 3p 2 Si has 10 core electrons and 4 valence electron Hence screening effect experienced , =10×1=10 Substituting the values of Z and in equation 1 we get Zeff=14-10=4 The atomic number and electronic configurations of Cl is as given below Cl- Z=17 Electronic configuration is 1s 2s 2p 3s 3p 5 Cl has 10 core electrons Hence screening effect experienced =10×1=10 Substituting the values of Z and in equation 1 we get Zeff=17-10=7 b)Case 2-Using slaters rule Slaters rule In slaters rule the electrons are arranged into a sequence of groups in order of increasing principal quantum number n, and for equal n in order of increasing azimuthal quantum number l. [1s] [2s,2p] [3s,3p] [3d] [4s,4p] [4d] [4f] [5s, 5p] [5d] etc. The shielding constant for each group is formed as the sum of the following contributions: 1)All other electrons in the same group as the electron of interest shield to an extent of 0.35 nuclear charge units except 1s group, where the other electron contributes only 0.30. 2)If the group is of the [s, p] type, an amount of 0.85 from each electron (n-1) shell and an amount of 1.00 for each electron from (n-2) and lower shell. 3)If the group is of the [d] or [f] type then an amount of 1.00 for each electron lying closer to the atom than the group.This includes both i) electrons with a smaller principal quantum number than n and ii) electrons with principal quantum number n and a smaller azimuthal quantum number l. Applying slaters rule to Si and Cl Si- Z= 14 Electronic configuration is 1s 2s 2p 3s 3p 2 So value of =3× 0.35+8× 0.85+2× 1 =1.05+6.8+2 =9.85 Substituting the values of Z and in equation 1 we get Zeff=14-9.85=4.15 Cl- Z=17 2 2 6 2 5 Electronic configuration is 1s 2s 2p 3s 3p So value of = 6 × 0.35+8× 0.85+2× 1 =2.1+6.8+2 =10.9 Substituting the values of Z and in equation 1 we get Zeff=17-10.9 =6.10 c)Step 3 The value of zeff from detailed calculations are 4.29+ and 6.12+, respectively,Thus Slaters method give a more accurate value - 4.15 and 6.10 d)