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Aturan Slater ContohContoh yang diberikan dalam terbitan asal oleh Slater menggunakan atom besi dengan cas nukleus 26 dan konfigurasi elektron 1s22s22p63s23p63d64s2. Pemalar penskrinan, dan seterusnya cas nukleus berkesan bagi setiap elektron dikira seperti berikut:[1]
4 s : 0.35 × 1 + 0.85 × 14 + 1.00 × 10 = 22.25 ⇒ Z e f f ( 4 s ) = 3.75 3 d : 0.35 × 5 + 1.00 × 18 = 19.75 ⇒ Z e f f ( 3 d ) = 6.25 3 s , 3 p : 0.35 × 7 + 0.85 × 8 + 1.00 × 2 = 11.25 ⇒ Z e f f ( 3 s , 3 p ) = 14.75 2 s , 2 p : 0.35 × 7 + 0.85 × 2 = 4.15 ⇒ Z e f f ( 2 s , 2 p ) = 21.85 1 s : 0.30 × 1 = 0.30 ⇒ Z e f f ( 1 s ) = 25.7 {\displaystyle {\begin{matrix}4s&:0.35\times 1&+&0.85\times 14&+&1.00\times 10&=&22.25&\Rightarrow &Z_{\mathrm {eff} }(4s)=3.75\\3d&:0.35\times 5&&&+&1.00\times 18&=&19.75&\Rightarrow &Z_{\mathrm {eff} }(3d)=6.25\\3s,3p&:0.35\times 7&+&0.85\times 8&+&1.00\times 2&=&11.25&\Rightarrow &Z_{\mathrm {eff} }(3s,3p)=14.75\\2s,2p&:0.35\times 7&+&0.85\times 2&&&=&4.15&\Rightarrow &Z_{\mathrm {eff} }(2s,2p)=21.85\\1s&:0.30\times 1&&&&&=&0.30&\Rightarrow &Z_{\mathrm {eff} }(1s)=25.7\end{matrix}}}Perhatikan bahawa cas nuklear berkesan dikira dengan menolak pemalar penskrinan daripada nombor atom, 26.
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WikiPedia: Aturan Slater http://adsabs.harvard.edu/abs/1930PhRv...36...57S http://adsabs.harvard.edu/abs/1963JChPh..38.2686C http://adsabs.harvard.edu/abs/1967JChPh..47.1300C //doi.org/10.1063%2F1.1712084 //doi.org/10.1063%2F1.1733573 //doi.org/10.1103%2FPhysRev.36.57 http://astrophysics.fic.uni.lodz.pl/100yrs/pdf/04/...