Saved Bookmarks
| 1. |
The binding energy of the valance electron in a Li atom in the states 2S and 2P is equal to 5.39 and 3.54 eV respectively. Find the Rydberg corrections for s and p terms of the atom. |
|
Answer» <P> SOLUTION :From the RYDBERG formula we write`E_(n)= -( ħR)/((n+alpha_(l)^(2))` we use ` ħR= 13.6eV`. Then for n= 2 state `5.39= -(13.6)/((2+alpha_(0))^(2)), l=0(S)` state `alpha_(0)~~ -0.41` for `p` state `3.54= -(13.6)/((2+alpha_(1))^(2))` `alpha_(1)=-0.039`. |
|