Find the binding energy of a valence electrons in the ground state of a `Li` atom if the wavelength of the first line of the sharp series is known to be equal to `lambda_(1)=813nm` and the short-wave cut-off wavelength of that series to `lambda_(2)=350nm`.

Explanation

For the first line of the sharp series `(3Srarr2P)` in a `Li` atom
`(2piħC)/(lambda_(1))=-(ħR)/((3+alpha_(0))^(2))+(ħR)/((2+alpha_(1))^(2))`
For the short wave cut-off wave-length of the same series
`(2piħc)/(lambda_(2))=(ħR)/((2+alpha_(1))^(2))`
From these two equations we get on substraction
`3+alpha_(0)=sqrt(ħR//(2piħc(lambda_(1)lambda_(2)))/(lambda_(1)lambda_(2)))`
`=sqrt((Rlambda_(1)lambda_(2))/(2piCDeltalambda)),Delta lambda=lambda_(1)-lambda_(2)`
Thus in the ground state, the binding energy of the electron is
`E_(b)=(ħR)/((2+alpha_(0))^(2))=ħR//(sqrt(Rlambda_(1)lambda_(2)/(2picDeltalambda)-1))^(2)=5.32eV`