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`.

Find the binding energy of a valence electrons in the ground state of

1.	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`.
Answer» 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`

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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`.

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