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 If the binding energy of the electron in a hydrogen atom is 13.6 eV, the energy required to remove the electron from the first excited state of Li++ is :

Option: 1

122.4 eV
 

 


Option: 2

 30.6 eV


Option: 3

13.6 eV

 


Option: 4

 3.4 eV


Answers (1)

best_answer

Energy required to remove an electron according to Bohr model is

\Delta E=E_o.Z^2\left ( \frac{1}{n^{2}} -\frac{1}{\infty ^{2}}\right )

Here electron is in first excited state \therefore n=2

z=3, E_{o}=13.6 ev

\therefore\Delta E=13.6\times 9\times \frac{1}{4}=30.6 ev

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