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 Li2+ is

Question

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 Li2+ is (A) 30.6 eV (B) 13.6 eV (C) 3.4 eV (D) 122.4 eV

Tardigrade Admin · Accepted Answer

E=-Z2(13.6/n2)eV For first excited state E2=-32× (13.6/4) =-30.6eV Ionization energy for first excited state of Li2+ is 30.6 eV.

Q. 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 Li2+ is

30.6 eV

13.6 eV

3.4 eV

122.4 eV

$E = - Z^{2} \frac{13.6}{n ^{2}} e V$ For first excited state $E_{2} = - 3^{2} \times \frac{13.6}{4}$ $= - 30.6 e V$ Ionization energy for first excited state of $L i^{2 +}$ is 30.6 eV.

Q. 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 Li2+ is

30.6 eV

13.6 eV

3.4 eV

122.4 eV

E=−Z2n213.6​eV For first excited state E2​=−32×413.6​ =−30.6eV Ionization energy for first excited state of Li2+ is 30.6 eV.

$E = - Z^{2} \frac{13.6}{n ^{2}} e V$ For first excited state $E_{2} = - 3^{2} \times \frac{13.6}{4}$ $= - 30.6 e V$ Ionization energy for first excited state of $L i^{2 +}$ is 30.6 eV.