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

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

Tardigrade Admin · Accepted Answer

The energy of the first excited state of Li++ is E2=(Z2E0/n2)=(32×13.6/22)=30.6 eV.

Q. If the binding energy of the electron in a hydrogen atom is $13.6 e V$ , the energy required to remove the electron from the first excited state of $L i^{++}$ is

$30.6 e V$

$13.6 e V$

$3.4 e V$

$122.4 e V$

The energy of the first excited state of $L i^{++}$ is
$E_{2} = \frac{Z ^{2} E _{0}}{n ^{2}} = \frac{3 ^{2} \times 13.6}{2 ^{2}} = 30.6 e V .$

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

30.6eV

13.6eV

3.4eV

122.4eV