The energy of an electron in first Bohr orbit of H - atoms is -13.6 eV. The possible energy value of electron in the excited state of Li2+ is

Question

The energy of an electron in first Bohr orbit of H - atoms is -13.6 eV. The possible energy value of electron in the excited state of Li2+ is (A) - 122.4 eV (B)  -30.6 eV (C) 30.6 eV (D) - 13.6 eV

Tardigrade Admin · Accepted Answer

En=(E1/n2) × Z2 For Li 2+, the excited state, n=2 and Z=3 ∴ En =(-13.6/22) ×(3)2 =(-13.6/4) × 9=-30.6 eV

Q. The energy of an electron in first Bohr orbit of $H -$ atoms is $- 13.6 e V$ . The possible energy value of electron in the excited state of $L i^{2 +}$ is

$- 122.4 e V$

$- 30.6 e V$

$30.6 e V$

$- 13.6 e V$

$E_{n} = \frac{E _{1}}{n ^{2}} \times Z^{2}$
For $L i^{2 +}$ , the excited state, $n = 2$ and $Z = 3$
$∴ E_{n} = \frac{- 13.6}{2 ^{2}} \times (3)^{2}$
$= \frac{- 13.6}{4} \times 9 = - 30.6 e V$

Q. The energy of an electron in first Bohr orbit of H− atoms is −13.6eV. The possible energy value of electron in the excited state of Li2+ is

−122.4eV

−30.6eV

30.6eV

−13.6eV