Question

The energy of an electron in first Bohr orbit of H - atom is -13.6 eV. The energy value of electron in the excited state of $L{i}^{2+}$ is :

• A. $-27.2eV$
• B. $30.6eV$
• C. $-30.6eV$
• D. $27.2eV$

Answer 1

Answer: (C)

$-30.6eV$

Given:
Energy of an electron in the first Bohr orbit, $E_o=-13.6ev$

$E_n=E_0\times \frac{Z^2}{n^2}$
$E_2=-13.6\times \frac{3^2}{2^2}$
$E_2=-30.6eV$

Hence, the correct option is C.