Question

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

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

Answer 1

Answer: (C)

Energy of $e^{-}$ in the excited state of $L{i}^{+2}$

$E=-13.6\frac{Z^2}{n^2}$

$=-13.6\times \frac{{\left(3\right)}^2}{{\left(2\right)}^2}eV$

$=-\frac{9}{4}\times 13.6eV$

$=-30.6eN$