Question

The energy of the ground electronic state of hydrogen atom is -13.6 eV. The energy of the first excited stae will be :

• A. – 52.4 eV
• B. – 27.2 eV
• C. – 68 eV
• D. – 3.4 eV

Answer 1

The correct option is D – 3.4 eV

$E=-\frac{13.6}{n^2}n=2$
$=-\frac{13.6}{n^2}=-3.4eV$