Question

Energy of H- atom in the ground state is -13.6 eV, hence energy in the $2^{nd}$ excited state is

• A. -6.8 eV
• B. -3.4 eV
• C. -1.51eV
• D. -4.53 eV

Answer 1

The correct option is C -1.51eV

For H atom $E_n=\frac{-13.6}{n^2}eV$ For $2^{nd}$ excited state, n=3. Hence, $E_3=\frac{-13.6}{\left(3^2\right)}=-1.51eV$