Question

Let the potential energy of a hydrogen atom in the ground state be zero. Then, its energy in the first excites state will be

• A. $10.2eV$
• B. $13.6eV$
• C. $23.8eV$
• D. $27.2eV$

Answer 1

Answer: (C)

$23.8eV$

For an electron:
$U=-2KE$
$E=U+KE=\frac{U}{2}$
Energy in first excited state, $n=2$
$E(n=1)=-13.6eV$
$U(n=1)=2E=-27.2eV$
$E(n=2)=\frac{-13.6}{n^2}=-3.4eV$
If $U(n=1)$ is taken as the reference value
$E^{\prime }(n=2)=E(n=2)-U(n=1)=23.8eV$