Question

The energy of a hydrogen atom in the first excited state, if the potential energy is assumed to be zero in the ground state is

• A. $23.8eV$
• B. $21.2eV$
• C. $19.1eV$
• D. $13.6eV$

Answer 1

The correct option is A $23.8eV$

We know that, in ground state, $n=1$
Total energy = - kinetic energy $=\frac{\text{potential energy}}{2}$
$\Rightarrow T.E=-K.E=\frac{PE}{2}$
$\Rightarrow P.E=2[T.E]=2[-13.6eV]=-27.2eV$
We have assumed this energy to be zero ie. potential energy is increased by $27.2eV$. Since, the kinetic energy will remain unchanged in all energy states whereas the potential energy is increased by $27.2eV$ and hence the total energy in all states will increase by $27.2eV$.
$\Rightarrow$ for 1st excited state i.e., $n=2$
$E_2=-3.4eV$ [previously]
and now,
$E_2=[-3.4+27.2]eV=23.8eV$