The ground state energy of the hydrogen atom is 13.6 eV. Calculate
(i)the kinetic energy of the electron in the `1^(st)` excited state.
(ii)the potential energy of the electron in the `3^(rd)` excited state.
(iii)frequency of the photon emitted if the electron jumps from the `3^(rd)` excited state of `1^(st)` excited state.

Explanation

To solve the problem step by step, we will address each part of the question systematically.

### Given:
- Ground state energy of hydrogen atom, \( E_1 = -13.6 \, \text{eV} \)
- The formula for the energy of an electron in the \( n^{th} \) orbit:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
...