As per Bohr model for hydrogen atom,energy of $n^{th}$ electron orbital is given by $\dfrac{-13.6}{n^2}$ eV.
Energy of electron in the first orbit will be $-13.6$ eV and in second orbit will be $-3.4$ eV. For transition, the electron requires $10.2$ eV to be given. This will increase the potential energy of the electron but the potential energy of the nucleus should also increase as two oppositely charge particle separation increases.
From where does that nucleus acquire that energy?
A quick correction, the $E=-13.6eV/n^2$ is the total energy not the potential energy. The total energy is the sum of the kinetic energy $T$ and potential energy $V$, and the kinetic and potential energy are related by the virial theorem $2T = -V$.
Anyhow, this total energy is the energy of the whole system, not just the electron. In the hydrogen atom we tend to assume the proton is fixed because it is so much heavier than the electron, but this is only approximately true. The motions of the electron and proton are correlated and it's the position of their mutual centre of mass that is fixed.
So adding energy to the atom affects both the electron and the proton and the energy, or at least the potential energy, can't be assigned to just the electron or just the proton.
The potential energy is that of the system, not just electron. The potential energy of two charges is just $-q_1q_2/4\pi \epsilon_0 r$, it isn't the sum of two terms.
