Physics - Modern Physics Question with Solution | TestHub

The energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n=-\frac{13.6}{n^2}$ eV.

For the ground state ($n=1$), the energy is $E_1=-13.6$ eV.

For the next higher state ($n=2$), the energy is $E_2=-\frac{13.6}{2^2}=-3.4$ eV.

The energy required to excite the electron from the ground state to the first excited state is $\Delta E=E_2-E_1$.

Substituting the values, we get $\Delta E=-3.4-(-13.6)=10.2$ eV.

Answer: 10.2 eV