1. Balmer series condition

   In the hydrogen spectrum, the Balmer series corresponds to transitions ending at $n = 2.$

   So, for a hydrogen atom to emit a Balmer photon, the electron must first be excited from the ground state $n=1$ to at least some level $n \ge 3$, so that it can later make a transition to $n=2$.

2. Minimum excitation needed

   The minimum such level is $n=3.$

   Therefore, the minimum energy required is the energy needed to raise the electron from $n=1$ to $n=3$.

3. Energy levels of hydrogen

   The energy of the $n$th level is $E_n = -\frac{13.6}{n^2}\,\text{eV}.$

   Hence, $E_1 = -13.6\,\text{eV}$ $E_3 = -\frac{13.6}{9}\,\text{eV} \approx -1.51\,\text{eV}.$

4. Required energy

   $\Delta E = E_3 - E_1$ $\Delta E = \left(-\frac{13.6}{9}\right) - (-13.6)$ $\Delta E = 13.6\left(1 - \frac{1}{9}\right)$ $\Delta E = 13.6 \cdot \frac{8}{9}$ $\Delta E \approx 12.1\,\text{eV}.$

5. Option check

   • A: $13.6\,\text{eV}$ → energy for ionization from ground state, not correct
   • B: $1.5\,\text{eV}$ → too small
   • C: $12.1\,\text{eV}$ → correct
   • D: $1.9\,\text{eV}$ → not correct

Therefore, the correct answer is C.