Initial state of hydrogen atoms at $T=0\,\text{K}$

At $T=0\,\text{K}$, all hydrogen atoms are in the ground state: $n=1$ with energy $E_1=-13.6\,\text{eV}.$

Condition to observe Balmer series

Balmer series consists of spectral lines obtained when an electron falls to the level: $n=2.$

So, to observe any Balmer line, the atom must first be excited to some level $n\ge 3$, because then it can de-excite as $n\ge 3 \to 2$ and emit a Balmer photon.

Minimum excitation needed

The smallest such excitation is from the ground state to the first level above $n=2$ that can produce a Balmer transition, i.e. $n=1 \to n=3.$

Energy of the $n=3$ state is $E_3=-\frac{13.6}{3^2}=-\frac{13.6}{9}\,\text{eV}.$

Therefore, minimum excitation energy required is $\Delta E = E_3-E_1 = -\frac{13.6}{9}-(-13.6).$

$\Delta E = 13.6\left(1-\frac{1}{9}\right) = 13.6\cdot \frac{8}{9}.$

$\Delta E = 12.088\ldots\,\text{eV}.$

Relation with accelerating potential

An electron accelerated through a potential difference $V$ gains kinetic energy: $eV = \Delta E.$

Since $1\,\text{eV}$ corresponds to $1\,\text{V}$ for one electron, $V_{\min}=12.088\ldots\,\text{V}.$

Given form $\dfrac{\alpha}{10}\,\text{V}$

We have $\frac{\alpha}{10}=12.088\ldots$ so $\alpha = 120.88\ldots \approx 121.$

Final answer

$\boxed{\alpha=121}$