Hydrogen atom is bombarded with electrons accelerated through a potential different of $$V$$, which causes excitation of hydrogen atoms. If the experiment is being formed at $$T = 0$$ K. The minimum potential difference needed to observe any Balmer series lines in the emission spectra will be $$\frac{\alpha}{10}$$ V, where $$\alpha =$$ ______.

To observe Balmer series lines in the emission spectrum, the hydrogen atom must be excited to at least the $$n = 3$$ level, since Balmer series corresponds to transitions ending at $$n = 2$$ and the minimum transition is $$n = 3 \rightarrow n = 2$$.

The energy of the $$n$$th level of hydrogen is given by $$E_n = -\frac{13.6}{n^2} \text{ eV}$$. Therefore, the energy required to excite the hydrogen atom from $$n = 1$$ to $$n = 3$$ is calculated as $$\Delta E = E_3 - E_1 = -\frac{13.6}{9} - \left(-\frac{13.6}{1}\right) = 13.6\left(1 - \frac{1}{9}\right) = 13.6 \times \frac{8}{9} = \frac{108.8}{9} = 12.09 \text{ eV}.$$

At $$T = 0\text{ K}$$, all hydrogen atoms are in the ground state ($$n = 1$$). The electrons must be accelerated through a potential difference $$V$$ so that they gain at least $$12.09\text{ eV}$$ of kinetic energy. Since $$eV = 12.09\text{ eV}$$, it follows that $$V = 12.09\text{ V} \approx 12.1\text{ V}.$$

This potential difference is expressed as $$\frac{\alpha}{10}\text{ V}$$, so $$\frac{\alpha}{10} = 12.1 \quad\Rightarrow\quad \alpha = 121.$$

Hence, $$\alpha = 121$$.