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` = ________.

Answer

To solve the problem, we need to find the minimum potential difference required to excite a hydrogen atom to observe the Balmer series lines in its emission spectrum. The Balmer series corresponds to transitions from higher energy levels to the second energy level (n=2) in a hydrogen atom.

### Step-by-Step Solution:

1. **Understanding the Balmer Series**:
The Balmer series involves transitions from energy levels n ≥ 3 to n = 2. The minimum excitation energy needed for the first transition in the Balmer series is from n = 3 to n = 2.

2. **Energy Levels of Hydrogen Atom**:
The energy levels of a hydrogen atom are given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number.

3. **Calculating the Energy for n=3 and n=2**:
- For \( n = 3 \):
\[ E_3 = -\frac{13.6 \, \text{eV}}{3^2} = -\frac{13.6 \, \text{eV}}{9} \approx -1.51 \, \text{eV} \]
- For \( n = 2 \):
\[ E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} \]

4. **Finding the Minimum Excitation Energy**:
The minimum energy required to excite the electron from n = 3 to n = 2 is:
\[ E_{\text{excitation}} = E_2 - E_3 = (-3.4 \, \text{eV}) - (-1.51 \, \text{eV}) = -3.4 + 1.51 = -1.89 \, \text{eV} \]
However, we need the absolute value of energy difference:
\[ E_{\text{excitation}} = 1.89 \, \text{eV} \]

5. **Relating Energy to Potential Difference**:
The energy gained by an electron when accelerated through a potential difference \( V \) is given by:
\[ E = eV \]
where \( e \) is the charge of the electron (1 eV = 1 V for an electron). Therefore, the minimum potential difference \( V \) required is:
\[ V = E_{\text{excitation}} = 1.89 \, \text{V} \]

6. **Setting Up the Equation**:
According to the problem, the minimum potential difference needed to observe any Balmer series lines is given as:
\[ V = \frac{\alpha}{10} \, \text{V} \]
Setting the two expressions for \( V \) equal gives:
\[ 1.89 = \frac{\alpha}{10} \]

7. **Solving for \( \alpha \)**:
To find \( \alpha \):
\[ \alpha = 1.89 \times 10 = 18.9 \]
Rounding to the nearest whole number, we get:
\[ \alpha \approx 19 \]

### Final Answer:
\[ \alpha = 19 \]