A gas of monoatomic hydrogen is bombarded with a stream of electrons that have been accelerated from rest through a potential difference of `12.75 V`. In the emission spectrum, one can observe lines of

Options

A. Lyman series
B. Balmer series
C. Paschen series
D. Pfund series

Explanation

To solve the problem, we need to analyze the situation of a monoatomic hydrogen gas being bombarded by electrons that have been accelerated through a potential difference of 12.75 V. We will determine which emission lines can be observed in the spectrum as a result of this interaction.

### Step-by-Step Solution:

1. **Calculate the Kinetic Energy of the Electrons:**
The kinetic energy (KE) gained by the electrons when accelerated through a potential difference (V) is given by the formula:
\[ KE = e \cdot V \]
where \( e \) is the charge of the electron (approximately \( 1.6 \times 10^{-19} \) coulombs) and \( V \) is the potential difference. Here, \( V = 12.75 \) V.
\[ KE = 1 \cdot 12.75 \, \text{eV} = 12.75 \, \text{eV} \]

2. **Determine the Energy Levels of Hydrogen:**
The energy levels of a hydrogen atom are given by the formula:
\[ E_n = -\frac{13.6}{n^2} \, \text{eV} \]
where \( n \) is the principal quantum number. The first few energy levels are:
- \( n=1: E_1 = -13.6 \, \text{eV} \)
- \( n=2: E_2 = -3.4 \, \text{eV} \)
- \( n=3: E_3 = -1.51 \, \text{eV} \)
- \( n=4: E_4 = -0.85 \, \text{eV} \)

3. **Determine the Excitation of the Hydrogen Atom:**
The electrons from the gas will absorb the kinetic energy of 12.75 eV. The maximum energy level that can be reached by the hydrogen atom is determined by the energy difference between the ground state and the excited state:
\[ E_{excited} = E_1 + KE = -13.6 \, \text{eV} + 12.75 \, \text{eV} = -0.85 \, \text{eV} \]
This corresponds to the energy level \( n=4 \).

4. **Identify Possible Transitions:**
The electron can transition from \( n=4 \) to lower energy levels. The possible transitions that can occur are:
- \( n=4 \) to \( n=3 \)
- \( n=4 \) to \( n=2 \)
- \( n=4 \) to \( n=1 \)

5. **Determine the Emission Spectrum:**
The transitions will result in the emission of photons, and the wavelengths of these photons correspond to the lines in the emission spectrum. The series of lines that can be observed include:
- **Lyman Series:** Transitions to \( n=1 \) (from \( n=4 \) to \( n=1 \))
- **Balmer Series:** Transitions to \( n=2 \) (from \( n=4 \) to \( n=2 \))
- **Paschen Series:** Transitions to \( n=3 \) (from \( n=4 \) to \( n=3 \))

The transitions from \( n=4 \) can produce lines in the Lyman, Balmer, and Paschen series.

### Conclusion:
The emission spectrum will show lines corresponding to transitions from \( n=4 \) to \( n=1 \), \( n=4 \) to \( n=2 \), and \( n=4 \) to \( n=3 \). Therefore, the observed lines will belong to the Lyman, Balmer, and Paschen series.