A sample of hydrogen is bombarded by electrons. Through what potential difference should the electrons be accelerated so that third line of Lyman series be emitted?

Options

A. `2.55 V`
B. `10.2 V`
C. `12.09 V`
D. `12.75 V`

Explanation

To determine the potential difference through which electrons should be accelerated to emit the third line of the Lyman series in hydrogen, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Transition Levels**:
- The Lyman series corresponds to transitions where the final energy level (n_f) is 1. For the third line of the Lyman series, the initial energy level (n_i) is 4. Thus, the transition is from n = 4 to n = 1.

2. **Calculate the Energy Levels**:
- The energy of an electron in the nth level of hydrogen is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
- For n = 4:
\[ E_4 = -\frac{13.6 \, \text{eV}}{4^2} = -\frac{13.6 \, \text{eV}}{16} = -0.85 \, \text{eV} \]
- For n = 1:
\[ E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV} \]

3. **Calculate the Energy Difference**:
- The energy difference (ΔE) for the transition from n = 4 to n = 1 is:
\[ \Delta E = E_f - E_i = E_1 - E_4 = (-13.6) - (-0.85) = -13.6 + 0.85 = -12.75 \, \text{eV} \]
- Since we are interested in the absolute value of energy difference:
\[ \Delta E = 12.75 \, \text{eV} \]

4. **Relate Energy to Potential Difference**:
- The kinetic energy gained by an electron when accelerated through a potential difference (V) is given by:
\[ KE = eV \]
- Here, \( e \) is the charge of the electron (approximately 1 eV when considering energy in electron volts).
- Therefore, we can set:
\[ eV = \Delta E \]
- Rearranging gives:
\[ V = \frac{\Delta E}{e} = \Delta E \, \text{(in volts)} \]

5. **Calculate the Required Potential Difference**:
- Since we have already calculated ΔE as 12.75 eV, the potential difference required is:
\[ V = 12.75 \, \text{V} \]

### Final Answer:
The potential difference through which the electrons should be accelerated is **12.75 volts**.