An electron collides with a hydrogen atom in its ground state and excites it to `n = 3` ,. The energy gives to hydrogen aton n this inclastic collision is [Neglect the recoiling of hydrogen atom]

Options

A. `10.2 eV`
B. `12.1 eV`
C. `12.5 eV`
D. none of these

Explanation

To solve the problem of determining the energy given to a hydrogen atom when an electron collides with it and excites it from the ground state (n=1) to the excited state (n=3), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Energy Levels**:
The energy levels of a hydrogen atom can be calculated using the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number.

2. **Calculate the Energy of the Ground State (n=1)**:
For the ground state (n=1):
\[ E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV} \]

3. **Calculate the Energy of the Excited State (n=3)**:
For the excited state (n=3):
\[ E_3 = -\frac{13.6 \, \text{eV}}{3^2} = -\frac{13.6 \, \text{eV}}{9} \approx -1.51 \, \text{eV} \]

4. **Determine the Energy Difference**:
The energy given to the hydrogen atom during the collision is the difference between the energy of the excited state and the ground state:
\[ \Delta E = E_3 - E_1 = (-1.51 \, \text{eV}) - (-13.6 \, \text{eV}) = -1.51 + 13.6 = 12.1 \, \text{eV} \]

5. **Conclusion**:
The energy given to the hydrogen atom in this inelastic collision is:
\[ \Delta E = 12.1 \, \text{eV} \]

### Final Answer:
The energy given to the hydrogen atom in this inelastic collision is **12.1 eV**.