If an electron in a hydrogen atom is moving with a kinetic energy of `5.45 xx10^(-19)` j then what will be the energy level for this electron ?

Options

A. 1
B. 2
C. 3
D. 6

Answer

To find the energy level of an electron in a hydrogen atom given its kinetic energy, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Kinetic Energy (KE) of the electron = \( 5.45 \times 10^{-19} \) J.

2. **Relate Kinetic Energy to Total Energy:**
- The total energy (E) of an electron in a hydrogen atom is given by the relation:
\[ E = - KE \]
- Therefore, the total energy will be:
\[ E = - 5.45 \times 10^{-19} \text{ J} \]

3. **Convert Total Energy from Joules to Electron Volts:**
- We know that \( 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \).
- To convert the total energy into electron volts:
\[ E = \frac{-5.45 \times 10^{-19} \text{ J}}{1.6 \times 10^{-19} \text{ J/eV}} = -3.40625 \text{ eV} \]

4. **Use the Formula for Total Energy in Hydrogen Atom:**
- The total energy of an electron in a hydrogen atom is also given by:
\[ E = -\frac{13.6 Z^2}{n^2} \text{ eV} \]
- For hydrogen, \( Z = 1 \), so:
\[ E = -\frac{13.6}{n^2} \text{ eV} \]

5. **Set the Two Expressions for Energy Equal:**
- We can set the two expressions for energy equal to each other:
\[ -\frac{13.6}{n^2} = -3.40625 \]

6. **Solve for n²:**
- Rearranging gives:
\[ \frac{13.6}{n^2} = 3.40625 \]
- Cross-multiplying to solve for \( n^2 \):
\[ n^2 = \frac{13.6}{3.40625} \approx 3.99 \]

7. **Calculate n:**
- Taking the square root gives:
\[ n \approx \sqrt{3.99} \approx 2 \]

8. **Conclusion:**
- The energy level for the electron in the hydrogen atom is \( n = 2 \).