The ionization of H-atom is :

Options

A. 13.6 J
B. 13.6 eV
C. 1 eV
D. 10.2 eV

Answer

To find the ionization energy of a hydrogen atom, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Ionization Energy**:
- Ionization energy is the energy required to remove an electron from an atom in its ground state to a point where the electron is free (at infinity).

2. **Using the Formula**:
- The ionization energy for a hydrogen atom can be calculated using the formula:
\[ E = -\frac{2 \pi^2 m k^2 Z^2 e^4}{n^2 h^2} \]
- For hydrogen, \( Z = 1 \) (since it has one proton) and we consider the ground state where \( n = 1 \).

3. **Substituting Values**:
- The constants involved are:
- \( k = \frac{1}{4 \pi \epsilon_0} \)
- \( m \) is the mass of the electron (\( 9.11 \times 10^{-31} \) kg)
- \( e \) is the charge of the electron (\( 1.6 \times 10^{-19} \) C)
- \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \) J·s)

4. **Calculating the Energy**:
- Plugging in the values:
\[ E = -\frac{2 \pi^2 (9.11 \times 10^{-31}) \left(\frac{1}{4 \pi \epsilon_0}\right)^2 (1)^2 (1.6 \times 10^{-19})^4}{(1)^2 (6.63 \times 10^{-34})^2} \]
- After calculating, we find:
\[ E \approx -21.45 \times 10^{-19} \text{ J} \]

5. **Converting Joules to Electron Volts**:
- To convert the energy from joules to electron volts, we use the conversion factor \( 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \):
\[ E \approx \frac{21.45 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 13.4 \text{ eV} \]

6. **Final Result**:
- Therefore, the ionization energy of the hydrogen atom is approximately **13.4 eV**. The closest answer from the options provided is **13.6 eV**, which is a common approximation used in textbooks.

### Conclusion:
The ionization energy of the hydrogen atom is approximately **13.6 eV**.