The ionization energy of hydrogen atom in the ground state is

Options

A. 13.6 MeV
B. 13.6 eV
C. 13.6 Joule
D. Zero

Explanation

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

### Step-by-Step Solution:

1. **Identify the atomic number and electronic configuration of hydrogen:**
- The atomic number (Z) of hydrogen is 1.
- The electronic configuration of hydrogen is \(1s^1\).

2. **Determine the principal quantum number (n):**
- For hydrogen in the ground state, the principal quantum number \(n\) is 1.

3. **Use the formula for the energy of an electron in a hydrogen atom:**
- The formula for the energy of an electron in a hydrogen atom is given by:
\[ E = -\frac{13.6 \, Z^2}{n^2} \text{ eV} \]
- Here, \(Z\) is the atomic number and \(n\) is the principal quantum number.

4. **Substitute the values of Z and n into the formula:**
- For hydrogen, \(Z = 1\) and \(n = 1\):
\[ E = -\frac{13.6 \times 1^2}{1^2} = -13.6 \text{ eV} \]

5. **Understand the concept of ionization energy:**
- Ionization energy is the energy required to remove an electron from an atom in its ground state to infinity (where the electron is no longer bound to the nucleus).

6. **Calculate the ionization energy:**
- The ionization energy (IE) can be calculated as:
\[ \text{IE} = E_{\text{final}} - E_{\text{initial}} \]
- Here, \(E_{\text{final}} = 0\) eV (when the electron is at infinity) and \(E_{\text{initial}} = -13.6\) eV (the energy of the electron in the ground state):
\[ \text{IE} = 0 - (-13.6) = 13.6 \text{ eV} \]

7. **Conclusion:**
- Therefore, the ionization energy of the hydrogen atom in the ground state is **13.6 eV**.