The energy of the ground state of hydrogen atom is `-13.6 ` eV. The energy of the photon emitted in the transiton from n=4 to n=2 is

Options

A. 13.6 eV
B. 2.55 eV
C. 2.4 eV
D. 1.66 eV

Explanation

To find the energy of the photon emitted during the transition from n=4 to n=2 in a hydrogen atom, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Energy Levels**: The energy of an electron in a hydrogen atom at a principal quantum number \( n \) is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( E_n \) is the energy at level \( n \).

2. **Calculate the Energy at n=4**:
\[ E_4 = -\frac{13.6 \, \text{eV}}{4^2} = -\frac{13.6 \, \text{eV}}{16} = -0.85 \, \text{eV} \]

3. **Calculate the Energy at n=2**:
\[ E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} \]

4. **Determine the Energy of the Photon Emitted**: The energy of the photon emitted during the transition from n=4 to n=2 is given by the difference in energy levels:
\[ E_{\text{photon}} = E_4 - E_2 \]
Substituting the values we calculated:
\[ E_{\text{photon}} = (-0.85 \, \text{eV}) - (-3.4 \, \text{eV}) = -0.85 + 3.4 = 2.55 \, \text{eV} \]

### Final Answer:
The energy of the photon emitted in the transition from n=4 to n=2 is **2.55 eV**.

---