An electron is in the ground state of a hydrogen atom. A photon is absorbed by the atom and the electron is excited to the n= 2 state. What is the energy in eV of the photon?

Answer

To find the energy of the photon absorbed by the hydrogen atom that excites the electron from the ground state (n=1) to the first excited state (n=2), we can use the formula for the energy levels of a hydrogen atom.

### Step-by-Step Solution:

1. **Identify the Initial and Final States:**
- The electron is in the ground state, which corresponds to \( n_{initial} = 1 \).
- The electron is excited to the second energy level, which corresponds to \( n_{final} = 2 \).

2. **Use the Energy Level Formula:**
The energy levels of a hydrogen atom are given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( E_n \) is the energy of the level \( n \).

3. **Calculate the Energy of the Initial State (n=1):**
\[ E_{initial} = E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV} \]

4. **Calculate the Energy of the Final State (n=2):**
\[ E_{final} = E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} \]

5. **Calculate the Change in Energy (ΔE):**
The energy absorbed by the electron (which corresponds to the energy of the photon) is given by the difference in energy between the final and initial states:
\[ \Delta E = E_{final} - E_{initial} \]
Substituting the values:
\[ \Delta E = (-3.4 \, \text{eV}) - (-13.6 \, \text{eV}) = -3.4 \, \text{eV} + 13.6 \, \text{eV} = 10.2 \, \text{eV} \]

6. **Final Answer:**
The energy of the photon absorbed by the hydrogen atom is \( 10.2 \, \text{eV} \).