To solve the problem of determining the orbit to which the electron of a hydrogen atom will be excited when a photon beam of energy 12.1 eV is incident on it, we can follow these steps:
### Step-by-Step Solution:
1. **Understand the Energy Levels of Hydrogen Atom**:
The energy levels of a hydrogen atom are given by the formula:
\[
E_n = -\frac{13.6 \, \text{eV}}{n^2}
\]
where \( n \) is the principal quantum number (1, 2, 3, ...).
2. **Calculate the Energy of the Ground State (n=1)**:
For \( n=1 \):
\[
E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV}
\]
3. **Calculate the Energy of the First Excited State (n=2)**:
For \( n=2 \):
\[
E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -3.4 \, \text{eV}
\]
4. **Calculate the Energy of the Second Excited State (n=3)**:
For \( n=3 \):
\[
E_3 = -\frac{13.6 \, \text{eV}}{3^2} = -1.51 \, \text{eV}
\]
5. **Calculate the Energy Differences Between Levels**:
- Energy difference between \( n=1 \) and \( n=2 \):
\[
\Delta E_{1 \to 2} = E_2 - E_1 = -3.4 \, \text{eV} - (-13.6 \, \text{eV}) = 10.2 \, \text{eV}
\]
- Energy difference between \( n=1 \) and \( n=3 \):
\[
\Delta E_{1 \to 3} = E_3 - E_1 = -1.51 \, \text{eV} - (-13.6 \, \text{eV}) = 12.09 \, \text{eV}
\]
6. **Determine the Excitation Level**:
The energy of the incident photon is 12.1 eV. The closest energy level transition that can be achieved with this energy is from \( n=1 \) to \( n=3 \), which requires 12.09 eV. Since 12.1 eV is slightly more than 12.09 eV, it is sufficient to excite the electron to the third orbit.
7. **Conclusion**:
Therefore, the electron of the hydrogen atom will be excited to the third orbit (n=3).
### Final Answer:
The orbit to which the electron of the hydrogen atom will be excited is **n = 3** (third orbit).
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