A photon beam of energy 12.1eV is incident on a hydrogen atom. The orbit to which electron of H-atom be excited is

Options

A. `2^(nd)`
B. `3^(rd)`
C. `4^(th)`
D. `5^(th)`

Answer

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 1: Understand the Energy Levels of Hydrogen
The energy levels of a hydrogen atom can be described by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number (1, 2, 3, ...).

### Step 2: Calculate the Energy of the First Few Levels
- For \( n = 1 \):
\[ E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV} \]
- For \( n = 2 \):
\[ E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -3.4 \, \text{eV} \]
- For \( n = 3 \):
\[ E_3 = -\frac{13.6 \, \text{eV}}{3^2} = -1.51 \, \text{eV} \]
- For \( n = 4 \):
\[ E_4 = -\frac{13.6 \, \text{eV}}{4^2} = -0.85 \, \text{eV} \]
- For \( n = 5 \):
\[ E_5 = -\frac{13.6 \, \text{eV}}{5^2} = -0.544 \, \text{eV} \]

### Step 3: Determine the Energy Differences Between Levels
To find out which orbit the electron can be excited to, we need to calculate the energy differences between the levels:
- From \( n = 1 \) to \( n = 2 \):
\[ \Delta E_{1 \to 2} = E_2 - E_1 = -3.4 \, \text{eV} - (-13.6 \, \text{eV}) = 10.2 \, \text{eV} \]
- From \( n = 1 \) to \( n = 3 \):
\[ \Delta E_{1 \to 3} = E_3 - E_1 = -1.51 \, \text{eV} - (-13.6 \, \text{eV}) = 12.1 \, \text{eV} \]
- From \( n = 1 \) to \( n = 4 \):
\[ \Delta E_{1 \to 4} = E_4 - E_1 = -0.85 \, \text{eV} - (-13.6 \, \text{eV}) = 12.75 \, \text{eV} \]

### Step 4: Compare the Photon Energy with Energy Differences
The energy of the photon beam is given as 12.1 eV. This matches the energy difference calculated for the transition from \( n = 1 \) to \( n = 3 \).

### Conclusion
Since the photon energy of 12.1 eV corresponds to the transition from the first orbit (ground state) to the third orbit, the electron of the hydrogen atom will be excited to the **third orbit**.

### Final Answer
The orbit to which the electron of the hydrogen atom will be excited is the **third orbit** (n = 3).

---