To solve the problem, we need to analyze the energy levels of the hydrogen atom and the energy of the electrons being fired at it. Here’s a step-by-step breakdown of the solution:
### Step 1: Understand the Energy Levels of Hydrogen
The energy levels of the hydrogen atom can be calculated using the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number.
- For \( n = 1 \) (ground state):
\[ E_1 = -\frac{13.6}{1^2} = -13.6 \, \text{eV} \]
- For \( n = 2 \):
\[ E_2 = -\frac{13.6}{2^2} = -3.4 \, \text{eV} \]
- For \( n = 3 \):
\[ E_3 = -\frac{13.6}{3^2} = -1.51 \, \text{eV} \]
### Step 2: Determine the Ionization Potential
The ionization potential of hydrogen is given as 13.6 eV. This means that to remove an electron from the ground state (n=1), an energy of 13.6 eV is required.
### Step 3: Analyze the Energy of the Incoming Electrons
The energy of the incoming electrons is 12.1 eV. Since this energy is less than the ionization potential (13.6 eV), the electrons cannot ionize the hydrogen atoms. However, they can excite the electrons to higher energy levels.
### Step 4: Determine Possible Excitation Levels
The electrons can be excited from the ground state (n=1) to higher levels. The maximum energy available for excitation is:
\[ 12.1 \, \text{eV} - (-13.6 \, \text{eV}) = 12.1 + 13.6 = 25.7 \, \text{eV} \]
The energy differences between the levels are:
- From \( n=1 \) to \( n=2 \):
\[ E_{12} = E_2 - E_1 = -3.4 - (-13.6) = 10.2 \, \text{eV} \]
- From \( n=1 \) to \( n=3 \):
\[ E_{13} = E_3 - E_1 = -1.51 - (-13.6) = 12.09 \, \text{eV} \]
Since 12.1 eV is just slightly less than 12.09 eV, the electrons can excite the hydrogen atom to the \( n=3 \) state.
### Step 5: Calculate the Wavelengths of Emission
When the electron returns to a lower energy state, it emits a photon. We calculate the wavelengths for the transitions:
1. **From \( n=3 \) to \( n=1 \)**:
\[ E_{31} = E_3 - E_1 = -1.51 - (-13.6) = 12.09 \, \text{eV} \]
\[ \lambda_1 = \frac{hc}{E_{31}} \]
\[ \lambda_1 = \frac{(6.626 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{12.09 \times 1.6 \times 10^{-19} \, \text{J}} \]
\[ \lambda_1 \approx 102.6 \, \text{nm} \]
2. **From \( n=3 \) to \( n=2 \)**:
\[ E_{32} = E_2 - E_3 = -3.4 - (-1.51) = 1.89 \, \text{eV} \]
\[ \lambda_2 = \frac{hc}{E_{32}} \]
\[ \lambda_2 = \frac{(6.626 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{1.89 \times 1.6 \times 10^{-19} \, \text{J}} \]
\[ \lambda_2 \approx 657.5 \, \text{nm} \]
3. **From \( n=2 \) to \( n=1 \)**:
\[ E_{21} = E_2 - E_1 = -3.4 - (-13.6) = 10.2 \, \text{eV} \]
\[ \lambda_3 = \frac{hc}{E_{21}} \]
\[ \lambda_3 = \frac{(6.626 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{10.2 \times 1.6 \times 10^{-19} \, \text{J}} \]
\[ \lambda_3 \approx 121.6 \, \text{nm} \]
### Conclusion
The wavelengths corresponding to the transitions are approximately:
- \( \lambda_1 \approx 102.6 \, \text{nm} \) (from \( n=3 \) to \( n=1 \))
- \( \lambda_2 \approx 657.5 \, \text{nm} \) (from \( n=3 \) to \( n=2 \))
- \( \lambda_3 \approx 121.6 \, \text{nm} \) (from \( n=2 \) to \( n=1 \))
