To solve the problem of how many different lines are possible in the resulting spectrum of a hydrogen atom excited by monochromatic radiation of wavelength 975 Å, and to calculate the longest wavelength amongst them, we can follow these steps:
### Step 1: Calculate the energy of the incident photon
The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h \) (Planck's constant) = \( 6.63 \times 10^{-34} \, \text{J s} \)
- \( c \) (speed of light) = \( 3.00 \times 10^{8} \, \text{m/s} \)
- \( \lambda \) = \( 975 \, \text{Å} = 975 \times 10^{-10} \, \text{m} \)
Substituting the values:
\[ E = \frac{(6.63 \times 10^{-34} \, \text{J s})(3.00 \times 10^{8} \, \text{m/s})}{975 \times 10^{-10} \, \text{m}} \]
Calculating this gives:
\[ E \approx 2.04 \times 10^{-19} \, \text{J} \]
To convert this energy into electron volts (eV), we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ E \approx \frac{2.04 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 12.75 \, \text{eV} \]
### Step 2: Determine the maximum energy level (n) the electron can reach
The ground state energy of the hydrogen atom is given by:
\[ E_n = -\frac{13.6}{n^2} \, \text{eV} \]
Setting the energy of the photon equal to the energy difference from the ground state:
\[ 12.75 = 13.6 \left( \frac{1}{1^2} - \frac{1}{n^2} \right) \]
Solving for \( n \):
\[ 12.75 = 13.6 \left( 1 - \frac{1}{n^2} \right) \]
\[ \frac{12.75}{13.6} = 1 - \frac{1}{n^2} \]
\[ \frac{1}{n^2} = 1 - \frac{12.75}{13.6} \]
\[ \frac{1}{n^2} = \frac{0.85}{13.6} \]
\[ n^2 \approx 4 \]
\[ n \approx 4 \]
### Step 3: Determine the possible transitions
The electron can transition from \( n = 4 \) to lower energy levels \( n = 3, 2, 1 \). The possible transitions are:
1. \( 4 \rightarrow 3 \)
2. \( 4 \rightarrow 2 \)
3. \( 4 \rightarrow 1 \)
4. \( 3 \rightarrow 2 \)
5. \( 3 \rightarrow 1 \)
6. \( 2 \rightarrow 1 \)
Thus, there are a total of 6 different lines possible in the resulting spectrum.
### Step 4: Calculate the longest wavelength
The longest wavelength corresponds to the transition with the smallest energy difference, which is the transition from \( n = 4 \) to \( n = 3 \).
The energy difference for this transition is:
\[ E_{4 \rightarrow 3} = E_4 - E_3 = -\frac{13.6}{4^2} - \left(-\frac{13.6}{3^2}\right) \]
\[ = -\frac{13.6}{16} + \frac{13.6}{9} \]
\[ = 13.6 \left( \frac{1}{9} - \frac{1}{16} \right) \]
Calculating this gives:
\[ = 13.6 \left( \frac{16 - 9}{144} \right) = 13.6 \left( \frac{7}{144} \right) \approx 0.661 \, \text{eV} \]
Now, using the energy to find the wavelength:
\[ \lambda_{max} = \frac{hc}{E_{min}} \]
Substituting the values:
\[ \lambda_{max} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^{8})}{0.661 \times 1.6 \times 10^{-19}} \]
Calculating this gives:
\[ \lambda_{max} \approx 18807 \, \text{Å} \]
### Final Answers
- **Number of different lines possible**: 6
- **Longest wavelength**: 18807 Å
