To solve the problem, we need to follow these steps:
### Step 1: Determine the energy of the incoming radiation
The energy of the incoming radiation can be calculated using the formula:
\[
E = \frac{hc}{\lambda}
\]
Where:
- \( E \) = energy of the photon
- \( h \) = Planck's constant \( (6.626 \times 10^{-34} \, \text{J s}) \)
- \( c \) = speed of light \( (3.00 \times 10^{8} \, \text{m/s}) \)
- \( \lambda \) = wavelength of the radiation in meters
Given that the wavelength \( \lambda = 975 \, \text{Å} = 975 \times 10^{-10} \, \text{m} \), we can substitute the values into the formula.
### Step 2: Calculate the energy
Substituting the values into the energy formula:
\[
E = \frac{(6.626 \times 10^{-34} \, \text{J s})(3.00 \times 10^{8} \, \text{m/s})}{975 \times 10^{-10} \, \text{m}}
\]
Calculating this gives:
\[
E \approx 2.04 \times 10^{-19} \, \text{J}
\]
### Step 3: Determine the energy levels of the hydrogen atom
The energy levels of a hydrogen atom are given by the formula:
\[
E_n = -\frac{13.6 \, \text{eV}}{n^2}
\]
To find out how many energy levels can be reached with the energy \( E \) calculated, we need to convert the energy from joules to electron volts (1 eV = \( 1.6 \times 10^{-19} \, \text{J} \)):
\[
E \approx \frac{2.04 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 1.275 \, \text{eV}
\]
### Step 4: Determine the maximum principal quantum number (n)
We need to find the maximum principal quantum number \( n \) that can be reached from the ground state (n=1):
\[
E_1 = -13.6 \, \text{eV}
\]
\[
E_n = -\frac{13.6}{n^2} \, \text{eV}
\]
Setting up the equation:
\[
E_n - E_1 = 1.275 \, \text{eV}
\]
\[
-\frac{13.6}{n^2} + 13.6 = 1.275
\]
\[
\frac{13.6}{n^2} = 12.325
\]
\[
n^2 = \frac{13.6}{12.325} \approx 1.103
\]
Since \( n \) must be a whole number, the maximum \( n \) is 2.
### Step 5: Determine the possible transitions
The possible transitions from the ground state (n=1) to n=2 are:
1. n=1 to n=2
2. n=2 to n=1 (emission)
3. n=2 to n=3 (if n=3 is reached)
The possible transitions are:
- n=1 to n=2
- n=2 to n=1
- n=2 to n=3
- n=3 to n=2
- n=3 to n=1
Thus, the number of different lines possible in the resulting spectrum is 3 (from n=2 to n=1, n=3 to n=2, and n=3 to n=1).
### Step 6: Calculate the longest wavelength
The longest wavelength corresponds to the transition with the smallest energy difference, which is from n=2 to n=1.
Using the formula for the energy difference between two levels:
\[
\Delta E = E_1 - E_2 = 13.6 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 13.6 \left( 1 - \frac{1}{4} \right) = 13.6 \times \frac{3}{4} = 10.2 \, \text{eV}
\]
Now, convert this energy difference back to wavelength using:
\[
\lambda = \frac{hc}{\Delta E}
\]
Substituting the values:
\[
\lambda = \frac{(6.626 \times 10^{-34} \, \text{J s})(3.00 \times 10^{8} \, \text{m/s})}{10.2 \times 1.6 \times 10^{-19} \, \text{J}}
\]
Calculating this gives:
\[
\lambda \approx 121.6 \, \text{nm} = 12160 \, \text{Å}
\]
### Final Answer
- Number of different lines possible: 3
- Longest wavelength: 12160 Å
