For a certain hypothetical one electron atom, the wavelength `(in Å)` for the spectral lines for transitions originating at n=p and terminating at n=1 are given by `lambda = (1500 p^2)/(p^2 - 1), where p = 2,3,4`
(a)Find the wavelength of the least energetic and the most energetic photons in this series.
(b) Construct an energy level diagram for this element showing the energies of the lowest three levels.
(c ) What is the ionization potential fo this elelment?

Explanation

To solve the problem step by step, we will break it down into parts (a), (b), and (c) as per the question.

### Part (a): Finding the Wavelength of the Least Energetic and Most Energetic Photons

1. **Given Formula**: The wavelength for the spectral lines is given by:
\[ \lambda = \frac{1500 p^2}{p^2 - 1} \]
where \( p = 2, 3, 4 \).

2. **Calculate Wavelength for Each Value of p**:
- For \( p = 2 \):
\[ \lambda_2 = \frac{1500 \times 2^2}{2^2 - 1} = \frac{1500 \times 4}{4 - 1} = \frac{6000}{3} = 2000 \, \text{Å} \]
- For \( p = 3 \):
\[ \lambda_3 = \frac{1500 \times 3^2}{3^2 - 1} = \frac{1500 \times 9}{9 - 1} = \frac{13500}{8} = 1687.5 \, \text{Å} \]
- For \( p = 4 \):
\[ \lambda_4 = \frac{1500 \times 4^2}{4^2 - 1} = \frac{1500 \times 16}{16 - 1} = \frac{24000}{15} = 1600 \, \text{Å} \]

3. **Identify Least and Most Energetic Photons**:
- The least energetic photon corresponds to the longest wavelength:
- \( \lambda_{\text{max}} = 2000 \, \text{Å} \) (for \( p = 2 \))
- The most energetic photon corresponds to the shortest wavelength:
- \( \lambda_{\text{min}} = 1600 \, \text{Å} \) (for \( p = 4 \))

### Part (b): Constructing an Energy Level Diagram

1. **Energy Calculation**:
- The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
- Where \( h = 4.135667696 \times 10^{-15} \, \text{eV s} \) and \( c = 3 \times 10^{10} \, \text{cm/s} \).

2. **Calculate Energies**:
- For \( \lambda_2 = 2000 \, \text{Å} \):
\[ E_2 = \frac{1240 \, \text{eV} \cdot \text{Å}}{2000 \, \text{Å}} = 0.62 \, \text{eV} \]
- For \( \lambda_3 = 1687.5 \, \text{Å} \):
\[ E_3 = \frac{1240 \, \text{eV} \cdot \text{Å}}{1687.5 \, \text{Å}} \approx 0.735 \, \text{eV} \]
- For \( \lambda_4 = 1600 \, \text{Å} \):
\[ E_4 = \frac{1240 \, \text{eV} \cdot \text{Å}}{1600 \, \text{Å}} = 0.775 \, \text{eV} \]

3. **Energy Level Diagram**:
- \( E_1 = -8.25 \, \text{eV} \) (ground state)
- \( E_2 = -7.63 \, \text{eV} \) (calculated from transitions)
- \( E_3 = -7.52 \, \text{eV} \)

### Part (c): Ionization Potential

1. **Ionization Potential**:
- The ionization potential is the energy required to remove an electron from the ground state to infinity:
\[ \text{Ionization Potential} = |E_1| = 8.25 \, \text{eV} \]

### Final Answers
- (a) Least energetic photon: \( 2000 \, \text{Å} \); Most energetic photon: \( 1600 \, \text{Å} \)
- (b) Energy levels: \( E_1 = -8.25 \, \text{eV}, E_2 = -7.63 \, \text{eV}, E_3 = -7.52 \, \text{eV} \)
- (c) Ionization potential: \( 8.25 \, \text{eV} \)