The energy levels of a hypothetical one electron atom are given by `E_n = -(18.0)/(n^2) eV`
where n = 1,2,3,…. (a) Compute the four lowest energy levels and construct the energy levels diagram. (b) What is the first excitation potential (c ) What wavelength (Å) can be emitted when these atoms in the ground state are bombarged by electrons that have been accelerated through a potential difference of 16.2 V? (e) what is the photoelectric threshold wavelength of this atom?

Explanation

Let's break down the solution step by step for the given problem.

### Part (a): Compute the four lowest energy levels and construct the energy levels diagram.

1. **Energy Level Formula**: The energy levels of the hypothetical one-electron atom are given by:
\[ E_n = -\frac{18.0}{n^2} \text{ eV} \]
where \( n = 1, 2, 3, \ldots \)

2. **Calculate the Energy Levels**:
- For \( n = 1 \):
\[ E_1 = -\frac{18.0}{1^2} = -18.0 \text{ eV} \]
- For \( n = 2 \):
\[ E_2 = -\frac{18.0}{2^2} = -\frac{18.0}{4} = -4.5 \text{ eV} \]
- For \( n = 3 \):
\[ E_3 = -\frac{18.0}{3^2} = -\frac{18.0}{9} = -2.0 \text{ eV} \]
- For \( n = 4 \):
\[ E_4 = -\frac{18.0}{4^2} = -\frac{18.0}{16} = -1.125 \text{ eV} \]

3. **Energy Level Diagram**:
- Draw a vertical line representing energy levels.
- Label the levels as follows:
- \( E_1 = -18.0 \text{ eV} \)
- \( E_2 = -4.5 \text{ eV} \)
- \( E_3 = -2.0 \text{ eV} \)
- \( E_4 = -1.125 \text{ eV} \)

### Part (b): What is the first excitation potential?

1. **First Excitation Potential**: The first excitation potential is the energy difference between the first and second energy levels:
\[ \text{Excitation Potential} = E_2 - E_1 = (-4.5) - (-18.0) = 13.5 \text{ eV} \]

### Part (c): What wavelength (Å) can be emitted when these atoms in the ground state are bombarded by electrons that have been accelerated through a potential difference of 16.2 V?

1. **Energy of Accelerated Electrons**: The energy gained by an electron when accelerated through a potential difference \( V \) is given by:
\[ E = eV = 16.2 \text{ eV} \]

2. **Possible Transitions**:
- The maximum energy difference that can be achieved is \( E_4 - E_1 \):
\[ E_4 - E_1 = (-1.125) - (-18.0) = 16.875 \text{ eV} \]
- Since \( 16.875 \text{ eV} > 16.2 \text{ eV} \), transitions from \( n=1 \) to \( n=3 \) are possible.

3. **Calculate Wavelengths**:
- For \( \lambda_{32} \):
\[ \lambda_{32} = \frac{12375}{E_3 - E_2} = \frac{12375}{(-2.0) - (-4.5)} = \frac{12375}{2.5} = 4950 \text{ Å} \]
- For \( \lambda_{31} \):
\[ \lambda_{31} = \frac{12375}{E_3 - E_1} = \frac{12375}{(-2.0) - (-18.0)} = \frac{12375}{16.0} = 773 \text{ Å} \]
- For \( \lambda_{21} \):
\[ \lambda_{21} = \frac{12375}{E_2 - E_1} = \frac{12375}{(-4.5) - (-18.0)} = \frac{12375}{13.5} = 917 \text{ Å} \]

### Part (d): What is the photoelectric threshold wavelength of this atom?

1. **Photoelectric Threshold Energy**: The photoelectric threshold energy corresponds to the first excitation potential:
\[ E_{\text{threshold}} = 13.5 \text{ eV} \]

2. **Calculate the Threshold Wavelength**:
\[ \lambda_{\text{threshold}} = \frac{12375}{E_{\text{threshold}}} = \frac{12375}{13.5} = 915.74 \text{ Å} \]

### Summary of Answers:
- (a) Energy levels: \( E_1 = -18.0 \text{ eV}, E_2 = -4.5 \text{ eV}, E_3 = -2.0 \text{ eV}, E_4 = -1.125 \text{ eV} \)
- (b) First excitation potential: \( 13.5 \text{ eV} \)
- (c) Wavelengths emitted: \( 4950 \text{ Å}, 773 \text{ Å}, 917 \text{ Å} \)
- (d) Photoelectric threshold wavelength: \( 915.74 \text{ Å} \)