(a) A gas of hydrogen atoms is their ground state is bombarded by electrons with kinetic energy 12.5 ev. What emitted wavelengths would you expect to see?
(b) What if the electrons were replaced by photons of same anergy?.

Explanation

To solve the given problem step by step, we will analyze both parts (a) and (b) separately.

### Part (a): Emitted Wavelengths from Bombarding Electrons

1. **Identify the Energy Levels of Hydrogen**:
The energy levels of hydrogen atoms in the ground state can be expressed using the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number.

2. **Determine the Energy Difference**:
Given that the kinetic energy of the bombarding electrons is 12.5 eV, we set up the equation:
\[ E_n - E_1 = 12.5 \, \text{eV} \]
where \( E_1 = -13.6 \, \text{eV} \) (ground state energy).

3. **Substituting the Energy Levels**:
Substitute \( E_n \) into the equation:
\[ -\frac{13.6}{n^2} - (-13.6) = 12.5 \]
Simplifying gives:
\[ -\frac{13.6}{n^2} + 13.6 = 12.5 \]
\[ \frac{13.6}{n^2} = 1.1 \]

4. **Solving for n**:
Rearranging gives:
\[ n^2 = \frac{13.6}{1.1} \approx 12.36 \]
Taking the square root:
\[ n \approx 3.51 \]
Since \( n \) must be an integer, we consider \( n = 3 \).

5. **Possible Transitions**:
The possible transitions from \( n = 3 \) are:
- \( n = 3 \) to \( n = 2 \)
- \( n = 3 \) to \( n = 1 \)
- \( n = 2 \) to \( n = 1 \)

6. **Calculating Wavelengths**:
Using the Rydberg formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R = 1.097 \times 10^7 \, \text{m}^{-1} \).

- **For \( n = 3 \) to \( n = 1 \)**:
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( 1 - \frac{1}{9} \right) = 1.097 \times 10^7 \times \frac{8}{9} \]
Calculate \( \lambda \):
\[ \lambda \approx 102.55 \, \text{nm} \]

- **For \( n = 3 \) to \( n = 2 \)**:
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{4} - \frac{1}{9} \right) = 1.097 \times 10^7 \left( \frac{5}{36} \right) \]
Calculate \( \lambda \):
\[ \lambda \approx 656.33 \, \text{nm} \]

- **For \( n = 2 \) to \( n = 1 \)**:
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( 1 - \frac{1}{4} \right) = 1.097 \times 10^7 \times \frac{3}{4} \]
Calculate \( \lambda \):
\[ \lambda \approx 121.54 \, \text{nm} \]

### Summary of Part (a):
The emitted wavelengths expected are approximately:
- 102.55 nm (from \( n = 3 \) to \( n = 1 \))
- 121.54 nm (from \( n = 2 \) to \( n = 1 \))
- 656.33 nm (from \( n = 3 \) to \( n = 2 \))

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### Part (b): Emitted Wavelengths from Photons

1. **Understanding Photon Energy**:
If the electrons are replaced by photons with the same energy (12.5 eV), the energy transfer is complete. The photon energy is given by:
\[ E = h \nu = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant and \( c \) is the speed of light.

2. **Finding the Principal Quantum Number**:
The energy difference remains the same as in part (a), leading to the same calculation for \( n \):
\[ n \approx 3.51 \quad \text{(round to 3)} \]

3. **Possible Transitions**:
The transitions remain the same as in part (a), leading to the same wavelengths.

### Summary of Part (b):
The emitted wavelengths would also be the same as in part (a):
- 102.55 nm (from \( n = 3 \) to \( n = 1 \))
- 121.54 nm (from \( n = 2 \) to \( n = 1 \))
- 656.33 nm (from \( n = 3 \) to \( n = 2 \))

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