A hydrogen atom in a having a binding of `0.85eV`makes transition to a state with excited energy `10.2eV`(a) identify the quantum number n of theupper and the lower energy state involved in the transition (b) Find the wavelength of the emitted radiation

Explanation

To solve the problem step by step, we will break it down into two parts as specified in the question.

### Part (a): Identify the quantum number n of the upper and lower energy states involved in the transition.

1. **Understanding Binding Energy**: The binding energy of a hydrogen atom is given as `0.85 eV`. This means that the energy of the electron in this state is `-0.85 eV` (since binding energy is negative in the context of bound states).

2. **Energy Levels of Hydrogen Atom**: The energy levels of a hydrogen atom can be calculated using the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number.

3. **Finding the Quantum Number for the Lower State**:
- We set \( E_n = -0.85 \, \text{eV} \):
\[ -0.85 = -\frac{13.6}{n^2} \]
- Rearranging gives:
\[ n^2 = \frac{13.6}{0.85} \approx 16 \]
- Thus, \( n = 4 \).

4. **Finding the Quantum Number for the Upper State**:
- The excited energy state is given as `10.2 eV`. In terms of energy, this is:
\[ E = -13.6 + 10.2 = -3.4 \, \text{eV} \]
- We set \( E_n = -3.4 \, \text{eV} \):
\[ -3.4 = -\frac{13.6}{n^2} \]
- Rearranging gives:
\[ n^2 = \frac{13.6}{3.4} = 4 \]
- Thus, \( n = 2 \).

5. **Conclusion for Part (a)**:
- The lower energy state (binding energy of 0.85 eV) corresponds to \( n = 4 \).
- The upper energy state (excited energy of 10.2 eV) corresponds to \( n = 2 \).

### Part (b): Find the wavelength of the emitted radiation.

1. **Using the Rydberg Formula**: The wavelength of the emitted radiation during the transition can be calculated using the Rydberg formula:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
where \( R_H \) is the Rydberg constant for hydrogen, approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \), \( n_f \) is the final state, and \( n_i \) is the initial state.

2. **Substituting Values**:
- Here, \( n_f = 2 \) and \( n_i = 4 \):
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]
- Calculating the fractions:
\[ \frac{1}{2^2} = \frac{1}{4}, \quad \frac{1}{4^2} = \frac{1}{16} \]
- Thus:
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{4} - \frac{1}{16} \right) = 1.097 \times 10^7 \left( \frac{4 - 1}{16} \right) = 1.097 \times 10^7 \left( \frac{3}{16} \right) \]

3. **Calculating**:
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{3}{16} \approx 2.06 \times 10^6 \, \text{m}^{-1} \]
- Therefore:
\[ \lambda = \frac{1}{2.06 \times 10^6} \approx 4.87 \times 10^{-7} \, \text{m} = 487 \, \text{nm} \]

4. **Conclusion for Part (b)**:
- The wavelength of the emitted radiation is approximately \( 487 \, \text{nm} \).

### Final Answers:
- (a) \( n = 4 \) (lower state), \( n = 2 \) (upper state)
- (b) Wavelength \( \lambda \approx 487 \, \text{nm} \)