An electron in `C^(5+)` ion during the transition from `n = 3` to `n = 1` emits light of wavelength

Options

A. `2.85 nm`
B. `3.6 nm`
C. `8.7 nm`
D. `9.8 nm`

Answer

To find the wavelength of light emitted during the transition of an electron in the \( C^{5+} \) ion from \( n = 3 \) to \( n = 1 \), we can use the Rydberg formula for hydrogen-like atoms:

\[ \frac{1}{\lambda} = Z^2 R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Where:
- \( \lambda \) is the wavelength of emitted light,
- \( Z \) is the atomic number of the ion,
- \( R \) is the Rydberg constant (\( R \approx 1.097 \times 10^7 \, \text{m}^{-1} \)),
- \( n_1 \) is the lower energy level (1 in this case),
- \( n_2 \) is the higher energy level (3 in this case).

### Step-by-Step Solution:

1. **Identify the values:**
- For carbon (\( C \)), the atomic number \( Z = 6 \).
- Transition from \( n_2 = 3 \) to \( n_1 = 1 \).

2. **Substitute the values into the Rydberg formula:**
\[ \frac{1}{\lambda} = 6^2 \times 1.097 \times 10^7 \left( \frac{1}{1^2} - \frac{1}{3^2} \right) \]

3. **Calculate \( Z^2 \):**
\[ Z^2 = 6^2 = 36 \]

4. **Calculate the difference in the fractions:**
\[ \frac{1}{1^2} - \frac{1}{3^2} = 1 - \frac{1}{9} = 1 - 0.1111 = 0.8889 \]

5. **Combine the results:**
\[ \frac{1}{\lambda} = 36 \times 1.097 \times 10^7 \times 0.8889 \]

6. **Calculate \( 36 \times 1.097 \times 0.8889 \):**
\[ 36 \times 1.097 \approx 39.492 \]
\[ 39.492 \times 0.8889 \approx 35.104 \]

7. **Final calculation for \( \frac{1}{\lambda} \):**
\[ \frac{1}{\lambda} = 35.104 \times 10^7 \, \text{m}^{-1} \]

8. **Calculate \( \lambda \):**
\[ \lambda = \frac{1}{35.104 \times 10^7} \approx 2.85 \times 10^{-8} \, \text{m} \]

9. **Convert to nanometers:**
\[ \lambda \approx 2.85 \, \text{nm} \quad (\text{since } 1 \, \text{nm} = 10^{-9} \, \text{m}) \]

### Final Answer:
The wavelength of light emitted during the transition from \( n = 3 \) to \( n = 1 \) in the \( C^{5+} \) ion is approximately \( 2.85 \, \text{nm} \).