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 follow these steps:

### Step 1: Determine the Energy of the Transition
The energy of the transition can be calculated using the formula:
\[ E = 13.6 \, Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( Z \) is the atomic number of carbon, which is 6.
- \( n_1 = 1 \) (final state)
- \( n_2 = 3 \) (initial state)

### Step 2: Substitute Values into the Energy Formula
Substituting the values into the formula:
\[ E = 13.6 \times 6^2 \left( \frac{1}{1^2} - \frac{1}{3^2} \right) \]
Calculating \( Z^2 \):
\[ Z^2 = 6^2 = 36 \]
Now, calculate the fractions:
\[ \frac{1}{1^2} - \frac{1}{3^2} = 1 - \frac{1}{9} = \frac{9}{9} - \frac{1}{9} = \frac{8}{9} \]
Now substituting back:
\[ E = 13.6 \times 36 \times \frac{8}{9} \]

### Step 3: Calculate the Energy
Calculating \( E \):
\[ E = 13.6 \times 36 \times \frac{8}{9} = 13.6 \times 4 \times 8 = 13.6 \times 32 = 435.2 \, \text{eV} \]

### Step 4: Convert Energy to Joules
To convert electron volts to joules, use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ E = 435.2 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 6.9632 \times 10^{-17} \, \text{J} \]

### Step 5: Use the Energy to Find Wavelength
Using the formula \( E = \frac{hc}{\lambda} \), where:
- \( h = 6.626 \times 10^{-34} \, \text{J s} \) (Planck's constant)
- \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light)

Rearranging for \( \lambda \):
\[ \lambda = \frac{hc}{E} \]

### Step 6: Substitute Values to Calculate Wavelength
Substituting the values:
\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{J s}) \times (3 \times 10^8 \, \text{m/s})}{6.9632 \times 10^{-17} \, \text{J}} \]
Calculating \( \lambda \):
\[ \lambda = \frac{1.9878 \times 10^{-25}}{6.9632 \times 10^{-17}} \approx 2.85 \times 10^{-9} \, \text{m} \]

### Step 7: Convert Wavelength to Nanometers
To convert meters to nanometers:
\[ \lambda = 2.85 \times 10^{-9} \, \text{m} = 2.85 \, \text{nm} \]

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

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