Find the wavelength of light that may excite an electron in the valence band of diamond to the conduction band. The energy gap is 5.50 eV

Options

A. 226 nm
B. 312 nm
C. 432 nm
D. 550 nm

Explanation

To find the wavelength of light that can excite an electron from the valence band to the conduction band in diamond, we can use the relationship between energy and wavelength given by the formula:

\[ E = \frac{hc}{\lambda} \]

Where:
- \( E \) is the energy gap (in electron volts, eV),
- \( h \) is Planck's constant (\( 4.135667696 \times 10^{-15} \) eV·s),
- \( c \) is the speed of light (\( 3 \times 10^8 \) m/s),
- \( \lambda \) is the wavelength (in nanometers, nm).

Given that the energy gap \( E \) is 5.50 eV, we can rearrange the formula to solve for \( \lambda \):

\[ \lambda = \frac{hc}{E} \]

### Step-by-Step Solution:

1. **Identify Constants**:
- Planck's constant \( h = 4.135667696 \times 10^{-15} \) eV·s
- Speed of light \( c = 3 \times 10^8 \) m/s
- Energy gap \( E = 5.50 \) eV

2. **Convert Planck's constant and speed of light to compatible units**:
- We can use the value of \( hc \) in eV·nm, which is \( 1240 \) eV·nm.

3. **Substitute values into the formula**:
\[ \lambda = \frac{1240 \text{ eV·nm}}{5.50 \text{ eV}} \]

4. **Calculate the wavelength**:
\[ \lambda = \frac{1240}{5.50} \approx 226.36 \text{ nm} \]

5. **Round the result**:
- The wavelength can be rounded to \( 226 \text{ nm} \).

### Final Answer:
The wavelength of light that may excite an electron in the valence band of diamond to the conduction band is approximately **226 nm**.