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. The energy gap (E_g) given is 5.50 eV.

### Step-by-Step Solution:

1. **Understand the Energy-Wavelength Relationship**:
The energy of a photon can be expressed in terms of its wavelength using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \(E\) is the energy of the photon,
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \(c\) is the speed of light (\(3.00 \times 10^8 \, \text{m/s}\)),
- \(\lambda\) is the wavelength in meters.

2. **Convert Energy from eV to Joules**:
Since we are given the energy in electron volts (eV), we can convert it to joules. The conversion factor is:
\[ 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \]
Therefore, the energy in joules is:
\[ E = 5.50 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV} = 8.83 \times 10^{-19} \, \text{J} \]

3. **Rearranging the Formula to Solve for Wavelength**:
Rearranging the energy-wavelength formula to solve for \(\lambda\):
\[ \lambda = \frac{hc}{E} \]

4. **Substituting the Values**:
Now, substituting the values of \(h\), \(c\), and \(E\):
\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{Js})(3.00 \times 10^8 \, \text{m/s})}{8.83 \times 10^{-19} \, \text{J}} \]

5. **Calculating the Wavelength**:
Performing the calculation:
\[ \lambda = \frac{1.9878 \times 10^{-25} \, \text{Js m/s}}{8.83 \times 10^{-19} \, \text{J}} \approx 2.25 \times 10^{-7} \, \text{m} \]
Converting meters to nanometers (1 m = \(10^9\) nm):
\[ \lambda \approx 225 \, \text{nm} \]

6. **Final Answer**:
The wavelength of light that may excite an electron in the valence band of diamond to the conduction band is approximately:
\[ \lambda \approx 226 \, \text{nm} \]