To find the maximum wavelength that a photodetector can detect given its band gap energy, we can follow these steps:
### Step-by-Step Solution:
1. **Understand the relationship between energy and wavelength**: The energy of a photon is given by the equation:
\[
E = h \nu
\]
where \( E \) is the energy, \( h \) is Planck's constant, and \( \nu \) is the frequency of the photon. The frequency can also be related to the wavelength (\( \lambda \)) by the equation:
\[
\nu = \frac{c}{\lambda}
\]
where \( c \) is the speed of light. Therefore, we can express the energy in terms of wavelength:
\[
E = \frac{hc}{\lambda}
\]
2. **Substitute the given band gap energy**: The band gap energy of the semiconductor is given as \( 0.73 \, \text{eV} \). We need to convert this energy into joules because Planck's constant and the speed of light are typically in SI units. The conversion factor is:
\[
1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}
\]
Thus,
\[
E = 0.73 \, \text{eV} = 0.73 \times 1.6 \times 10^{-19} \, \text{J} = 1.168 \times 10^{-19} \, \text{J}
\]
3. **Use the formula to find the maximum wavelength**: Rearranging the energy equation to solve for wavelength gives:
\[
\lambda = \frac{hc}{E}
\]
Now, substitute the values for \( h \) (Planck's constant) and \( c \) (speed of light):
- \( h = 6.626 \times 10^{-34} \, \text{J s} \)
- \( c = 3 \times 10^8 \, \text{m/s} \)
Substituting these values into the equation:
\[
\lambda = \frac{(6.626 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{1.168 \times 10^{-19} \, \text{J}}
\]
4. **Calculate the wavelength**:
\[
\lambda = \frac{1.9878 \times 10^{-25} \, \text{J m}}{1.168 \times 10^{-19} \, \text{J}} \approx 1.700 \times 10^{-7} \, \text{m}
\]
5. **Convert the wavelength to angstroms**: Since \( 1 \, \text{angstrom} = 10^{-10} \, \text{m} \):
\[
\lambda \approx 1.700 \times 10^{-7} \, \text{m} = 1700 \, \text{nm} = 17000 \, \text{angstroms}
\]
6. **Final result**: The maximum wavelength that the photodetector can detect is approximately \( 17030 \, \text{angstroms} \).
### Final Answer:
The maximum wavelength the photodetector can detect is **17030 angstroms**.
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