Problem 23
Question
The semiconductor GaP has a band gap of \(2.2 \mathrm{eV}\). Green LEDs are made from pure GaP. What wavelength of light would be emitted from an LED made from GaP?
Step-by-Step Solution
Verified Answer
The wavelength of light emitted from a GaP-based green LED is approximately \(565 \, nm\).
1Step 1: Identify the given values
We are given the band gap energy of GaP semiconductor as \(2.2 \mathrm{eV}\).
2Step 2: Convert energy to joules
To work with the wavelength formula, we need the energy value in joules. To convert electron volts (eV) to joules (J), use the following conversion factor: \(1\, electronvolt (eV) = 1.6 \times 10^{-19}\,Joules\).
So, we have:
\(E = 2.2 \, eV \times \frac{1.6 \times 10^{-19} \, Joules}{1 \, eV}\)
\(E = 3.52 \times 10^{-19} \, Joules\)
3Step 3: Apply the photon energy-wavelength formula
The energy of a photon is related to its wavelength by the following formula:
\(E = h \frac{c}{\lambda}\)
Where,
\(E\) is the energy of the photon,
\(h\) is the Planck's constant (\(6.63 \times 10^{-34} \, Js\)),
\(c\) is the speed of light (\(3 \times 10^8 \, m/s\)),
\(\lambda\) is the wavelength of the photon.
4Step 4: Solve for the wavelength
Rearrange the photon energy-wavelength formula to solve for the wavelength:
\(\lambda = h \frac{c}{E}\)
Now, plug in the values of \(E\), \(h\), and \(c\) into the equation to get \(\lambda\):
\(\lambda = \frac{(6.63 \times 10^{-34} \, Js)(3 \times 10^8 \, m/s)}{3.52 \times 10^{-19} \, Joules}\)
5Step 5: Calculate the wavelength
Solve for the wavelength, \(\lambda\):
\(\lambda = \frac{(6.63 \times 10^{-34} \, Js)(3 \times 10^8 \, m/s)}{3.52 \times 10^{-19} \, Joules} = 5.65 \times 10^{-7} \, m\)
6Step 6: Convert wavelength to nanometers
Finally, to express the wavelength in nanometers, convert meters to nanometers by using the conversion factor (\(1\,m = 1 \times 10^{9}\,nm\)):
\(\lambda = 5.65 \times 10^{-7} \, m \times \frac{1 \times 10^{9} \, nm}{1 \, m}\)
\(\lambda = 565 \, nm\)
So, the wavelength of light emitted from a GaP-based green LED is approximately \(565 \, nm\).
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