a) To find the band edge in eV, we can use the formula:
E = hν, where E is the energy, h is the Planck's constant (4.13567 x 10^-15 eV·s), and ν is the frequency (5 x 10^14 Hz).
E = (4.13567 x 10^-15 eV·s)(5 x 10^14 s^-1)
E ≈ 2.07 eV
So, the band edge of this material is approximately 2.07 eV.
b) To find the wavelength at which the material becomes transparent, we can use the formula:
λ = c/ν, where λ is the wavelength, c is the speed of light (3 x 10^8 m/s), and ν is the frequency (5 x 10^14 Hz).
λ = (3 x 10^8 m/s) / (5 x 10^14 s^-1)
λ ≈ 599.6 x 10^-9 m
Converting to nanometers, the wavelength is approximately 599.6 nm. Therefore, the material is expected to be transparent at wavelengths above 599.6 nm.
A material exhibits an optical band edge at ν=5∗1014 Hz (s-1).
a) What is the band edge (in eV) of this material?
Answer:2.07; E=hc
Above what wavelength do you expect the material to be transparent? Express your answer in nm.
Answer: 599.6: landa= C/V y pasa a nanometros.
1 answer
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