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.
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.
To convert the frequency to energy in electron volts (eV), you can use the formula:
E = hν
where E is the energy, h is Planck's constant (6.63 x 10^-34 J s), and ν is the frequency.
Given the frequency ν = 5 x 10^14 Hz, we can calculate the energy using the formula:
E = (6.63 x 10^-34 J s) * (5 x 10^14 Hz)
E = 3.315 x 10^-19 J
To convert this energy to electron volts, we can use the conversion factor: 1 eV = 1.602 x 10^-19 J.
So, the band edge of this material in electron volts is:
E = (3.315 x 10^-19 J) / (1.602 x 10^-19 J/eV)
E = 2.07 eV
Therefore, the band edge of this material is 2.07 eV.
To find the wavelength at which the material is 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.
Given the frequency ν = 5 x 10^14 Hz, we can calculate the wavelength using the formula:
λ = (3 x 10^8 m/s) / (5 x 10^14 Hz)
λ = 6 x 10^-7 m
To convert this wavelength to nanometers, we multiply it by 10^9:
λ = 6 x 10^-7 m * 10^9 nm/m
λ = 600 nm
Therefore, the material is expected to be transparent above a wavelength of 600 nm.
To find the band edge in eV, we can use the equation E = hc/λ, where E is the energy, h is Planck's constant (6.626 x 10^-34 J.s), c is the speed of light (3 x 10^8 m/s), and λ is the wavelength.
First, let's convert the frequency to wavelength using the equation ν = c/λ. Rearranging the equation, we get λ = c/ν.
λ = c/ν = (3 x 10^8 m/s) / (5 x 10^14 Hz) = 6 x 10^-7 m = 600 nm (approximately)
Now we can use this wavelength to find the energy in eV.
E = hc/λ = (6.626 x 10^-34 J.s) * (3 x 10^8 m/s) / (6 x 10^-7 m) = 3.313 x 10^-19 J
To convert this energy to eV, we can use the conversion factor 1 eV = 1.602 x 10^-19 J.
E (in eV) = (3.313 x 10^-19 J) / (1.602 x 10^-19 J/eV) = 2.07 eV (approximately)
Therefore, the band edge of the material is 2.07 eV.
To determine above what wavelength the material will be transparent, we need to consider that transparency generally occurs when the energy of incoming photons is not enough to excite the electrons in the material.
For a transparent material, the band edge typically corresponds to the minimum energy level required for the electrons to be excited. Since we already know the band edge is 2.07 eV, we need to find the corresponding wavelength.
Using the equation E = hc/λ and rearranging for λ, we get λ = hc/E.
λ = (6.626 x 10^-34 J.s) * (3 x 10^8 m/s) / (2.07 eV * 1.602 x 10^-19 J/eV) = 5.996 x 10^-7 m = 599.6 nm (approximately).
Therefore, we expect the material to be transparent above a wavelength of 599.6 nm.