The threshold wavelength of a photo diode is 750 nm.
Calculate the energy gap in the photo diode in electron
volts
To calculate the energy gap (E) in electron volts (eV), we can use the formula:
E = (hc) / λ
Where:
- h is Planck's constant (approximately 6.62607015 × 10^-34 Js)
- c is the speed of light (approximately 2.998 × 10^8 m/s)
- λ is the wavelength in meters
First, we need to convert the threshold wavelength from nanometers (nm) to meters (m):
750 nm = 750 × 10^-9 m = 7.5 × 10^-7 m
Now we can substitute the values into the formula:
E = (6.62607015 × 10^-34 Js × 2.998 × 10^8 m/s) / (7.5 × 10^-7 m)
E ≈ 2.648 × 10^-19 J
To convert the energy from joules (J) to electron volts (eV), we can use the conversion factor:
1 eV = 1.60219 × 10^-19 J
So, the energy gap in the photo diode is approximately:
E ≈ (2.648 × 10^-19 J) / (1.60219 × 10^-19 J/eV)
E ≈ 1.65 eV
Therefore, the energy gap in the photo diode is approximately 1.65 electron volts.