The threshold wavelength of a photo diode is 750 nm.

Question

Calculate the energy gap in the photo diode in electron
volts

Answer 1

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.