Question

If a semiconductor photodiode can detect a photon with a maximum wavelength of 400 nm, then its band gap energy is: Planck's constant h

(a) 1.55 eV
(b) 2.78 eV
(c) 3.10 eV
(d) 4.13 eV

Answer 1

To determine the band gap energy of a semiconductor photodiode that can detect photons at a maximum wavelength of 400 nm, we use the equation E = hc/λ with hc = 1240 eV·nm, resulting in a band gap energy of 3.10 eV, which is option (c).

Step-by-step explanation:

The question is asking for the band gap energy of a semiconductor photodiode that can detect photons with a maximum wavelength of 400 nm. To find the energy of a photon with this wavelength, we use the equation E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength.

Given that hc is equal to 1240 eV·nm, we calculate the energy as follows: E = 1240 eV·nm / 400 nm = 3.10 eV. Therefore, the correct answer is (c) 3.10 eV, which is the band gap energy corresponding to a maximum detectable wavelength of 400 nm for the semiconductor photodiode.