Question

The maximum wavelength of photons that can be detected by a photo-diode, which is made of a semiconductor of band gap $2eV$ is about :

• A. $620nm$
• B. $700nm$
• C. $740nm$
• D. $860nm$
• E. $1240nm$

Answer 1

Answer: (A)

$620nm$

The wavelength $\lambda$ (in Angstrom) of a photon of energy $E$ (in $eV$) is given by :

$\lambda E=12400$, very nearly.

Therefore, $\lambda =12400/E$

[The above expression can be easily obtained by remembering that a photon of energy $1eV$ has wavelength $12400\mathring{A}$ and the energy is inversely proportional to the wave length].

Since $E=2eV$ we have $\lambda =12400/2=6200\mathring{A}=620nm$.

Photons with wavelength greater than $640nm$ will have energy less than $2eV$ so that they will be unable to produce electron-hole pairs in the semiconductor of band gap $2eV$.