Skip to content
Exercises · 14.7

Q.A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 6000 nm?

Odisha ChseTextbookSubjective· 2mImportance★★★★★est
30% · 11/37 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Compute the photon energy of the given wavelength and compare it with the band gap; since Ephoton≪EgE_{photon} \ll E_g, the diode cannot detect this wavelength (its cutoff wavelength is only about 444 nm).

Step 1 — Photon energy at λ=6000 nm\lambda = 6000\ \text{nm}

E=hcλ=(6.626×10−34 J⋅s)(3×108 m/s)6000×10−9 m=3.31×10−20 JE = \frac{hc}{\lambda} = \frac{(6.626\times10^{-34}\ \text{J·s})(3\times10^{8}\ \text{m/s})}{6000\times10^{-9}\ \text{m}} = 3.31\times10^{-20}\ \text{J}

Converting to eV (1 eV=1.6×10−19 J1\ \text{eV} = 1.6\times10^{-19}\ \text{J}):

E=3.31×10−201.6×10−19≈0.207 eVE = \frac{3.31\times10^{-20}}{1.6\times10^{-19}} \approx 0.207\ \text{eV}

Step 2 — Compare with the band gap

A photodiode can only generate an electron-hole pair (and hence produce a photocurrent) when the incident photon's energy is at least equal to the semiconductor's band gap, Ephoton≥EgE_{photon} \geq E_g. Here Eg=2.8 eVE_g = 2.8\ \text{eV}, more than 13 times larger than the 0.207 eV0.207\ \text{eV} carried by a 6000 nm photon.

Step 3 — Maximum detectable wavelength (cutoff)

The longest wavelength this photodiode can detect corresponds to Ephoton=EgE_{photon} = E_g: …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.