Exams › NEET › Physics

A p-n photodiode is fabricated from a semiconductor with a band gap of 2.5 eV. It can detect a signal of wavelength

1. 4000 nm
2. 6000 nm
3. 4000 Å
4. 6000 Å

Correct answer: 4000 Å

Solution

The energy of a photon is given by E = hc/λ. For a band gap of 2.5 eV, the maximum wavelength it can detect is λ = hc/E = (12400 Å·eV) / 2.5 eV = 4960 Å. Thus, the photodiode can detect wavelengths shorter than 4960 Å, and 4000 Å is within this range.

Related NEET Physics questions

• An α-particle moves in a circular path of radius 0.83 cm in the presence of a magnetic field of 0.25 Wb/m². The wavelength associated with the particle will be:
• For photoelectric emission from certain metal the cut-off frequency is ν. If radiation of frequency 2ν impinges on the metal plate, the maximum possible velocity of the emitted electron will be (m is the electron mass)
• Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The wavelength of the emitted electron is
• A source of light is placed at a distance of 50 cm from a photocell and the stopping potential is found to be V₀. If the distance between the light source and photocell is made 25 cm, the new stopping potential will be
• A photoelectric surface is illuminated successively by monochromatic light of wavelength λ and λ/2. If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface of the material is (h = Planck's constant, c = speed of light)
• A certain metallic surface is illuminated with monochromatic light of wavelength λ. The stopping potential for photoelectric current for this light is 3V₀. If the same surface is illuminated with light of wavelength 2λ, the stopping potential is V₀. The threshold wavelength for this surface for photo-electric effect is

⚔️ Practice NEET Physics free + battle 1v1 →