Three photodiodes : D1, D2 and D3 are made of semiconductors having band gaps of 2.5 eV, 1.8eV and 3 eV respectively. Which one will be able to detect light of wavelength 6000 Å? D1D2D1 & D2D1, D2, D3

To answer this question, we first need to understand the relationship between the energy of a photon and its wavelength. This relationship is given by the equation:

E = hc/λ

where: E is the energy of the photon, h is Planck's constant (approximately 6.63 x 10^-34 Joule seconds), c is the speed of light (approximately 3 x 10^8 meters per second), and λ is the wavelength of the light.

The energy E is also related to the band gap energy of the semiconductor material. A photodiode can detect light if the energy of the photons is greater than or equal to the band gap energy of the semiconductor.

First, we need to convert the wavelength from Angstroms to meters. 1 Angstrom is 10^-10 meters, so 6000 Angstroms is 6000 x 10^-10 = 6 x 10^-7 meters.

Next, we substitute the values into the equation to find the energy of the photon:

E = (6.63 x 10^-34 J.s)(3 x 10^8 m/s) / (6 x 10^-7 m) E = 3.315 x 10^-19 Joules

To compare this with the band gap energies, we need to convert the energy to electronvolts (eV). 1 eV is approximately 1.6 x 10^-19 Joules, so:

E = 3.315 x 10^-19 J / (1.6 x 10^-19 J/eV) E = 2.07 eV

Therefore, the photodiodes made of semiconductors with band gaps less than or equal to 2.07 eV will be able to detect the light. These are D1 (2.5 eV) and D2 (1.8 eV). So, the answer is D1 & D2.