Learn

Saturday, 31 October

JEE Main 2020 · PhysicsMultiple choiceSingle correctEasyCalculation

JEE Main 3 September 2020, Shift 2, Physics Q19

Question 19 of 75 in this shift, Physics question 19 of 25.

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=6.63×10−34h = 6.63 \times 10^{-34} J.s. Speed of light c=3×108c = 3 \times 10^{8} m/s
  1. (1)3.1 eVOfficial answer
  2. (2)1.1 eV
  3. (3)2.0 eV
  4. (4)1.5 eV

Official answer

Option 1

NTA final key (Sep 2020).