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Saturday, 31 October

JEE Main 2021 · PhysicsMultiple choiceSingle correctEasyCalculation

JEE Main 26 August 2021, Shift 1, Physics Q7

Question 7 of 90 in this shift, Physics question 7 of 30, Section A.

In a photoelectric experiment ultraviolet light of wavelength 280 nm is used with lithium cathode having work function ϕ=2.5\phi=2.5 eV. If the wavelength of incident light is switched to 400 nm, find out the change in the stopping potential. (h=6.63×10−34h=6.63\times10^{-34} Js, c=3×108 ms−1c=3\times10^{8}\ \mathrm{ms^{-1}})
  1. (1)1.3 VOfficial answer
  2. (2)1.9 V
  3. (3)0.6 V
  4. (4)1.1 V

Official answer

Option 1

NTA final key.