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Friday, 9 October

JEE Main 2020 · PhysicsNumerical answerNumerical valueMediumMulti-step

JEE Main 2 September 2020, Shift 1, Physics Q25

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

When radiation of wavelength λ\lambda is used to illuminate a metallic surface, the stopping potential is VV. When the same surface is illuminated with radiation of wavelength 3λ3\lambda, the stopping potential is V4\frac{V}{4}. If the threshold wavelength for the metallic surface is nλn\lambda then value of nn will be ______.

Official answer

9.00

NTA final key (Sep 2020).