JEE Advanced 2015 · Paper 1 · Physics Q18

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PhysicsMultiple correct+4 / -2Question 18 of 60

For photo-electric effect with incident photon wavelength $\lambda$ , the stopping potential is $V_0$ . Identify the correct variation(s) of $V_0$ with $\lambda$ and $1/\lambda$ .

V₀ against λ: a curve falling from high values and decreasing (concave up) to zero at a finite λ.

V₀ against λ: a straight line falling to zero at a finite λ.

V₀ against 1/λ: a straight line rising from a positive intercept on the 1/λ axis.

V₀ against 1/λ: a curve rising from a positive intercept on the 1/λ axis, concave up.

Answer and worked solution

Official answer: A, C

Einstein's equation. $eV_0=\dfrac{hc}\lambda-\phi$ .

Against $1/\lambda$ : a straight line with positive slope, cutting the axis at the threshold $1/\lambda_0$ . That is graph (C).
Against $\lambda$ : a decreasing hyperbola-like curve, concave up, reaching zero at $\lambda_0$ . That is graph (A).

(A), (C)

Topics:Photoelectric effect

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JEE Advanced 2015 · Paper 1, Physics question 18