The stopping potentials for light of frequencies ν_1 and ν_2 on a metal are V_1 and V_2. Planck's constant is given by

(C) (e(V_1 - V_2))/(ν_1 - ν_2). eV_1 = hν_1 - φ and eV_2 = hν_2 - φ. Subtracting: h = (e(V_1 - V_2))/(ν_1 - ν_2), the slope of the V_0–ν graph times e.

Dual Nature of Radiation and Matter question for NEET / JEE Main, with solution

Q 12-11-023NEETJEE MainMedium

The stopping potentials for light of frequencies $\nu_1$ and $\nu_2$ on a metal are $V_1$ and $V_2$. Planck's constant is given by

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Dual Nature of Radiation and Matter question for NEET / JEE Main, with solution

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