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NEET UG PHYSICS MCQ: DUAL NATURE OF RADIATION AND MATTER

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Application +4 | -1
Wave equation of light used in PEE at a given point is given by E=E0[sin⁡5ωt+cos⁡15ωt]\mathrm{E}=\mathrm{E}_0[\sin 5 \omega \mathrm{t}+\cos 15 \omega \mathrm{t}] here t\mathrm{t} is instant, If work function of electron emitter is ϕ\phi then value of stopping potential will be:-
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Correct option: B
he(15ω2π)−ϕe\frac{\mathrm{h}}{\mathrm{e}}\left(\frac{15 \omega}{2 \pi}\right)-\frac{\phi}{\mathrm{e}}
Option B (he(15ω2π)−ϕe\frac{\mathrm{h}}{\mathrm{e}}\left(\frac{15 \omega}{2 \pi}\right)-\frac{\phi}{\mathrm{e}}) is correct because the stopping potential is determined by the highest frequency component of the incident light, which is 15ω15\omega.

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