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The following graph shown the variation of stopping potential V0 with the frequency v of the incident radiation for two photosensitive metals P and Q: (i) Explain which metal has smaller threshold wavelengths.(ii) Explain, giving reason, which metal emits photoelectrons having smaller kinetic energy. (iii) If the distance between the light source and metal P is doubled, how will the stopping potential change? (i) Suppose the frequency of incident radiations of metal Q and P be v0 and v0 ’ respectively.  E = hv0 (iii) Stopping potential remains unaffected because the value of stopping potential for a given metal surface does not depend on the intensity of the incident radiation. It depends on the frequency of incident radiation.
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CONCEPT:
λ = h/ √ 2mev
h = 6.63 x 10-34 me = mass of an electron v = potential difference EXPLANATIONS: we know Plancks constant (h) = 6.63 x 10-34. Volt = 100 v Mass of electron (me ) = 9.1 x 10-31.. Putting all these values in λ = h/√ 2mev 6.63 x 10-34/√ 2 x 9.1 x 10-31 x 100 0.123 nm Option 1 is correct. India’s #1 Learning Platform Start Complete Exam Preparation
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