(i) The graphs below show variation of stopping potential versus frequency of incident radiation…
(i) The graphs below show variation of stopping potential versus frequency of incident radiation for metals A and B.
(i)[2.0]
The graphs below show variation of stopping potential versus frequency of incident radiation for metals A and B.
UV radiation of appropriate wavelength is allowed to fall on both the metals. Which metal will emit photoelectrons with higher maximum kinetic energy ($E_{max}$)? Give a reason.
(ii)[1.0]
State the conclusion that was drawn from Davisson-Germer's experiment.
Answer
Answer (i)
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Metal A will emit photoelectrons with higher maximum kinetic energy.
Reason: From the graph, Metal A has a lower threshold frequency than Metal B ($f_{0A} < f_{0B}$), which means Metal A has a smaller work function ($W_{0} = hf_{0}$). According to Einstein's photoelectric equation, $E_{\max} = hf - W_{0}$; hence, for the same incident frequency $f$, a smaller work function results in a higher maximum kinetic energy.
Answer (ii)
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Davisson-Germer's experiment verified the de Broglie hypothesis and confirmed the wave nature of electrons (matter waves).
From ISC 2026 Physics Paper 1, question 17.