The graphs below show variation of stopping potential versus frequency of incident radiation for…
Physics20262 marksShort answer
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.
Answer
Answer
AI
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.
