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(C) The energy of an incident photon is given by $E = \frac{hc}{\lambda}$.Using the approximation $E \approx \frac{12400}{\lambda (	ext{in } \mathrin

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$A$ and $B$ only

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(C) The energy of an incident photon is given by $E = \frac{hc}{\lambda}$.Using the approximation $E \approx \frac{12400}{\lambda (	ext{in } \mathring{A})} eV$,we get:$E = \frac{12400}{4100} \approx 3.02\, eV$.Photoelectric emission occurs if the energy of the incident photon is greater than or equal to the work function $(\phi)$ of the metal.For metal $A$: $\phi_A = 1.92\, eV$. Since $3.02\, eV > 1.92\, eV$,metal $A$ will emit photoelectrons.For metal $B$: $\phi_B = 2.0\, eV$. Since $3.02\, eV > 2.0\, eV$,metal $B$ will emit photoelectrons.For metal $C$: $\phi_C = 5.0\, eV$. Since $3.02\, eV Therefore,only metals $A$ and $B$ will emit photoelectrons.

The work functions for metals $A, B$ and $C$ are respectively $1.92\, eV, 2.0\, eV$ and $5\, eV$. According to Einstein's photoelectric equation,which of these metals will emit photoelectrons for an incident radiation of wavelength $4100\, \mathring{A}$?

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