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(C) The energy of an incident photon is given by the formula $E = \frac{hc}{\lambda}$.Using the approximation $hc \approx 12400\, eV\cdot\mathring{A}$

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$A$ and $B$ only

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(C) The energy of an incident photon is given by the formula $E = \frac{hc}{\lambda}$.Using the approximation $hc \approx 12400\, eV\cdot\mathring{A}$,we get:$E = \frac{12400\, eV\cdot\mathring{A}}{4100\, \mathring{A}} \approx 3.02\, eV$.Photoelectric emission occurs if the energy of the incident photon is greater than 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,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,the metals which will emit photoelectrons for a radiation of wavelength $4100\, \mathring{A}$ is/are:

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