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(D) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Given $h = 6.63 	imes 10^{-34} J s$, $c = 3 	imes 10^8 m/s$, and $\lambda

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$A$ and $B$

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(D) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Given $h = 6.63 	imes 10^{-34} J s$, $c = 3 	imes 10^8 m/s$, and $\lambda = 4100 	imes 10^{-10} m$.$E = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}{4100 	imes 10^{-10}} J = 4.85 	imes 10^{-19} J$.To convert this energy into electron-volts $(eV)$, divide by $e = 1.6 	imes 10^{-19} C$:$E = \frac{4.85 	imes 10^{-19}}{1.6 	imes 10^{-19}} eV \approx 3.03 eV$.Photoelectric emission occurs if the incident photon energy is greater than the work function $(\Phi)$ of the metal.For metal $A$: $\Phi_A = 1.92 eV For metal $B$: $\Phi_B = 2.0 eV For metal $C$: $\Phi_C = 5 eV > 3.03 eV$ (No emission).Therefore, metals $A$ and $B$ will emit photoelectrons.

The work functions for metals $A$, $B$, and $C$ are $1.92 eV$, $2.0 eV$, and $5 eV$ respectively. The metal(s) which will emit photoelectrons for incident radiation of wavelength $4100 Å$ is/are $[h=6.63 \times 10^{-34} J s, e=1.6 \times 10^{-19} C, c=3 \times 10^8 m/s]$.

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