Monochromatic light of wavelength lambda = 47 | vedclass

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(D) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Using $hc \approx 12400 \ 	ext{eV} \cdot \mathring{A}$, we get $E = \frac{

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$D$ only

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(D) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Using $hc \approx 12400 \ 	ext{eV} \cdot \mathring{A}$, we get $E = \frac{12400}{4770} \ 	ext{eV} \approx 2.6 \ 	ext{eV}$.For photoelectric emission to occur, the energy of the incident photon must be greater than or equal to the work function $(\phi)$ of the metal $(E \ge \phi)$.Comparing $E = 2.6 \ 	ext{eV}$ with the work functions:For metal $A$: $2.6 \ 	ext{eV} For metal $B$: $2.6 \ 	ext{eV} For metal $C$: $2.6 \ 	ext{eV} For metal $D$: $2.6 \ 	ext{eV} > 2.3 \ 	ext{eV}$ (Emission occurs)Therefore, electrons will be emitted only from metal $D$.

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