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(D) According to Einstein's photoelectric equation,the kinetic energy $E$ of an emitted photoelectron is given by $E = \frac{hc}{\lambda} - \phi$,wher

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$E \propto \frac{1}{\lambda}, N \propto I$

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(D) According to Einstein's photoelectric equation,the kinetic energy $E$ of an emitted photoelectron is given by $E = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function of the material. This shows that $E$ depends only on the wavelength $\lambda$ of the incident light,specifically $E \propto \frac{1}{\lambda}$.Intensity $I$ of light is proportional to the number of photons incident per unit area per unit time. Since each photon interacts with one electron to cause emission,the number of emitted photoelectrons $N$ is directly proportional to the intensity $I$ of the incident light $(N \propto I)$.

$A$ light of wavelength $\lambda$ and intensity $I$ falls on a photosensitive material. If $N$ photoelectrons are emitted,each with kinetic energy $E$,then:

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