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(D) According to the photoelectric effect,the energy of a photon is given by $E = \frac{hc}{\lambda}$.For photoemission to occur,the energy of the inc

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(D) According to the photoelectric effect,the energy of a photon is given by $E = \frac{hc}{\lambda}$.For photoemission to occur,the energy of the incident photon must be greater than or equal to the work function $\Phi$ of the metal surface,i.e.,$E \ge \Phi$.This implies $\frac{hc}{\lambda} \ge \Phi$,or $\lambda \le \frac{hc}{\Phi} = \lambda_0$,where $\lambda_0$ is the threshold wavelength.If the wavelength $\lambda$ of the incident light is increased beyond the threshold wavelength $\lambda_0$,the energy of the photons becomes less than the work function,and no photoelectrons are emitted.Consequently,the photoelectric current $I$ becomes zero for all $\lambda > \lambda_0$.As $\lambda$ decreases from a value greater than $\lambda_0$ towards $\lambda_0$,the current remains zero until $\lambda = \lambda_0$,and then it increases as $\lambda$ decreases further (assuming constant intensity).Among the given options,the graph that shows current $I$ decreasing to zero as $\lambda$ increases towards the threshold wavelength is represented by graph $D$.

The anode voltage of a photocell is kept fixed. The wavelength $\lambda$ of the light falling on the cathode is gradually changed. The plate current $I$ of the photocell varies as follows:

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