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(D) According to the photoelectric effect,the energy of a photon is given by $E = \frac{hc}{\lambda}$.As the wavelength $\lambda$ of the incident ligh

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(D) According to the photoelectric effect,the energy of a photon is given by $E = \frac{hc}{\lambda}$.As the wavelength $\lambda$ of the incident light increases,the energy of the incident photons decreases.Photoelectric emission occurs only if the energy of the incident photon is greater than or equal to the work function $\phi$ of the metal surface.This implies that there exists a maximum wavelength,known as the threshold wavelength $\lambda_0$,such that for all $\lambda > \lambda_0$,the energy of the photons is insufficient to eject electrons from the cathode.Consequently,the photocurrent $I$ will be zero for all $\lambda > \lambda_0$.As $\lambda$ increases from a small value towards $\lambda_0$,the number of emitted photoelectrons (and thus the current $I$) generally decreases or remains constant depending on the intensity,but it must drop to zero at $\lambda = \lambda_0$.Among the given options,the graph that shows the current decreasing and eventually becoming zero at a specific wavelength is represented by option $(d)$.

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:

Similar Questions

Which of the following support the quantum nature of the $EM$ radiations?
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$(B)$ Compton effect
$(C)$ Doppler effect
$(D)$ Field effect

Assertion : Photoelectric effect demonstrates the wave nature of light.
Reason : The number of photoelectrons is proportional to the frequency of light.

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