From the photoelectric effect experiment,sele | vedclass

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(D) According to Einstein's photoelectric equation,$K_{max} = h
u - \Phi_0$,where $K_{max}$ is the maximum kinetic energy,$h
u$ is the energy of the

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The saturation current increases as the intensity of incident light increases.

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(D) According to Einstein's photoelectric equation,$K_{max} = h
u - \Phi_0$,where $K_{max}$ is the maximum kinetic energy,$h
u$ is the energy of the incident photon,and $\Phi_0$ is the work function of the metal.$1$. The wave theory of light fails to explain the photoelectric effect because it cannot account for the instantaneous emission of electrons or the existence of a threshold frequency.$2$. $K_{max}$ depends on the frequency of incident light $(
u)$,not its intensity.$3$. The stopping potential $(V_s)$ is given by $eV_s = K_{max} = h
u - \Phi_0$. Thus,it depends on both the frequency of incident light and the work function of the metal.$4$. The saturation current is directly proportional to the number of photoelectrons emitted per second,which is directly proportional to the intensity of the incident light. Therefore,as intensity increases,the saturation current increases.

From the photoelectric effect experiment,select the correct statement.

Similar Questions

Light of frequency $1.5 \nu_0$ (where $\nu_0$ is the threshold frequency) is incident on a photosensitive material. If the frequency is halved and the intensity is doubled,what happens to the photocurrent?

The magnitude of saturation photoelectric current depends upon

The electrons are emitted in the photoelectric effect from a metal surface:

From the following graph of photo-current against collector plate potential,for two different intensities of light $I_{1}$ and $I_{2}$,one can conclude:

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