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

Photoelectric effect is produced by a point source on a small metal plate. Which of the following curves represents the saturation photoelectric current as a function of the distance between the source and the metal?

How are the frequency and intensity of incident radiation changed in the experiment of the photoelectric effect?

Assertion : The photoelectrons produced by a monochromatic light beam incident on a metal surface have a spread in their kinetic energies.
Reason : The work function of the metal varies as a function of depth from the surface.

If the intensity of incident radiation in a photocell is increased, the stopping potential

For a given source,with an increase in intensity of incident radiation,the . . . . . .

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