The kinetic energy with which electrons are e | vedclass

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Question

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

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Independent of the intensity of illumination

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(A) According to Einstein's photoelectric equation,$K_{max} = h
u - \phi$,where $K_{max}$ is the maximum kinetic energy of the emitted photoelectrons,$h$ is Planck's constant,$
u$ is the frequency of incident radiation,and $\phi$ is the work function of the metal.From this equation,it is clear that the kinetic energy of photoelectrons depends only on the frequency of the incident radiation $(
u)$ and the work function of the metal $(\phi)$.It is completely independent of the intensity of the incident light,which only affects the number of photoelectrons emitted per unit time.

The kinetic energy with which electrons are emitted from a metal surface due to the photoelectric effect is:

Similar Questions

Which of the following graphs represents the variation of stopping potential $(V_0)$ with the frequency $(\nu)$ of incident light for a photoelectric cell?

If the frequency of light falling on a photosensitive material doubles, which of the following is true?

When a photosensitive surface is irradiated by light of wavelength $\lambda_1$ and $\lambda_2$,the maximum kinetic energies of the emitted photoelectrons are $E_1$ and $E_2$ respectively. The work function of the photosensitive surface is:

Two light beams of different frequencies,having energies $1 \ eV$ and $2.5 \ eV$ respectively,are incident on a metal surface with a work function of $0.5 \ eV$. The ratio of the maximum kinetic energies of the emitted photoelectrons is:

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