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

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Independent of the intensity of illumination

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

Class 12 Physics Dual Nature of Radiation and matter Einstein's Photoelectric Equation and Energy Quantum of Radiation

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

A
Independent of the intensity of illumination
B
Independent of the frequency of light
C
Inversely proportional to the intensity of illumination
D
Directly proportional to the intensity of illumination

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Dual Nature of Radiation and matter

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Einstein's Photoelectric Equation and Energy Quantum of Radiation

In a photoelectric experiment,a graph of maximum kinetic energy $(KE_{\max})$ against the frequency of incident radiation $(\nu)$ is plotted. If $A$ and $B$ are the intercepts on the $X$ and $Y$ axis respectively,the Planck's constant is given by:

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Two identical photo-cathodes receive light of frequencies $f_1$ and $f_2$. If the velocities of the photo-electrons (of mass $m$) coming out are $v_1$ and $v_2$ respectively,then:

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Photoelectric emission is observed from a metallic surface for frequencies $v_1$ and $v_2$ of the incident light rays $(v_1 > v_2)$. If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of $1:k$,then the threshold frequency of the metallic surface is

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The maximum velocity of the photoelectrons emitted by a metal surface is $9 \times 10^5 \ m/s$. The value of the ratio of charge $(e)$ to mass $(m)$ of the photoelectron is $1.8 \times 10^{11} \ C/kg$. The value of the stopping potential in volts is:

EasyMHT CET 2025

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If light of wavelength $6600 \ \mathring A$ is incident on a metal surface with a work function of $2 \ eV$,what will be the maximum kinetic energy of the emitted photoelectrons?

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

Similar Questions

In a photoelectric experiment,a graph of maximum kinetic energy $(KE_{\max})$ against the frequency of incident radiation $(\nu)$ is plotted. If $A$ and $B$ are the intercepts on the $X$ and $Y$ axis respectively,the Planck's constant is given by:

Two identical photo-cathodes receive light of frequencies $f_1$ and $f_2$. If the velocities of the photo-electrons (of mass $m$) coming out are $v_1$ and $v_2$ respectively,then:

The maximum velocity of the photoelectrons emitted by a metal surface is $9 \times 10^5 \ m/s$. The value of the ratio of charge $(e)$ to mass $(m)$ of the photoelectron is $1.8 \times 10^{11} \ C/kg$. The value of the stopping potential in volts is:

If light of wavelength $6600 \ \mathring A$ is incident on a metal surface with a work function of $2 \ eV$,what will be the maximum kinetic energy of the emitted photoelectrons?

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