(A) Einstein's photoelectric equation is given by: $K{E_{\max }} = h\nu - \Phi$,where $\Phi = h{\nu _0}$ is the work function of the metal.Comparing t

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Is the same for all metals and independent of the intensity of the radiation

(A) Einstein's photoelectric equation is given by: $K{E_{\max }} = h\nu - \Phi$,where $\Phi = h{\nu _0}$ is the work function of the metal.Comparing this with the equation of a straight line $y = mx + c$,where $y = K{E_{\max }}$ and $x = \nu$ (frequency):$K{E_{\max }} = h\nu - h{\nu _0}$The slope $m$ of this line is equal to $h$ (Planck's constant).Since $h$ is a universal constant,the slope is the same for all metals and is independent of the intensity of the incident radiation.

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

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According to Einstein's photoelectric equation,the plot of the maximum kinetic energy of the emitted photoelectrons from a metal versus the frequency of the incident radiation gives a straight line whose slope:

AIEEE 2004 All AIEEE

A
Is the same for all metals and independent of the intensity of the radiation
B
Depends on the intensity of the radiation
C
Depends both on the intensity of the radiation and the metal used
D
Depends on the nature of the metals used

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Dual Nature of Radiation and Matter

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Einstein's Photoelectric Equation and Energy Quantum of Radiation

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When a photon of energy $hv$ is incident on an aluminum plate (work function $E_0$),photoelectrons with maximum kinetic energy $K$ are emitted. What will be the maximum kinetic energy of the emitted photoelectrons when a photon of energy $2hv$ is incident on the same aluminum plate?

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In the photoelectric effect,if the intensity of light is doubled,then the maximum kinetic energy of the photoelectrons will become:

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When the work function of a metal increases,the maximum kinetic energy of the emitted photoelectrons:

EasyMHT CET 2020

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$A$ beam of light of wavelength $400\,nm$ and power $1.55\,mW$ is directed at the cathode of a photoelectric cell. If only $10\%$ of the incident photons effectively produce photoelectrons,then find the current due to these electrons in $\mu A$. (Given: $hc = 1240\,eV\cdot nm$,$e = 1.6 \times 10^{-19}\,C$)

MediumAIIMS 2015

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When radiation of the wavelength $\lambda$ is incident on a metallic surface,the stopping potential is $4.8 \ V$. If the same surface is illuminated with radiation of double the wavelength,then the stopping potential becomes $1.6 \ V$. Then,the threshold wavelength for the surface is :

MediumTS EAMCET 2003

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According to Einstein's photoelectric equation,the plot of the maximum kinetic energy of the emitted photoelectrons from a metal versus the frequency of the incident radiation gives a straight line whose slope:

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When a photon of energy $hv$ is incident on an aluminum plate (work function $E_0$),photoelectrons with maximum kinetic energy $K$ are emitted. What will be the maximum kinetic energy of the emitted photoelectrons when a photon of energy $2hv$ is incident on the same aluminum plate?

In the photoelectric effect,if the intensity of light is doubled,then the maximum kinetic energy of the photoelectrons will become:

When the work function of a metal increases,the maximum kinetic energy of the emitted photoelectrons:

When radiation of the wavelength $\lambda$ is incident on a metallic surface,the stopping potential is $4.8 \ V$. If the same surface is illuminated with radiation of double the wavelength,then the stopping potential becomes $1.6 \ V$. Then,the threshold wavelength for the surface is :

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