(A) According to Einstein's photoelectric equation:$K_{max} = h\nu - W_0$Since $K_{max} = eV_0$,we have:$eV_0 = h\nu - W_0$$V_0 = \left( \frac{h}{e} \

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(A) According to Einstein's photoelectric equation:$K_{max} = h\nu - W_0$Since $K_{max} = eV_0$,we have:$eV_0 = h\nu - W_0$$V_0 = \left( \frac{h}{e} \right) \nu - \frac{W_0}{e}$Comparing this with the equation of a straight line $y = mx + c$,where $y = V_0$,$x = \nu$,$m = \frac{h}{e}$,and $c = -\frac{W_0}{e}$.The intercept on the negative $V_0$ axis is $OB = \frac{W_0}{e}$.Therefore,the work function $W_0 = OB \times e$.

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

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In an experiment on the photoelectric effect,the frequency $\nu$ of the incident light is plotted against the stopping potential $V_0$. The work function of the photoelectric surface is given by ($e$ is the electronic charge):

A
$OB \times e$ in $eV$
B
$OB$ in volt
C
$OA$ in $eV$
D
The slope of the line $AB$

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Class 12 Questions

Class 12

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Dual Nature of Radiation and matter

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Einstein's Photoelectric Equation and Energy Quantum of Radiation

When a metal surface is illuminated by light of wavelengths $400\; nm$ and $250\; nm$,the maximum velocities of the photoelectrons ejected are $v$ and $2v$ respectively. The work function of the metal is ($h =$ Planck's constant,$c =$ velocity of light in air).

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In a photoelectric experiment, the wavelength of the light incident on the metal is changed from $200 \, nm$ to $400 \, nm$. The decrease in the stopping potential is close to [Use $hc = 1240 \, eV \cdot nm$ where $h$ is Planck's constant and $c$ is the velocity of light]. (in $ \, V$)

MediumTS EAMCET 2022

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Two identical photocathodes receive light of frequencies $n_1$ and $n_2$. If the velocities of the emitted photoelectrons of mass $m$ are $V_1$ and $V_2$ respectively,then ($h=$ Planck's constant):

DifficultMHT CET 2024

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Two identical capacitors are arranged as shown. The work function of plate $1$ is $\phi \ eV$. The $emf$ of the battery is $\frac{\phi}{e}$. If the energy of the incident photon is $hv$,then what is the maximum kinetic energy of the $e^-$ reaching plate $2$?

Medium

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The energy of the incident photon on a metal surface is $3W$ and then $5W$,where $W$ is the work function of that metal. The ratio of the maximum velocities of the emitted photoelectrons is:

EasyMHT CET 2020

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In an experiment on the photoelectric effect,the frequency $\nu$ of the incident light is plotted against the stopping potential $V_0$. The work function of the photoelectric surface is given by ($e$ is the electronic charge):

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When a metal surface is illuminated by light of wavelengths $400\; nm$ and $250\; nm$,the maximum velocities of the photoelectrons ejected are $v$ and $2v$ respectively. The work function of the metal is ($h =$ Planck's constant,$c =$ velocity of light in air).

In a photoelectric experiment, the wavelength of the light incident on the metal is changed from $200 \, nm$ to $400 \, nm$. The decrease in the stopping potential is close to [Use $hc = 1240 \, eV \cdot nm$ where $h$ is Planck's constant and $c$ is the velocity of light]. (in $ \, V$)

Two identical photocathodes receive light of frequencies $n_1$ and $n_2$. If the velocities of the emitted photoelectrons of mass $m$ are $V_1$ and $V_2$ respectively,then ($h=$ Planck's constant):

Two identical capacitors are arranged as shown. The work function of plate $1$ is $\phi \ eV$. The $emf$ of the battery is $\frac{\phi}{e}$. If the energy of the incident photon is $hv$,then what is the maximum kinetic energy of the $e^-$ reaching plate $2$?

The energy of the incident photon on a metal surface is $3W$ and then $5W$,where $W$ is the work function of that metal. The ratio of the maximum velocities of the emitted photoelectrons is:

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