(B) Given: Work function $\phi = 4.5\,eV$,Wavelength $\lambda = 200\,nm$.Energy of incident photon $E = \frac{hc}{\lambda} = \frac{1240\,eV\cdot nm}{2

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The maximum kinetic energy of the photoelectrons reaching the collector is $3.7\,eV$.

(B) Given: Work function $\phi = 4.5\,eV$,Wavelength $\lambda = 200\,nm$.Energy of incident photon $E = \frac{hc}{\lambda} = \frac{1240\,eV\cdot nm}{200\,nm} = 6.2\,eV$.Maximum kinetic energy of emitted photoelectrons $K_{max} = E - \phi = 6.2\,eV - 4.5\,eV = 1.7\,eV$.The collector plate is at a potential of $+2.0\,V$ relative to the emitter,meaning it accelerates the electrons.The minimum kinetic energy of electrons reaching the collector is $K_{min} = K_{initial} + qV = 0 + 2.0\,eV = 2.0\,eV$.The maximum kinetic energy of electrons reaching the collector is $K_{max}' = K_{max} + qV = 1.7\,eV + 2.0\,eV = 3.7\,eV$.Thus,option $B$ is correct.

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

Difficult

In a photoelectric experiment,the collector plate is at $2.0\,V$ with respect to the emitter plate made of copper $(\phi = 4.5\,eV)$. The emitter is illuminated by a source of monochromatic light of wavelength $200\,nm$.

A
The minimum kinetic energy of the photoelectrons reaching the collector is $0$.
B
The maximum kinetic energy of the photoelectrons reaching the collector is $3.7\,eV$.
C
If the polarity of the battery is reversed,then the answer to part $A$ will be $0$.
D
If the polarity of the battery is reversed,then the answer to part $B$ will be $1.7\,eV$.

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Dual Nature of Radiation and matter

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Einstein's Photoelectric Equation and Energy Quantum of Radiation

The maximum kinetic energy of the emitted photoelectrons from a photosensitive material of work function $\phi$,when light of frequency $\nu$ is incident on it,is $E$. If the frequency of the incident light is $3\nu$,the maximum kinetic energy of the emitted photoelectrons is:

EasyAP EAMCET 2023

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Light described at a place by the equation $\varepsilon = (100 \text{ V/m}) [\sin(5 \times 10^{15} \text{ s}^{-1})t + \sin(8 \times 10^{15} \text{ s}^{-1})t]$ falls on a metal surface having a work function of $2 \text{ eV}$. Calculate the maximum kinetic energy of the photoelectrons in $\text{eV}$.

Difficult

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In a photoelectric experiment,light of energy $2.48 eV$ irradiates a photosensitive material. The stopping potential was measured to be $0.5 V$. The work function of the photosensitive material is: (in $eV$)

DifficultJEE MAIN 2024

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Radiation of wavelength $300 \ nm$ and intensity $100 \ W \ m^{-2}$ falls on the surface of a photosensitive material. If $2 \%$ of the incident photons produce photoelectrons,the number of photoelectrons emitted from an area of $2 \ cm^2$ of the surface per second is nearly:

DifficultAP EAMCET 2021

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When a photon of energy $4.0 \; eV$ strikes the surface of a metal $A$,the ejected photoelectrons have maximum kinetic energy $T_{A} \; eV$ and de-Broglie wavelength $\lambda_{A}$. The maximum kinetic energy of photoelectrons liberated from another metal $B$ by a photon of energy $4.50 \; eV$ is $T_{B} = (T_{A} - 1.5) \; eV$. If the de-Broglie wavelength of these photoelectrons is $\lambda_{B} = 2 \lambda_{A}$,then the work function of metal $B$ is ............. $eV$.

MediumJEE MAIN 2020

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In a photoelectric experiment,the collector plate is at $2.0\,V$ with respect to the emitter plate made of copper $(\phi = 4.5\,eV)$. The emitter is illuminated by a source of monochromatic light of wavelength $200\,nm$.

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The maximum kinetic energy of the emitted photoelectrons from a photosensitive material of work function $\phi$,when light of frequency $\nu$ is incident on it,is $E$. If the frequency of the incident light is $3\nu$,the maximum kinetic energy of the emitted photoelectrons is:

In a photoelectric experiment,light of energy $2.48 eV$ irradiates a photosensitive material. The stopping potential was measured to be $0.5 V$. The work function of the photosensitive material is: (in $eV$)

Radiation of wavelength $300 \ nm$ and intensity $100 \ W \ m^{-2}$ falls on the surface of a photosensitive material. If $2 \%$ of the incident photons produce photoelectrons,the number of photoelectrons emitted from an area of $2 \ cm^2$ of the surface per second is nearly:

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