(B) Let the work functions of metals $A$ and $B$ be $\phi_A$ and $\phi_B$ respectively. Given $\frac{\phi_A}{\phi_B} = \frac{1}{2}$,so $\phi_B = 2\phi

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(B) Let the work functions of metals $A$ and $B$ be $\phi_A$ and $\phi_B$ respectively. Given $\frac{\phi_A}{\phi_B} = \frac{1}{2}$,so $\phi_B = 2\phi_A$.According to Einstein's photoelectric equation,the maximum kinetic energy $K$ is given by $K = h\nu - \phi$.For metal $A$: $K_A = hf - \phi_A$ ---$(1)$For metal $B$: $K_B = h(2f) - \phi_B = 2hf - 2\phi_A$ ---$(2)$From equation $(2)$,we can factor out $2$: $K_B = 2(hf - \phi_A)$.Substituting equation $(1)$ into this,we get $K_B = 2K_A$.Therefore,the ratio $\frac{K_A}{K_B} = \frac{1}{2}$.

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

Medium

The ratio of work functions of two metals $A$ and $B$ is $1:2$. If radiations of frequencies $f$ and $2f$ are incident on the surfaces of $A$ and $B$ respectively,the ratio of the maximum kinetic energies of the emitted photoelectrons will be: (Given $f >$ threshold frequency of $A$ and $2f >$ threshold frequency of $B$)

A
$1:1$
B
$1:2$
C
$1:3$
D
$1:4$

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Class 12 Questions

Class 12

Physics Questions

Chapter

Dual Nature of Radiation and matter

Subtopic

Einstein's Photoelectric Equation and Energy Quantum of Radiation

Ultraviolet light of wavelength $300 \ nm$ and intensity $1.0 \ W/m^2$ falls on the surface of a photosensitive material. If $1\%$ of the incident photons produce photoelectrons,then the number of photoelectrons emitted from an area of $1.0 \ cm^2$ of the surface is nearly

Medium

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$V$ (stopping potential) is plotted against $\frac{1}{\lambda}$,where $\lambda$ is the wavelength of incident radiation,for two metals.

Medium

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Light of wavelength $\lambda$ which is less than threshold wavelength is incident on a photosensitive material. If incident wavelength is decreased so that emitted photoelectrons are moving with the same velocity,then stopping potential will:

MediumMHT CET 2016

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When radiation of wavelength $\lambda$ is used to illuminate a metallic surface,the stopping potential is $V.$ When the same surface is illuminated with radiation of wavelength $3 \lambda,$ the stopping potential is $\frac{V}{4}.$ If the threshold wavelength for the metallic surface is $n \lambda,$ then the value of $n$ will be......

MediumJEE MAIN 2020

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All electrons ejected from a metallic surface by incident light of wavelength $400 \,nm$ travelled $1 \,m$ in the direction of a uniform electric field of $2 \,N/C$ and came to rest. The work function of the surface is (in $\,eV$)

MediumAP EAMCET 2019

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Ultraviolet light of wavelength $300 \ nm$ and intensity $1.0 \ W/m^2$ falls on the surface of a photosensitive material. If $1\%$ of the incident photons produce photoelectrons,then the number of photoelectrons emitted from an area of $1.0 \ cm^2$ of the surface is nearly

$V$ (stopping potential) is plotted against $\frac{1}{\lambda}$,where $\lambda$ is the wavelength of incident radiation,for two metals.

Light of wavelength $\lambda$ which is less than threshold wavelength is incident on a photosensitive material. If incident wavelength is decreased so that emitted photoelectrons are moving with the same velocity,then stopping potential will:

All electrons ejected from a metallic surface by incident light of wavelength $400 \,nm$ travelled $1 \,m$ in the direction of a uniform electric field of $2 \,N/C$ and came to rest. The work function of the surface is (in $\,eV$)

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