(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by $K_{\max} = h\nu - h\nu_0$,where $\nu_0$ is the t

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(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by $K_{\max} = h\nu - h\nu_0$,where $\nu_0$ is the threshold frequency.For frequency $\nu_1$,$K_1 = h(\nu_1 - \nu_0) \dots (i)$For frequency $\nu_2$,$K_2 = h(\nu_2 - \nu_0) \dots (ii)$Given the ratio $\frac{K_1}{K_2} = \frac{1}{k}$,we have:$\frac{h(\nu_1 - \nu_0)}{h(\nu_2 - \nu_0)} = \frac{1}{k}$$k(\nu_1 - \nu_0) = \nu_2 - \nu_0$$k\nu_1 - k\nu_0 = \nu_2 - \nu_0$$k\nu_1 - \nu_2 = k\nu_0 - \nu_0$$k\nu_1 - \nu_2 = \nu_0(k - 1)$$\nu_0 = \frac{k\nu_1 - \nu_2}{k - 1}$

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

Medium

Photoelectrons are emitted from a metal surface for frequencies $\nu_1$ and $\nu_2$. If the ratio of the maximum kinetic energy of the emitted photoelectrons is $1 : k$,then the threshold frequency of the metal surface is:

A
$\frac{\nu_1 - \nu_2}{k - 1}$
B
$\frac{k\nu_1 - \nu_2}{k - 1}$
C
$\frac{k\nu_2 - \nu_1}{k - 1}$
D
$\frac{\nu_2 - \nu_1}{k}$

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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

Which theory of light can explain the photoelectric effect?

Easy

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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 $B$)

Medium

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When a photosensitive surface is illuminated with light of wavelength $\lambda$,the stopping potential is $V$. When the same surface is illuminated by light of wavelength $2\lambda$,the stopping potential is $V/3$. The threshold wavelength for the surface is:

Difficult

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In case of photoelectric emission from a certain metal, the cutoff frequency is $\nu$. If radiation of frequency $3\nu$ is incident on the metal plate, the maximum possible velocity of the emitted electrons will be ($m$ = mass of electron, $h$ = Planck's constant).

DifficultMHT CET 2026

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The work function of caesium is $2.14 \ eV$. Find the wavelength of the incident light if the photocurrent is brought to zero by a stopping potential of $0.60 \ V$. (Answer in $nm$)

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Photoelectrons are emitted from a metal surface for frequencies $\nu_1$ and $\nu_2$. If the ratio of the maximum kinetic energy of the emitted photoelectrons is $1 : k$,then the threshold frequency of the metal surface is:

Similar Questions

Which theory of light can explain the photoelectric effect?

When a photosensitive surface is illuminated with light of wavelength $\lambda$,the stopping potential is $V$. When the same surface is illuminated by light of wavelength $2\lambda$,the stopping potential is $V/3$. The threshold wavelength for the surface is:

In case of photoelectric emission from a certain metal, the cutoff frequency is $\nu$. If radiation of frequency $3\nu$ is incident on the metal plate, the maximum possible velocity of the emitted electrons will be ($m$ = mass of electron, $h$ = Planck's constant).

The work function of caesium is $2.14 \ eV$. Find the wavelength of the incident light if the photocurrent is brought to zero by a stopping potential of $0.60 \ V$. (Answer in $nm$)

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