(A) According to Einstein's photoelectric equation, the maximum kinetic energy $K_{max}$ is given by $K_{max} = h\nu - h\nu_0$, where $\nu$ is the fre

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$\frac{K\nu_1 - 3\nu_2}{K - 3}$

(A) According to Einstein's photoelectric equation, the maximum kinetic energy $K_{max}$ is given by $K_{max} = h\nu - h\nu_0$, where $\nu$ is the frequency of incident light and $\nu_0$ is the threshold frequency.For the first case: $K_{max1} = h\nu_1 - h\nu_0$.For the second case: $K_{max2} = h\nu_2 - h\nu_0$.The ratio is given as $\frac{K_{max1}}{K_{max2}} = \frac{3}{K}$.Substituting the expressions: $\frac{h\nu_1 - h\nu_0}{h\nu_2 - h\nu_0} = \frac{3}{K}$.$K(\nu_1 - \nu_0) = 3(\nu_2 - \nu_0)$.$K\nu_1 - K\nu_0 = 3\nu_2 - 3\nu_0$.$K\nu_1 - 3\nu_2 = K\nu_0 - 3\nu_0$.$K\nu_1 - 3\nu_2 = \nu_0(K - 3)$.Therefore, the threshold frequency $\nu_0 = \frac{K\nu_1 - 3\nu_2}{K - 3}$.

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

Difficult

Photoelectric emission is observed from a metallic surface for frequencies $\nu_1$ and $\nu_2$ of the incident light rays $(\nu_1 > \nu_2)$. If the ratio of the maximum kinetic energy of the photoelectrons emitted in the first case to that in the second case is $3 : K$, then the threshold frequency of the metallic surface is:

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A
$\frac{K\nu_1 - 3\nu_2}{K - 3}$
B
$\frac{K\nu_1 - \nu_2}{K - 1}$
C
$\frac{3\nu_2 - K\nu_1}{3 - K}$
D
$\frac{K\nu_2 - 3\nu_1}{K - 3}$

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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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If the stopping potential of photoelectrons emitted by a source placed at a distance of $0.2 \ m$ from a photocell is $0.6 \ V$,then the stopping potential of photoelectrons emitted by the same source placed at a distance of $0.6 \ m$ from the photocell will be .......... $V$.

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The graph shows the variation of stopping potential $V_o$ with the frequency $\nu$ of the incident radiation for three photosensitive metals $X_1, X_2$ and $X_3$. Which metal will emit photoelectrons with greater kinetic energy, for the same wavelength of incident radiation?

MediumJEE MAIN 2026

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In a photoelectric cell,the wavelength of incident light is changed from $4000 \, \mathring{A}$ to $3600 \, \mathring{A}$. The change in stopping potential will be ............. $V$.

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In the experiment of photoelectric effect, $K_{max}$ for a photoelectron is $K_0$. If the frequency of light is doubled, then $K_{max}$ is $3K_0$. What would be the value of $K_{max}$ if the frequency of light is tripled? (in terms of $K_0$)

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The maximum kinetic energies of photoelectrons emitted are $K_1$ and $K_2$ when lights of wavelengths $\lambda_1$ and $\lambda_2$ are incident on a metallic surface. If $\lambda_1 = 3 \lambda_2$,then:

MediumMHT CET 2023

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If the stopping potential of photoelectrons emitted by a source placed at a distance of $0.2 \ m$ from a photocell is $0.6 \ V$,then the stopping potential of photoelectrons emitted by the same source placed at a distance of $0.6 \ m$ from the photocell will be .......... $V$.

The graph shows the variation of stopping potential $V_o$ with the frequency $\nu$ of the incident radiation for three photosensitive metals $X_1, X_2$ and $X_3$. Which metal will emit photoelectrons with greater kinetic energy, for the same wavelength of incident radiation?

In a photoelectric cell,the wavelength of incident light is changed from $4000 \, \mathring{A}$ to $3600 \, \mathring{A}$. The change in stopping potential will be ............. $V$.

In the experiment of photoelectric effect, $K_{max}$ for a photoelectron is $K_0$. If the frequency of light is doubled, then $K_{max}$ is $3K_0$. What would be the value of $K_{max}$ if the frequency of light is tripled? (in terms of $K_0$)

The maximum kinetic energies of photoelectrons emitted are $K_1$ and $K_2$ when lights of wavelengths $\lambda_1$ and $\lambda_2$ are incident on a metallic surface. If $\lambda_1 = 3 \lambda_2$,then:

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