(D) According to Einstein's photoelectric equation, the maximum kinetic energy $K_{max}$ is given by $K_{max} = h v - \Phi_0$, where $h$ is Planck's c

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(D) According to Einstein's photoelectric equation, the maximum kinetic energy $K_{max}$ is given by $K_{max} = h v - \Phi_0$, where $h$ is Planck's constant, $v$ is the frequency of incident light, and $\Phi_0 = h v_0$ is the work function ($v_0$ is the threshold frequency).For frequency $v_1$, $K_1 = h v_1 - h v_0 = h(v_1 - v_0)$.For frequency $v_2$, $K_2 = h v_2 - h v_0 = h(v_2 - v_0)$.Given the ratio $K_1 : K_2 = 1 : k$, we have $\frac{K_1}{K_2} = \frac{1}{k}$.Substituting the expressions: $\frac{h(v_1 - v_0)}{h(v_2 - v_0)} = \frac{1}{k}$.This simplifies to $k(v_1 - v_0) = v_2 - v_0$.Expanding the terms: $k v_1 - k v_0 = v_2 - v_0$.Rearranging to solve for $v_0$: $k v_1 - v_2 = k v_0 - v_0 = v_0(k - 1)$.Therefore, $v_0 = \frac{k v_1 - v_2}{k - 1}$.

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

Medium

Photoelectric emission is observed from a metallic surface for frequencies $v_1$ and $v_2$ of the incident light rays $(v_1 > v_2)$. If the maximum kinetic energies of the photoelectrons emitted in the two cases are in the ratio of $1:k$, then the threshold frequency of the metallic surface is:

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A
$\frac{k v_2 - v_1}{k - 1}$
B
$\frac{v_2 - v_1}{k}$
C
$\frac{v_1 - v_2}{k - 1}$
D
$\frac{k v_1 - v_2}{k - 1}$

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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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When a certain metal 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 $\frac{V}{3}$. The threshold wavelength for the surface is:

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Light of energy $E$ falls normally on a metal of work function $\frac{E}{3}$. The kinetic energies $K$ of the photoelectrons are

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$A$ photon of energy $6 \ eV$ is incident on a metal surface. The work function of the metal is $2 \ eV$. The minimum reverse potential required to stop the emission of electrons is ........ $V$.

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In a photoelectric experiment, the slope of the graph drawn between stopping potential $(V_s)$ along the $y$-axis and the frequency $(\nu)$ of incident radiation along the $x$-axis is (Planck's constant $h = 6.6 \times 10^{-34} \text{ Js}$)

MediumTS EAMCET 2025

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Match the temperature of a black body given in List-$I$ with an appropriate statement in List-$II$, and choose the correct option.
[Given: Wien's constant as $2.9 \times 10^{-3} \, m-K$ and $\frac{hc}{e}=1.24 \times 10^{-6} \, V-m$ ]

List-$I$ List-$II$

$(P)$ $2000 \, K$ $(1)$ The radiation at peak wavelength can lead to emission of photoelectrons from a metal of work function $4 \, eV$

$(Q)$ $3000 \, K$ $(2)$ The radiation at peak wavelength is visible to human eye.

$(R)$ $5000 \, K$ $(3)$ The radiation at peak emission wavelength will result in the widest central maximum of a single slit diffraction.

$(S)$ $10000 \, K$ $(4)$ The power emitted per unit area is $1/16$ of that emitted by a blackbody at temperature $6000 \, K$.

$(5)$ The radiation at peak emission wavelength can be used to image human bones.

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List-$I$	List-$II$
$(P)$ $2000 \, K$	$(1)$ The radiation at peak wavelength can lead to emission of photoelectrons from a metal of work function $4 \, eV$
$(Q)$ $3000 \, K$	$(2)$ The radiation at peak wavelength is visible to human eye.
$(R)$ $5000 \, K$	$(3)$ The radiation at peak emission wavelength will result in the widest central maximum of a single slit diffraction.
$(S)$ $10000 \, K$	$(4)$ The power emitted per unit area is $1/16$ of that emitted by a blackbody at temperature $6000 \, K$.
	$(5)$ The radiation at peak emission wavelength can be used to image human bones.

Photoelectric emission is observed from a metallic surface for frequencies $v_1$ and $v_2$ of the incident light rays $(v_1 > v_2)$. If the maximum kinetic energies of the photoelectrons emitted in the two cases are in the ratio of $1:k$, then the threshold frequency of the metallic surface is:

Similar Questions

When a certain metal 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 $\frac{V}{3}$. The threshold wavelength for the surface is:

Light of energy $E$ falls normally on a metal of work function $\frac{E}{3}$. The kinetic energies $K$ of the photoelectrons are

$A$ photon of energy $6 \ eV$ is incident on a metal surface. The work function of the metal is $2 \ eV$. The minimum reverse potential required to stop the emission of electrons is ........ $V$.

In a photoelectric experiment, the slope of the graph drawn between stopping potential $(V_s)$ along the $y$-axis and the frequency $(\nu)$ of incident radiation along the $x$-axis is (Planck's constant $h = 6.6 \times 10^{-34} \text{ Js}$)

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