(B) The energy of a photon is given by $E = \frac{hc}{\lambda}$.Given for $\lambda_1 = 10,000\;\mathring A$,$E_1 = 1.23\; eV$.For $\lambda_2 = 5000\;\

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(B) The energy of a photon is given by $E = \frac{hc}{\lambda}$.Given for $\lambda_1 = 10,000\;\mathring A$,$E_1 = 1.23\; eV$.For $\lambda_2 = 5000\;\mathring A$,the energy $E_2$ is:$E_2 = E_1 \times \frac{\lambda_1}{\lambda_2} = 1.23 \times \frac{10,000}{5,000} = 1.23 \times 2 = 2.46\; eV$.According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max} = eV_s = E_2 - \phi$,where $\phi$ is the work function.Given stopping potential $V_s = 1.36\; V$,so $K_{max} = 1.36\; eV$.Substituting the values: $1.36 = 2.46 - \phi$.Therefore,$\phi = 2.46 - 1.36 = 1.10\; eV$.

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

Difficult

Given that a photon of light of wavelength $10,000\;\mathring A$ has an energy equal to $1.23\; eV$. When light of wavelength $5000\;\mathring A$ and intensity $I_0$ falls on a photoelectric cell,the saturation current is $0.40 \times 10^{-6}\; A$ and the stopping potential is $1.36\; V$; then the work function is ..... $eV$.

A
$0.43$
B
$1.10$
C
$1.36$
D
$2.47$

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Dual Nature of Radiation and Matter

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Einstein's Photoelectric Equation and Energy Quantum of Radiation

When light of wavelength '$\lambda$' is incident on a photosensitive surface, the stopping potential is '$V$'. When a light of wavelength $1.5\lambda$ is incident on the same surface, the stopping potential is '$\frac{V}{4}$'. Threshold wavelength for the surface is

DifficultMHT CET 2026

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In the photoelectric effect,the kinetic energy $(K.E.)$ of electrons emitted from the metal surface depends upon:

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The threshold frequency of a photosensitive material is $v$. When photons of frequency $2v$ are incident on the material,photoelectrons are emitted with a maximum linear momentum $P$. To get photoelectrons with maximum linear momentum $2P$,the frequency of the incident photons is: (in $v$)

MediumAP EAMCET 2023

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Photoelectrons are emitted from a photosensitive surface for light of wavelengths $\lambda_{1} = 360 \ nm$ and $\lambda_{2} = 600 \ nm$. What is the ratio of the work functions for the lights of wavelengths $\lambda_{1}$ and $\lambda_{2}$?

EasyMHT CET 2020

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In a photoelectric effect experiment,the maximum kinetic energy of emitted photoelectrons is $K_0$. If the frequency of the incident radiation is increased by a factor of $n_1$,the new maximum kinetic energy becomes $n_2K_0$. Find the work function of the metal.

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When light of wavelength '$\lambda$' is incident on a photosensitive surface, the stopping potential is '$V$'. When a light of wavelength $1.5\lambda$ is incident on the same surface, the stopping potential is '$\frac{V}{4}$'. Threshold wavelength for the surface is

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The threshold frequency of a photosensitive material is $v$. When photons of frequency $2v$ are incident on the material,photoelectrons are emitted with a maximum linear momentum $P$. To get photoelectrons with maximum linear momentum $2P$,the frequency of the incident photons is: (in $v$)

Photoelectrons are emitted from a photosensitive surface for light of wavelengths $\lambda_{1} = 360 \ nm$ and $\lambda_{2} = 600 \ nm$. What is the ratio of the work functions for the lights of wavelengths $\lambda_{1}$ and $\lambda_{2}$?

In a photoelectric effect experiment,the maximum kinetic energy of emitted photoelectrons is $K_0$. If the frequency of the incident radiation is increased by a factor of $n_1$,the new maximum kinetic energy becomes $n_2K_0$. Find the work function of the metal.

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