(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

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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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Class 12

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Dual Nature of Radiation and matter

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Einstein's Photoelectric Equation and Energy Quantum of Radiation

The metallic surface is illuminated with monochromatic light of wavelength $\lambda$ and the stopping potential for the photoelectric current is $5V_0$. When the same metallic surface is illuminated with light of wavelength $2\lambda$,the stopping potential is $V_0$. What is the threshold wavelength for the surface?

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The wavelengths of light incident on a photocell are $400 \ nm$ and $250 \ nm$. The velocities of the emitted photoelectrons are $v$ and $2v$ respectively. What is the work function of the metal?

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$A$ metal surface is illuminated by light of a given intensity and frequency, causing photoelectric emission. If the intensity of the incident light is reduced to $1/4$ of its original value, what will be the kinetic energy of the emitted photoelectrons?

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When electromagnetic radiation is incident on a metallic surface,the maximum kinetic energy of the photoelectrons depends on .......

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The work function of a metal is $1 \ eV$. If light of wavelength $3000 \ \mathring{A}$ is incident on the surface of this metal, the maximum velocity of the emitted photoelectrons is ........

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The wavelengths of light incident on a photocell are $400 \ nm$ and $250 \ nm$. The velocities of the emitted photoelectrons are $v$ and $2v$ respectively. What is the work function of the metal?

$A$ metal surface is illuminated by light of a given intensity and frequency, causing photoelectric emission. If the intensity of the incident light is reduced to $1/4$ of its original value, what will be the kinetic energy of the emitted photoelectrons?

When electromagnetic radiation is incident on a metallic surface,the maximum kinetic energy of the photoelectrons depends on .......

The work function of a metal is $1 \ eV$. If light of wavelength $3000 \ \mathring{A}$ is incident on the surface of this metal, the maximum velocity of the emitted photoelectrons is ........

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