(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by:$K_{\max} = E - \phi_0$Where $E$ is the energy of

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(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by:$K_{\max} = E - \phi_0$Where $E$ is the energy of the incident photon and $\phi_0$ is the work function.Given $E = 8 \ eV$.The work function $\phi_0 = h f_0$,where $f_0 = 1.6 \times 10^{15} \ Hz$.Converting $\phi_0$ to $eV$:$\phi_0 = \frac{6.6 \times 10^{-34} \times 1.6 \times 10^{15}}{1.6 \times 10^{-19}} \ eV$$\phi_0 = 6.6 \times 10^{-34+15+19} \ eV = 6.6 \times 10^0 \ eV = 6.6 \ eV$.Now,$K_{\max} = 8 \ eV - 6.6 \ eV = 1.4 \ eV$.

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

Medium

When a photon of energy $8 \ eV$ is incident on a metal surface with a threshold frequency of $1.6 \times 10^{15} \ Hz$,the maximum kinetic energy of the emitted photoelectrons is .......... $eV$. (Given: $h = 6.6 \times 10^{-34} \ Js, 1 \ eV = 1.6 \times 10^{-19} \ J$)

A
$0.8$
B
$1.4$
C
$2.8$
D
$4.2$

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Dual Nature of Radiation and matter

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Einstein's Photoelectric Equation and Energy Quantum of Radiation

When two monochromatic lights of frequency $v$ and $\frac{v}{2}$ are incident on a photoelectric metal,their stopping potentials are $\frac{V_{s}}{2}$ and $V_{s}$ respectively. The threshold frequency for this metal is:

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$V$ (stopping potential) is plotted against $\frac{1}{\lambda}$,where $\lambda$ is the wavelength of incident radiation,for two metals.

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The work function for a certain metal is $4.2 \; eV$. Will this metal give photoelectric emission for incident radiation of wavelength $330 \; nm$?

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$4 \ eV$ is the energy of the incident photon and the work function is $2 \ eV$. What is the stopping potential in $V$?

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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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When a photon of energy $8 \ eV$ is incident on a metal surface with a threshold frequency of $1.6 \times 10^{15} \ Hz$,the maximum kinetic energy of the emitted photoelectrons is .......... $eV$. (Given: $h = 6.6 \times 10^{-34} \ Js, 1 \ eV = 1.6 \times 10^{-19} \ J$)

Similar Questions

When two monochromatic lights of frequency $v$ and $\frac{v}{2}$ are incident on a photoelectric metal,their stopping potentials are $\frac{V_{s}}{2}$ and $V_{s}$ respectively. The threshold frequency for this metal is:

$V$ (stopping potential) is plotted against $\frac{1}{\lambda}$,where $\lambda$ is the wavelength of incident radiation,for two metals.

The work function for a certain metal is $4.2 \; eV$. Will this metal give photoelectric emission for incident radiation of wavelength $330 \; nm$?

$4 \ eV$ is the energy of the incident photon and the work function is $2 \ eV$. What is the stopping potential in $V$?

$A$ 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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