(C) The work function $\phi$ of the metal is given by $\phi = h\nu$,where $\nu$ is the cutoff (threshold) frequency.According to Einstein's photoelect

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(C) The work function $\phi$ of the metal is given by $\phi = h\nu$,where $\nu$ is the cutoff (threshold) frequency.According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ of the emitted electron is given by $K_{\max} = E - \phi$.Here,the incident frequency is $2\nu$,so the incident energy $E = h(2\nu) = 2h\nu$.Substituting the values,we get $K_{\max} = 2h\nu - h\nu = h\nu$.Since $K_{\max} = \frac{1}{2}mv_{\max}^2$,we have $\frac{1}{2}mv_{\max}^2 = h\nu$.Solving for $v_{\max}$,we get $v_{\max}^2 = \frac{2h\nu}{m}$,which implies $v_{\max} = \sqrt{\frac{2h\nu}{m}}$.

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

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For photoelectric emission from a certain metal,the cutoff frequency is $\nu$. If radiation of frequency $2\nu$ impinges on the metal plate,the maximum possible velocity of the emitted electron will be ($m$ is the electron mass).

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A
$\sqrt{\frac{h\nu}{2m}}$
B
$\sqrt{\frac{h\nu}{m}}$
C
$\sqrt{\frac{2h\nu}{m}}$
D
$2\sqrt{\frac{h\nu}{m}}$

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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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For photoelectric emission from a certain metal,the cutoff frequency is $\nu$. If radiation of frequency $2\nu$ impinges on the metal plate,the maximum possible velocity of the emitted electron will be ($m$ is the electron mass).

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Photons of wavelength $\lambda$ emitted by a source of power $P$ are incident on a photocell. If the current produced in the cell is $I$,then the percentage of incident photons which produce current in the photocell is: (where $h$ is Planck's constant and $c$ is the speed of light in vacuum)

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