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The maximum kinetic energy of the emitted photoelectrons from a photosensitive material of work function $\phi$, when light of frequency ' $V$ ' incidents on it is ' $E$ '. If the frequency of the incident light is $3 \mathrm{v}$, the maximum kinetic energy of the emitted photoelectrons is
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The correct answer is:
$3 \mathrm{E}+2 \phi$
From the photoelectric equation
$\mathrm{E}=\mathrm{h} v-\phi$ ...(1)
$\mathrm{KE}_{\max }=3 \mathrm{~h} v-\phi$ ...(2)
Solving equation (1) and (2), we have
$\mathrm{KE}_{\max }=3 \mathrm{E}+2 \phi$
$\mathrm{E}=\mathrm{h} v-\phi$ ...(1)
$\mathrm{KE}_{\max }=3 \mathrm{~h} v-\phi$ ...(2)
Solving equation (1) and (2), we have
$\mathrm{KE}_{\max }=3 \mathrm{E}+2 \phi$
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