When photons of energy $h v$ fall on an aluminium plate (work function $E_{0})$ , photoelectrons of maximum kinetic energy $K$ are ejected. If the frequency of radiation is doubled, the maximum kinetic energy of the ejected photoelectrons will be

K + h v

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K + E_{0}

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2 K

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K

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Solution

The correct option is A $K + h v$
Let $K$ and $K^{'}$ be the maximum kinetic energy of photoelectrons for incident light of frequency $v$ and $2 v$ respectively.
According to Einstein's photoelectric equation,
$K = h v - E_{0}$ ...(1)
and $K^{'} = h (2 v) - E_{0}$ ...(2)
= $h v + h v - E_{0}$
= $h v + K$ (using (1))
So,
$K^{'}$ = $h v + K$

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