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For photoelectric emission from certain metal, the cut-off frequency is $ν$ . If radiation of frequency 2 $ν$
impinges on the metal plate, the maximum possible velocity of the emitted electron will be (m is the
electron mass) . [NEET 2013]

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$\sqrt{\frac{h v}{(2 m)}}$

$\sqrt{\frac{h v}{m}}$

$\sqrt{\frac{2 h v}{m}}$

$2 \sqrt{\frac{h v}{m}}$

answer is C.

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Detailed Solution

As, $(KE)_{\max} = \frac{1}{2} m v_{\max}^{2} = h v \Rightarrow v_{\max}^{2} = \frac{2 h v}{m}$

$∴ v_{\max} = \sqrt{\frac{2 h v}{m}}$

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