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The maximum velocity of the photoelectrons emitted from the surface is v when light of frequency n falls on a metal surface. If the incident frequency is increased to 3n, the maximum velocity of the ejected photoelectrons will be :

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More than $\sqrt{3} v$

Less than $\sqrt{3} v$

Equal to $\sqrt{3} v$

answer is C.

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Detailed Solution

$E - ϕ = K_{1}$

$3 E - ϕ = K_{2}$ $\Rightarrow K_{2} = 3 K_{1} + 2 ϕ$

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