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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Correct Answer

(C) more thanequal \(\sqrt 3 v\,\)

Step-by-step Solution

Detailed explanation

As the metal surface is same, work function \((\phi)\) is same for both the case.

Initially \(\mathrm{KE}_{\max }=\mathrm{nh}-\phi\) ..... \((i)\)

After increase

\(\mathrm{KE}_{\mathrm{max}}^{\prime}=3 \mathrm{nh}-\phi\) ..... \((ii)\)

For work function \(\phi-\) not to be \(-\,ve\) or zero.

\(v^{\prime}>\sqrt{3} v\)

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