JEE Main 2017 · 8 Apr · Shift 1 · Physics Q24

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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 :

less than

\sqrt3

more than

\sqrt3

equal to

\sqrt3

Answer and worked solution

Official answer: C

$\frac12mv^2=hn-\phi$ and $\frac12mv'^2=3hn-\phi=3\left(\frac12mv^2\right)+2\phi$ . So $v'^2>3v^2$ and $v'>\sqrt3v$ .

Topics:Photoelectric effect

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JEE Main 2017 · 8 Apr · Shift 1, Physics question 24