Q 12-11-176JEE MainJEE Main 2017 (8 Apr)Medium
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:
Answer: (A) more than $\sqrt3v$
$$\tfrac12mv^2 = hn - \phi,\qquad \tfrac12mv'^2 = 3hn - \phi = 3(hn - \phi) + 2\phi$$
$$\tfrac12mv'^2 = 3\left(\tfrac12mv^2\right) + 2\phi > 3\left(\tfrac12mv^2\right)$$
So $v'^2 > 3v^2$, i.e. $v' > \sqrt3v$.
Solution by Sreeraj P, M.Sc Physics