1. Use Einstein’s photoelectric equation

   The maximum kinetic energy of emitted photoelectrons is $K_{\max}=h\nu-\phi$ where $\phi$ is the work function of the metal.

2. For incident frequency $\nu=n$

   Given maximum speed is $v$, so $\frac12 mv^2=hn-\phi \quad ...(1)$

3. For incident frequency $\nu=3n$

   Let the new maximum speed be $v'$. Then $\frac12 m{v'}^2=3hn-\phi \quad ...(2)$

4. Compare with $\sqrt{3}\,v$

   From (1), $hn=\phi+\frac12 mv^2$

   Substitute into (2): $\frac12 m{v'}^2=3\left(\phi+\frac12 mv^2\right)-\phi$ $\frac12 m{v'}^2=2\phi+\frac32 mv^2$

   So, ${v'}^2=\frac{4\phi}{m}+3v^2$

   Since photoelectric emission occurs for frequency $n$, we must have $hn>\phi$, hence $\phi>0$ and therefore ${v'}^2>3v^2$

   Taking square roots, $v'>\sqrt{3}\,v$

5. Check options

   • A: less than $\sqrt3 v$ — false
   • B: $v$ — false
   • C: more than $\sqrt3 v$ — true
   • D: equal to $\sqrt3 v$ — false

Therefore, the correct option is C.