The frequency of the incident light falling on a photosensitive metal plate is doubled, the kinetic energy of the emitted photoelectron gets:

more than doubled

Let K₁ and K₂ be the kinetic energy of photoelectrons for incident light of frequency v and 2v respectively.

According to Einstein’s photoelectric equation

$hv=K_1+\phi_0$

and $h2v =K_2+\phi_0$

$∴2(K_1+\phi_0)=K_2+\phi_0$ or $K_2=2K_1+\phi_0$

It means the kinetic energy of the photoelectrons becomes more than doubled.