A given photosensitive metal has a work function equal to 0.75J.When a radiation of frequency $\nu$ is incident on it,the maximum kinetic energy of the emitted photoelectrons is 30J.If the frequency of incident radiation is doubled, then the maximum kinetic energy of the emitted photoelectron would be:

60.75J

$ K.E (K) = h\nu -\phi = 30J$

$ K' = 2h\nu - \phi = 2(30+\phi) - \phi = 60+\phi = 60.75J$