Light of frequency 4 _0 is incident on the metal of threshold...

According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectron is given by

K_{\text\max} = h\nu - h\nu_0

Given that the frequency of incident light is $\nu = 4\nu_0$ and the threshold frequency is $\nu_0$.

Substituting the values, we get

K_{\text\max} = h(4\nu_0) - h\nu_0

K_{\text\max} = 3h\nu_0

Answer: $3h\nu_0$