Photoelectric emission takes place from a certain metal at threshold frequency v . If the radiation of frequency 4 v is incident on the metal plate, the maximum velocity of the emitted photoelectrons will be ( m = mass of photoelectron, h = Planck's constant)

Step-by-step solution

According to Einstein’s photoelectric equation, hf = + K_ max , where h is Planck’s constant, f is the incident frequency, is the work function, and K_ max is the maximum kinetic energy of ejected electrons. Given the threshold frequency v , the work function is = hv . For incident frequency 4v , the photoelectric equation becomes h(4v) = hv + K_ max . Solving yields K_ max = 3hv . Since K_ max = 1 2 m v_ max ^2 , equating gives 1 2 m v_ max ^2 = 3hv . Solving for v_ max , we find v_ max ^2 = 6hv m , so v_ max = 6h