The correct option is D All of the above
(K.E)max= The energy of the most energetic photo electrons emitted from the surface.
(KE)max=E−ϕ0
Where; E= Energy of incident radiation =hcλ=hf where, λ is wavelength of incident radiation.
∴ Option (C) is true.
Also, ϕ0 is work function
∴ Option (B) is valid.
Now, ϕ0=hfo
Where, f0 is threshold frequency of the metal.
∴ option (A) is valid too.
Hence, Option (D) is the correct answer.
Why this question?This question tests the knowledge regarding the dependnce of most energetic particles emitted from metal target with respect to different options given.