The question is incomplete, the complete question is;
Light of frequency f falls on a metal surface and ejects electrons of maximum kinetic energy K by the photoelectric effect.
Part A If the frequency of this light is doubled, the maximum kinetic energy of the emitted electrons will be If the frequency of this light is doubled, the maximum kinetic energy of the emitted electrons will be
K/2.
K.
2K.
greater than 2K.
Answer:
2K
Explanation:
Given that the kinetic energy of photo electrons is given by;
K= E -Wo
Where;
K = kinetic energy
E= energy of incident photon
Wo = work function
But;
E= hf
Wo = fo
h= Plank's constant
f= frequency of incident photon
fo= Threshold frequency
So:
K= hf - hfo
Where the frequency of incident light is doubled;
K= 2hf - hfo
Hence, maximum kinetic energy of the emitted electrons in this case will be 2K
