The maximum kinetic energy of photoelectrons ejected from a metal surface is $y$ when irradiated wit

Question

The maximum kinetic energy of photoelectrons ejected from a metal surface is $y$ when irradiated with radiation of frequency $3
u_0$ (where $
u_0$ is the threshold frequency). If the frequency of the incident radiation is increased to $5
u_0$, what is the new maximum kinetic energy of the photoelectrons?

Professor PC Thomas Classes · Answer

(a) : According to Einstein's photoelectric equation, $KE_{\max} = h(
u - 
u_0)$. For the first case, the frequency is $
u = 3
u_0$. Thus, $y = h(3
u_0 - 
u_0) = 2h
u_0$. For the second case, the frequency is increased to $
u = 5
u_0$. The new kinetic energy is $KE_{new} = h(5
u_0 - 
u_0) = 4h
u_0$. Comparing the two, we see that $KE_{new} = 2(2h
u_0) = 2y$.