Question

Photons of frequency u are incident on the surfaces of two metals A and B
of threshold frequencies 3/4 u and 2/3 u. respectively. The ratio of
maximum kinetic energy of electrons emitted from A to that from B is
(a) 2:3
(b) 4:3
(c) 3:4
(d) 3:2
which one option is correct?
and explain .​

Answer 1

Answer:

The ratio of maximum kinetic energy of electrons

then the answer is c

Answer 2

The ratio of maximum kinetic energy of electrons emitted from A to that from B is equal to 3:4.

• The energy of the incident light (E) is equal to h.u
• The work functions of metals A and B are 3hu/4 and 2hu/3 respectively.
• The maximum kinetic energy of the photo electrons emmited is equal to the difference between the energy of the incident light and the work function of the metal.
• Hence, the maximum kinetic energy of photo electron fom metals A and B are hu/4 and hu/3 respectively.
• Therefore, the ratio is equal to 3:4.