# Two metallic plates A and B, each of area 5 x 10-4 m2 ,  are placed parallel to each other at a separation of 1 cm. Plate B carries a positive charge of 33. 7 x 10-12 C. A monochromatic beam of light, with photons of energy 5 eV each, starts falling on plate A at t = 0 so that 1016 photons fall on it per square meter per second. Assume that one photoelectron is emitted for every 106 incident photons. Also assume that all the emitted photoelectrons are collected by plate B and the work function of plate A remains constant at the value 2eV. Determine          (a) the number of photoelectrons emitted up to t = 10 s,(b) the magnitude of the electric field between the plates A and B at t = 10 s, and(c) the kinetic energy of the most energetic photoelectron emitted at t = 10 s when it reaches plate B.Neglect the time taken by the photoelectron to reach plate B. Take ε0 = 8.85 x 10-12 C2 / N –m2

Navjyot Kalra
10 years ago
Hello Student,
(a) Number of electron falling on the metal plate A = 1016 x (5 x 10-4)
∴ Number of photoelectrons emitted from metal plate A upto 10 seconds is
Ne = (5 x 104) x 1016/ 106 x 10 = 5 107
(b) Charge on plate B at t = 10 sec
Qb = 33.7 x 10-12 – 5 x 107 x 1. 6 x 10-19 = 25.7 x 10-12 C
Also Qa = 8 x 10-12 C
E = σB / 2ε0 – σ A / 2ε0 = 1 / 2A ε0 (Q­­B­ - QA)
= 17.7 x 10-12 / 5 x 10-4 x 8.85 x 10-12 = 2000 N/C
(c) K.E. of most energetic particles
= (hv - ϕ) + e (Ed) = 23 eV
Note : (hv - ϕ) is energy of photoelectrons due to light e (Ed) is the energy of photoelectrons due to work done by photoelectrons between the plates.
Thanks
Navjot Kalra