Q. Two metallic plates $A$ and $B,$ each of area $5 \times 10^{- 4}m^{2}$ are placed parallel to each other at a separation of $1\ cm.$ plate $B$ carries a positive charge of $33.7\ pc.$ A monochromatic beam of light. With photons of energy $5\ eV$ each, starts falling on plate $A$ at $t = 0,$ so that $10^{16}$ photons fall on it per square meter per second. Assume that one photoelectron is emitted for energy $10^{6}$ 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 $2\ eV.$ Electric field between the plates at the end of 10 seconds is
A) $2 \times 10^{3}N/C$
B) $10^{3}N/C$
C) $5 \times 10^{3}N/C$
D) Zero
Explanation
(b) Number of photoelectrons emitted up to $t = 10\ sec$ are
$n = \frac{(Number\ of\ photons/unit\ area/unit\ time) \times (Area \times Time)}{10^{6}}$
$= \frac{1}{10^{6}}{\lbrack(10)}^{16} \times (5 \times 10^{- 4}) \times (10)\rbrack = 5 \times 10^{7}$
At time $t = 10\ s$
Change on plate $A;q_{A} = + ne = 5 \times 10^{7} \times 1.6 \times 10^{- 19}$
$= 8 \times 10^{- 12}C = 8\ pC$
and charge on plate, $B;q_{B} = 33.7 - 8 = 25.7\ pc$
Electric field between the plates
$E = \frac{(q_{B} - q_{A})}{2\ \varepsilon_{0}A} = \frac{(25.7 - 8) \times 10^{- 12}}{2 \times 8.85 \times 10^{- 12} \times 5 \times 10^{- 4}} = 2 \times 10^{3}\frac{N}{C}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.