Question

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 $1cm$. Plate $B$ carries a positive charge of $33.7pc$. A monochromatic beam of light, with photons of energy $5eV$ each, starts falling on plate $A$ at $t=0$ so that ${10}^{16}$ photons fall on it per square metre per second. Assume that one photoelectron is emitted for every ${10}^6$ 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 $2eV$. Determine the number of photoelectrons emitted upto $t=10s$.

• A. $2\times {10}^{3}N/C$
• B. ${10}^{3}N/C$
• C. $5\times {10}^{3}N/C$
• D. $Zero$

Answer 1

Answer: (C)

$5\times {10}^{3}N/C$

Area of each plate = $5\times {10}^{-4}{m}^2$

Separation between plates = $1cm={10}^{-2}m$

$\frac{photonsincident}{area\times time}={10}^{16},t=10s$

$\therefore No.ofphotonsincident={10}^{16}\times area\times time={10}^{16}\times (5\times {10}^{-4})\times 10=5\times {10}^{13}$

$\therefore Requirednumberofphotoelectronsemitted=\frac{5\times {10}^{13}}{{10}^6}$

Number of photoelectrons = $5\times {10}^7$