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 $33.7\times 10^{-12}C$ 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 $t=10s$, find the kinetic energy of the most energetic photoelectron emitted at $t=10s$ when it reaches plate $B$.
(Neglect the time taken by the photoelectron to reach plate $B)$

• A. $23eV$
• B. $30eV$
• C. $15eV$
• D. $20eV$

Answer 1

Answer: (A)

Charge on $A,Q_A=\left(5\times 10^7\times 1.6\times 10^{-19}\right)$
$=8\times 10^{-12}$
Charge on $B$, $Q_B=\left(33.7-8\right)\times 10^{-12}C=25.7\times 10^{-12}C$
$\therefore E=\frac{{\sigma }_B}{2{\epsilon }_0}-\frac{{\sigma }_A}{2{\epsilon }_0}$ or $E=\frac{1}{2A{\epsilon }_0}\left(Q_B-Q_A\right)$
or $E=\frac{17.7\times 10^{-12}}{2\times \left(5\times 10^{-4}\right)\times \left(8.85\times 10^{-12}\right)}$
or $E=2000N/C$
Energy of photoelectrons on plate $B$
Energy $=E-W=\left(5-2\right)eV=3eV$
Increase in energy $=\left(Ed\right)eV=\left(2\times 10^3\right)\left(10^{-2}\right)eV=20eV$
$\therefore$ Energy of photoelectrons on plate
$B=\left(20+3\right)eV=23eV$