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 $1 \,cm$. Plate $B$ carries a positive charge of $33.7 \times 10^{-12}\, C$ A monochromatic beam of light, with photons of energy $5\, eV$ each, starts falling on plate $A$ at $t = 0$ so that $10^{16}$ 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 $2\, eV$. Determine the number of photoelectrons emitted upto $t=10\,s$, find 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 $B)$

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

Answer 1

Answer: (A)

Charge on $A, Q_{A}=\left(5\times10^{7}\times1.6\times10^{-19}\right)$
$=8\times10^{-12}$
Charge on $B$, $Q_{B}=\left(33.7-8\right)\times10^{-12}C=25.7\times10^{-12}C$
$\therefore E=\frac{\sigma_{B}}{2\varepsilon_{0}}-\frac{\sigma_{A}}{2\varepsilon_{0}}$ or $E=\frac{1}{2A\varepsilon_{0}}\left(Q_{B}-Q_{A}\right)$
or $E=\frac{17.7\times10^{-12}}{2\times\left(5\times10^{-4}\right)\times\left(8.85\times10^{-12}\right)}$
or $E=2000 N/ C$
Energy of photoelectrons on plate $B$
Energy $= E - W = \left(5 - 2\right) eV = 3\,eV$
Increase in energy $= \left(Ed\right) eV = \left(2 \times 10^{3}\right) \left(10^{-2}\right) eV = 20 \,eV$
$\therefore $ Energy of photoelectrons on plate
$B=\left(20+3\right)eV=23\,eV$