Question

Two metallic plates A and B each of area $5\times 10^{-4}{m}^2$, are placed parallel to each other at 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 mitted photoelectrons are collected by plate B and the work function of plate A remains constant at the value 2 eV. Determine :
(a) the number of photoelectrons emitted up to t=10 s
(b) the magnitude of the electric field between the plates A and Batt = 10s and
(c) The kinetic energy of the most energetic photoelectrons emitted at t = 10s when it reaches plate B
Neglect the time taken by the photoelectron to reach plate B.
(Take \varepsilon_0=8.85\times10^{-12}C^2/N-m^2)

Answer 1

Area of plates $A=5\times 10^{-4}{m}^2$
Distance between the plates $d=1cm=10^{-2}m$
(a) Number of photoelectrons emitted upto t = 10 s are
$n=\frac{(numberofphotonsfallinginunitareainunittime)\times (area\times time)}{10^6}$
=$\frac{1}{10^6}[(10{)}^{16}\times (5\times 10^{-4})\times (10)]=5.0\times 10^7$
(b) At time t=10s
charge on plate A, $q_A=+ne=(5.0\times 10^7)(1.6\times 10^{-19})$
$=8.0\times 10^{-12}$
and charge on plate B,
$q_B=(33.7\times 10^{-12}-8.0\times 10^{-12})$
$=25.7\times 10^{-12}C$
$\therefore$ Electric field between the plates $E=\frac{(q_3-q_a)}{2A{\epsilon }_0}$
or $E=\frac{(25.7-8.0)\times 10^{-12}}{2\times (5\times 10^{-4})(8.85\times 10^{-12})}=2\times 10^{3}N/C$
(c) Energy of photoelectrons at plate A
$=E-W=(5-2)eV=3eV$
Increase in energy of photoelectrons
$=(eEd)joule=(Ed)eV$
$=(2\times 10^3)(10^{-2})eV=20eV$
Energy of photoelectrons at plate B
$=(20+3)eV=23eV$