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Two metallic plates $A$ and $B$, each of area $5 ×10^{-4}m^2$ 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 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 $2\, eV$. Electric field between the plates at the end of $10$ seconds is
$2 × 10^3 N/C$
$10^3 N/C$
$5 ×10^3 N/C$
Zero
Solution
(a) Number of photoelectrons emitted up to $t = 10$ sec are
$n = \frac{\begin{array}{l}({\rm{Number\, of\, photons \,per\, unit \,area }}\\{\rm{ }}\,\,\,\,\,\,{\rm{per\, unit \,time)}} \times {\rm{(Area}} \times {\rm{Time)}}\end{array}}{{{\rm{1}}{{\rm{0}}^{\rm{6}}}}}$
$ = \frac{1}{{{{10}^6}}}[{(10)^{16}} \times (5 \times {10^{ – 4}}) \times (10)] = 5 \times {10^7}$
At time $t = 10$ sec
Charge on plate $A$ ; $q_A$ $= +ne = 5 × 10^7 ×1.6 × 10^{-19}$
$= 8 ×10^{-12}$ $C = 8\, pC$
and charge on plate $B$ ; $qB = 33.7 -8 = 25.7\, pc$
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}.$
