Two metallic plates A and B, each of area 5 x 10^–4m² 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 10¹⁶ photons fall on it per square meter per second. Assume that one photoelectron is emitted for every 10⁶ 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

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2 x 10³ N/C

10³N/C

5 x 10³N/C

Zero

answer is A.

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Detailed Solution

Number of photoelectrons emitted up to t = 10 sec are
$n\;=\;\frac{\left ( photons\;per\;unit\;area\;per\;unit\;time\right )\times\left ( Area\;\times\;Time \right )}{10^{6}}$
$= \frac{1}{{{{10}^6}}}[{(10)^{16}} \times (5 \times {10^{ - 4}}) \times (10)] = 5 \times {10^7}$
At time t = 10 sec

Charge on plate A ; q_A = +ne = 5 x 10⁷ x 1.6 x 10^–19
= 8 x 10^–12 C = 8 pC
and charge on plate B ; q_B = 33.7 – 8 = 25.7 pc
Electric field between the plates
$E = \frac{{({q_B} - {q_A})}}{{2\,{\varepsilon _0}A}} = \frac{{(25.7 - 8) \times {{10}^{ - 12}}}}{{2 \times 8.85 \times {{10}^{ - 12}} \times 5 \times {{10}^{ - 4}}}} = 2 \times {10^3}\frac{N}{C}.$