Dual Nature Of Light Question 119
Question: Two metallic plates A and B, each of area 5 ´ 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
Options:
A) 2 ´ 103 N/C
B) 103 N/C
C) 5 ´ 103 N/C
D) Zero
Show Answer
Answer:
Correct Answer: A
Solution:
Number of photoelectrons emitted up to t = 10 sec are $ n=\frac{\begin{align} & (\text{Number of photons per unit area } \ & \text{ }\text{per unit time)}\times \text{(Area}\times \text{Time)} \ }{1{{0}^{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 ; qA = +ne = 5 ´ 107 ´ 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}. $
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