Photoelectric EffectEasy
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 falls on it per square meter per second. Assume that one photo electron 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 of 2eV. 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
Solution
Number of photoelectrons emitted up to
t = 10 sec are
n =
=
[(10)16 × (5 × 10-4) × (10)] = 5 × 107
At time t = 10s
Charge on plate A = qA = + ne
= 5 × 107 × 1.6 × 10-19
= 8 × 10-12 C = 8pC
and charge on plate B; qs = 33.7 - 8 = 25.7 pc
Electric field batween the plates
E =
=
= 2 × 103 
t = 10 sec are
n =
=
[(10)16 × (5 × 10-4) × (10)] = 5 × 107At time t = 10s
Charge on plate A = qA = + ne
= 5 × 107 × 1.6 × 10-19
= 8 × 10-12 C = 8pC
and charge on plate B; qs = 33.7 - 8 = 25.7 pc
Electric field batween the plates
E =
=
= 2 × 103 Create a free account to view solution
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