Two metallic plates A and B, each of area 5 x 10^-4 m² 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

Difficult

Solution

Number of photoelectrons emitted up to t = 10 sec are
$n = \frac{(Number of photons per unit area per unit time \times (Area) \times (Time )}{10^{6}}$
$= \frac{1}{10^{6}} [(10)^{16} \times (5 \times 10^{- 4}) \times (10)] = 5 \times 10^{7}$
At time t =10s
Photoelectrons are ejected from plate A, so it will acquire positive charge.
$\begin{matrix} Charge on plate A : q_{A} = + n e = 5 \times 10^{7} \times 1.6 \times 10^{- 19} \\ = 8 \times 10^{- 12} C = 8 pC \end{matrix}$
Photoelectrons are absorbed by plate B, so its initial charge reduces.
Charge on plate $B : q_{B} = 33.7 - 8 = 25.7 pC$
Consider these plates as of capacitor.
Charge on sides facing each other $= \frac{q_{B} - q_{A}}{2}$
Electric field between the plates
$E = \frac{σ}{ε_{0}} = \frac{(q_{B} - q_{A})}{2 ε_{0} A} = \frac{(25.7 - 8) \times 10^{- 12}}{2 \times 8.85 \times 10^{- 12} \times 5 \times 10^{- 4}} = 2 \times 10^{3} \frac{N}{C}$

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