SATHEE: Modern Physics 2 Question 33

Modern Physics 2 Question 33

32. Two metallic plates $A$ and $B$ each of area $5 \times 10^{- 4} m^{2}$ , are placed parallel to each other at separation of $1 c m$ . Plate $B$ carries a positive charge of $33.7 \times 10^{- 12} C$ . A monochromatic beam of light, with photons of energy $5 e V$ each, starts falling on plate $A$ at $t = 0$ so that $10^{16}$ photons fall on it per square metre 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 $2 e V$ . Determine

$(2002, 5 M)$

(a) the number of photoelectrons emitted up to $t = 10 s$ ,

(b) the magnitude of the electric field between the plates $A$ and $B$ at $t = 10 s$ and

Neglect the time taken by the photoelectron to reach plate $B$ . (Take $ε_{0} = 8.85 \times 10^{- 12} C^{2} / N - m^{2}$ ).

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Solution:

Area of plates $A = 5 \times 10^{- 4} m^{2}$

Distance between the plates $d = 1 c m = 10^{- 2} m$

(a) Number of photoelectrons emitted upto $t = 10 s$ are

$\begin{aligned} (number of photons falling in unit \\ n = \frac{area in unit time) \times (area \times time)}{10^{6}} \\ = \frac{1}{10^{6}} [(10)^{16} \times (5 \times 10^{- 4}) \times (10)] = 5.0 \times 10^{7} \end{aligned}$

(b) At time, $t = 10 s$

Charge on plate $A, q_{A} = + n e = (5.0 \times 10^{7}) (1.6 \times 10^{- 19})$

$= 8.0 \times 10^{- 12} C$

and charge on plate $B$ ,

$\begin{aligned} q_{B} & = (33.7 \times 10^{- 12} - 8.0 \times 10^{- 12}) \\ = 25.7 \times 10^{- 12} C \end{aligned}$

$∴$ Electric field between the plates, $E = \frac{(q_{B} - q_{A})}{2 A ε_{0}}$

or $E = \frac{(25.7 - 8.0) \times 10^{- 12}}{2 \times (5 \times 10^{- 4}) (8.85 \times 10^{- 12})} = 2 \times 10^{3} N / C$ (c) Energy of photoelectrons at plate $A$