The number of electrons that appear on the surface of the cathode plate are given as K × 1 0 9 , when the jockey is connected at the end 'P' of the potentiometer wire. Assume that no radiation is falling on the plates. Find the K.

Voltage across wire = 20 × 8 + 2 8 = 16 V ; capacitance C = d ε 0 A n = e C V = 0.5 × 1 0 − 3 × 1.6 × 1 0 − 19 8.85 × 1 0 − 12 × 50 × 1 0 − 4 × 16 n = 8.85 × 1 0 9 , so K = 8.85 Answer 8.85

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Passage / Comprehension

An experimental setup of verification of photoelectric effect is shown in figure. The voltage across the electrodes is measured with the help of an ideal voltmeter, which can be varied by moving jockey 'J' on the potentiometer wire. The battery used in the potentiometer circuit is of 20 V and its internal resistance is 2 Ω. The resistance of 100 cm long potentiometer wire is 8 Ω. The photocurrent is measured with the help of an ideal ammeter. Two plates of potassium oxide of area 50 cm² at separation 0.5 mm are used in the vacuum tube. Photocurrent in the circuit is very small, so the potentiometer circuit can be treated as an independent circuit.

The number of electrons that appear on the surface of the cathode plate are given as $K \times 1 0^{9}$ , when the jockey is connected at the end 'P' of the potentiometer wire. Assume that no radiation is falling on the plates. Find the K.

Answer:

8.85

Solution:

Voltage across wire = $20 \times \frac{8}{8 + 2} = 16 V$ ; capacitance $C = \frac{ε _{0} A}{d}$

$n = \frac{C V}{e} = \frac{8.85 \times 1 0 ^{- 12} \times 50 \times 1 0 ^{- 4} \times 16}{0.5 \times 1 0 ^{- 3} \times 1.6 \times 1 0 ^{- 19}}$

$n = 8.85 \times 1 0^{9}$ , so K = 8.85

Answer 8.85

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