In a photoelectric experimental setup ultraviolet light of wavelength 350 nm is incident on emitter plate (E). The work function of the emitter plate is ϕ = 2.2 eV . AB is a uniform wire resistor havi

Question

In a photoelectric experimental setup ultraviolet light of wavelength 350 nm is incident on emitter plate (E). The work function of the emitter plate is $ϕ = 2.2 eV . AB$ is a uniform wire resistor having length $L = 100 cm$ . Resistance of the wire AB is $10Ω$ and the emf of the battery is $V = 10$ volt. The sliding contact J can be moved along the wire AB . When the sliding contact is placed at end B of the wire resistor the micro-ammeter shows a reading of $i = + 6 μ A$ . Assume that photoelectric current is very small compared to the current through the cell. Find the reading of the ammeter $(in μ A)$ when the slider is moved to end A of the wire.

TestHub · Accepted Answer

When slider is at A, the applied potential difference across the photoelectric cell is a retarding potential difference of 10 V . Energy of incident photon

$E = \frac{h c}{λ} = \frac{1240 eVnm}{350 nm} = 3.5 eV$

Maximum KE of emitted photoelectrons is $K E_{max} = \frac{h c}{λ} - ϕ = 3.5 - 2.2 = 1.3 eV ∴$ stopping potential difference $V_{s} = 1.3 V$ (a) The applied potential difference is $V (= 10 V) > V_{8} (= 1.3 V)$ Hence there is no photocurrent and reading of Ammeter is zero.

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JEE Advanced (2028)