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Question:

Explain giving reasons for the following: (a) Photoelectric current in a photocell increases with the increase in the intensity of the incident radiation. (b) The stopping potential (V₀) varies linearly with the frequency (ν) of the incident radiation for a given photosensitive surface with the slope remaining the same for different surfaces. (c) Maximum kinetic energy of the photo-electrons is independent of the intensity of incident radiation.

Solution:

A) Photoelectric current in a photocell increases with increases in intensity. The intensity of light is directly proportional to the number of photons incident per unit area per unit time. As per the equation, 1λ=φhc, when the intensity of light increases, the saturation current increases. This is because a larger number of photons now fall on the metal surface and hence a larger number of electrons interact with photons. The number of electrons emitted increases and hence the current increases.

B) We know, tanθ=hc/e. The slope of the photocurrent is the same for every surface because the slope represents Planck's constant (h) divided by the elementary charge (e), which are fundamental constants and independent of the material of the photosensitive surface. The stopping potential (V₀) is linearly related to the frequency (ν) by the equation: eV₀ = hν - φ, where φ is the work function of the material. A plot of V₀ vs ν will have a slope of h/e.

C) The maximum kinetic energy of a stream of photoelectrons is determined by measuring the stopping potential (the applied voltage needed to keep the photoelectrons trapped in the photoemissive surface). The maximum kinetic energy increases linearly with the frequency of the incident light above the threshold frequency and is independent of the intensity of the incident light. This is because the maximum kinetic energy of an emitted electron is determined by the energy of a single photon (hν) and the work function (φ) of the material: KEmax = hν - φ. Increasing the intensity increases the number of photons, but not the energy of individual photons.