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NEETPhysicsDual Nature of Matter

Monochromatic light of frequency and intensity I falls on a photosensitive surface, producing a stopping potential V₀ and a saturation current i_s . If the frequency of the incident light is doubled to 2 and its intensity is halved to I 2 , what will be the new stopping potential ( V₀' ) and the new saturation current ( i_s' )?

Options

  1. AV₀' = 2V₀ and i_s' = i_s 2
  2. BV₀' > 2V₀ and i_s' = i_s 2
  3. CV₀' = V₀ 2 and i_s' = 2i_s
  4. DV₀' > 2V₀ and i_s' = i_s

Correct answer

B. V₀' > 2V₀ and i_s' = i_s 2

Step-by-step solution

According to Einstein's photoelectric equation, eV₀ = h - . When frequency is doubled to 2 , the new stopping potential V₀' is given by: eV₀' = h(2 ) - = 2h - Substitute h = eV₀ + : eV₀' = 2(eV₀ + ) - = 2eV₀ + Since the work function is positive, eV₀' > 2eV₀ , which means V₀' > 2V₀ . Saturation current depends only on the intensity of incident light (number of photons per second) and is independent of frequency (provided > ₀ ). Halving the intensity halves the number of incident photons, thus halving the saturation

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