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If a photocell is illuminated with a radiation of 1240   A ∘ , the stopping potential is found to be 8   V ; then the work function of the emitter and the threshold wavelength are:

Options

  1. A2   eV ,   2000   A ∘
  2. B2   eV ,   6200   A ∘
  3. C2   eV ,   2480   A ∘
  4. D3   eV , 6200   A ∘

Correct answer

B. 2   eV ,   6200   A ∘

Step-by-step solution

The Einstein's equation for photoelectric effect : K E max = h c λ - W where, W = work function of metal, λ = wavelength of incident light. If V 0 be the stopping potential, then K E max = e V 0 So, e V 0 = h c λ - W W = h c λ - e V 0 = 6 . 62 × 10 - 34 × 3 × 10 8 1240 × 10 - 10 - 1 . 6 × 10 - 19 8 = 3 . 2 × 10 - 19 = 2   eV as 1   eV = 1 . 6 × 10 - 19 If λ 0 be the threshold wavelength, W = h c λ 0 λ 0 = h c W = 6 . 62 × 10 -

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