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In a photoelectric experiment, the relation between the applied potential difference V and the photoelectric current I was found to be as shown in the graph below. If the work function of the cathode plate is 28.8 eV and h = 6 .6 × 10 - 34 J s , the frequency of incident radiation would be nearly ( in s - 1 )

Options

  1. A0.436 × 10 18 Hz
  2. B0.436 × 10 17 Hz
  3. C0.775 × 10 16 Hz
  4. D0.775 × 10 15 Hz

Correct answer

C. 0.775 × 10 16 Hz

Step-by-step solution

For the photoelectric effect, hf = ϕ + KE max From the graph KE max = 3 .2 eV ∴ hf = ( 28 .8 + 3 .2 ) eV f = 32 h eV = 32 × 1 .6 × 10 − 19 6 .6 × 10 − 34 Hz = 0 .775 × 10 16 Hz

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