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JEE Main201910 Jan 2019Evening ShiftPhysicsDual Nature of MatterActual

A metal plate of area 1 × 10 - 4 m 2 is illuminated by a radiation of intensity 16 m i l l i W m 2 . The work function of the metal is 5 e V . The energy of the incident photons is 10 e V and only 10 % of it produces photo electrons. The number of emitted photo electron per second and their maximum energy, respectively, will be: 1 e V = 1.6 × 10 - 19 J

Options

  1. A10 14 and 10 e V
  2. B10 12 and 5 e V
  3. C10 11 and 5 e V
  4. D10 10 and 5 e V

Correct answer

C. 10 11 and 5 e V

Step-by-step solution

Einstein's photoelectric equation Maximum K E = h C λ - ϕ = 10 - 5 = 5   e V Intensity I = N p h C λ t A ⇒ N p t = I A h C λ and number of electrons per second. = N p t × 10 % = I A h C λ × 1 10 = 16 × 10 - 3 × 10 - 4 10 × 1.6 × 10 - 19 × 10 = 10 11 per sec

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