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The electric field of a light beam is given by E = 50 [1 + 0.5 (1 10¹⁵ t)] (5 10¹⁵ t) V/m where t is in seconds. If this light is incident on a metal surface of work function 2.0 eV , the stopping potential for the emitted photoelectrons is: (Take h 2 = 6.6 10⁻¹⁶ eV s )

Options

  1. A1.30 V
  2. B1.96 V
  3. C0.64 V
  4. D7.90 V

Correct answer

B. 1.96 V

Step-by-step solution

The given electric field equation represents an amplitude-modulated wave. We can expand it using the trigonometric identity 2 A B = (A+B) + (A-B) : E = 50 (5 10¹⁵ t) + 25 (1 10¹⁵ t) (5 10¹⁵ t) E = 50 (5 10¹⁵ t) + 12.5 (6 10¹⁵ t) + 12.5 (4 10¹⁵ t) The light beam consists of three distinct angular frequencies: ₁ = 4 10¹⁵ rad/s ₂ = 5 10¹⁵ rad/s ₃ = 6 10¹⁵ rad/s The stopping potential is determined by the maximum kinetic energy of the photoelectrons, which in turn is produced by the highest frequency component, _ max =

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