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NEETPhysicsDual Nature of Matter

Three metallic surfaces A, B, and C are illuminated by monochromatic light of frequency . The stopping potentials for the photoelectrons emitted from surfaces A and B are V_A and V_B respectively, where V_A > V_B > 0 . For surface C, no photoelectric emission is observed. If _A , _B , and _C are the threshold frequencies for metals A, B, and C respectively, which of the following relations is correct?

Options

  1. A_A > _B > _C
  2. B_B < _A < _C
  3. C_A < _B < _C
  4. D_A < _B < _C

Correct answer

C. _A < _B < _C

Step-by-step solution

According to Einstein's photoelectric equation, the stopping potential V is given by: eV = h - h _ th For photoelectric emission to occur, the incident frequency must be greater than the threshold frequency _ th . Since surfaces A and B emit photoelectrons, we have > _A and > _B . From the equation, for a given incident frequency , a larger stopping potential implies a smaller threshold frequency. Since V_A > V_B , it follows that _A For surface C, no photoelectric emission is observed, which means the incident fre

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