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The work function of a metal is 6.63 eV. The threshold frequency of the metal is:

Correct answer: A. 1.6 × 10^15 Hz

  • A. 1.6 × 10^15 Hz
  • B. 1.6 × 10^12 Hz
  • C. 6.63 × 10^-34 Hz
  • D. 1.6 × 10^-19 Hz

Explanation

The work function (W) is the minimum energy required to remove an electron from the surface of a metal. It is related to the threshold frequency (V) by the formula V = W/h, where h is Planck's constant (6.626 × 10^-34 J·s). To find the threshold frequency in Hertz, convert the work function from electronvolts to joules (1 eV = 1.602 × 10^-19 J), and then divide by Planck's constant:V = (6.63 eV × 1.602 × 10^-19 J/eV) / (6.626 × 10^-34 J·s) = 1.6 × 10^15 Hz.Option A is correct because it reflects this calculation. Options B, C, and D are incorrect because they either misrepresent the relationship between work function and threshold frequency or use incorrect values.

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About Quantum Theory and Radiation

Quantum theory treats electromagnetic radiation as packets of energy called photons, with energy proportional to frequency through E equals hf. The topic covers Planck's hypothesis, photon momentum, the photoelectric effect, threshold frequency, work function and wave particle duality, distinguishing quantised energy exchange from classical continuous radiation.

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