A photocathode having a work function of 2.8 eV is illuminated with monochromatic electromagnetic radiation whose photon energy is 4.0 eV.What is the threshold (cutoff)frequency for photoelectron production? (1 eV = 1.60 × 10^-19 J,h = 6.626 × 10^-34 J ∙ s) A)6.8 × 10¹⁴_Hz B)2.9 × 10¹⁴_Hz C)7.7 × 10¹⁴_Hz D)8.6 × 10¹⁴_Hz E)9.7 × 10¹⁴_Hz

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Quizplus · Accepted Answer

The threshold frequency is given by the equation:$$f_0 = \frac{\phi}{h}$$where:* $f_0$ is the threshold frequency* $\phi$ is the work function* $h$ is Planck's constantSubstituting the given values into the equation, we get:$$f_0 = \frac{2.8 	ext{ eV}}{6.626 	imes 10^{-34} 	ext{ J s}} = 6.8 	imes 10^{14} 	ext{ Hz}$$Therefore, the correct answer is A.

A Photocathode Having a Work Function of 2