What is the cutoff (threshold) frequency for a metal surface that has a work function of 5.42 eV? (1 eV = 1.60 × 10^-19 J, h = 6.626 × 10^-34 J ∙ s) A) 1.31 × $10^{15}$ Hz B) 2.01 × $10^{15}$ Hz C) 3.01 × $10^{15}$ Hz D) 5.02 × $10^{15 }$ Hz E) 6.04 × $10^{15 }$ Hz

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Quizplus · Accepted Answer

The cutoff frequency $ f_0 $ is given by $ f_0 = \frac{\phi}{h} $, where $ \phi $ is the work function in joules. Converting 5.42 eV to joules gives $ 5.42 \times 1.60 \times 10^{-19} $ J. Dividing by Planck's constant $ h = 6.626 \times 10^{-34} $ J∙s gives approximately $ 1.31 \times 10^{15} $ Hz.

What Is the Cutoff (Threshold) Frequency for a Metal Surface 101510^{15}1015

What Is the Cutoff (Threshold) Frequency for a Metal Surface $10^{15}$