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 is calculated using the formula $ f = \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 That 101510^{15}1015

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