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)

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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.

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