In a photoelectric effect experiment, the frequency of photons bombarding the surface is increased until photoelectrons just start to leave the surface. If this occurs at a frequency of 6.0 × 10¹⁴ Hz, what is the work function of the surface? A) 1.0 eV B) 1.5 eV C) 2.0 eV D) 2.5 eV E) 3.0 eV

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Quizplus · Accepted Answer

The work function ($\phi$) can be calculated using the equation $\phi = hf$, where $h$ is Planck's constant ($6.626 \times 10^{-34}$ J·s) and $f$ is the frequency of the incident light. Given $f = 6.0 \times 10^{14}$ Hz, $\phi = 6.626 \times 10^{-34} \text{ J·s} \times 6.0 \times 10^{14} \text{ Hz} = 3.976 \times 10^{-19}$ J. Converting joules to electronvolts (1 eV = $1.602 \times 10^{-19}$ J), $\phi \approx 2.48$ eV, which rounds to 2.5 eV, option D.

In a Photoelectric Effect Experiment, the Frequency of Photons Bombarding