The work function of a metal is 2.56 x 10-19J. Calculate the frequency of a photon whose energy is required to eject from the metal an electron with kinetic energy of 3.0ev. [1 eV = 1.6 x 10-19J, h = 6.6 x 10-34Js]
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Correct Answer: Option C
Explanation:
F = \(\frac{K.E + Wo}{h}\)
= \(\frac{3.0 \times 1.6 \times 10^{-19} + 2.56 \times 106{-19}}{6.6 \times 106{-34}}\)
- 1.12 x 10-15HZ
F = \(\frac{K.E + Wo}{h}\)
= \(\frac{3.0 \times 1.6 \times 10^{-19} + 2.56 \times 106{-19}}{6.6 \times 106{-34}}\)
- 1.12 x 10-15HZ