Photons of what minimum frequency are required to remove electrons from gold? Note: The work function for gold is 4.8 eV. A)7.3 × 10¹⁴ Hz B)1.2 × 10¹⁵ Hz C)3.8 × 10¹⁷ Hz D)6.5 × 10¹⁵ Hz E)4.6 × 10¹⁴ Hz

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The work function for gold is 4.8 eV. The minimum frequency of photons required to remove electrons from gold can be calculated using the equation:$$hf = \phi$$where:* h is Planck's constant (6.63 x 10^-34 J s)* f is the frequency of the photons (Hz)* $\phi$ is the work function (4.8 eV)Converting eV to J:$$4.8 	ext{ eV} = 4.8 	imes 1.6 	imes 10^{-19} 	ext{ J} = 7.68 	imes 10^{-19} 	ext{ J}$$Substituting the values into the equation:$$6.63 	imes 10^{-34} 	ext{ J s} 	imes f = 7.68 	imes 10^{-19} 	ext{ J}$$Solving for f:$$f = \frac{7.68 	imes 10^{-19} 	ext{ J}}{6.63 	imes 10^{-34} 	ext{ J s}} = 1.16 	imes 10^{15} 	ext{ Hz}$$Therefore, the minimum frequency of photons required to remove electrons from gold is 1.16 x 10^15 Hz, which corresponds to choice B.

Photons of What Minimum Frequency Are Required to Remove Electrons