Photons of what minimum frequency are required to remove electrons from gold? the work function for gold is 4.8 ev. 1 ev = 1.6 x 10 -19 j. h = 6.626 x 10 -34 j s. a. 7.3 x10^14 hz b. 6.5x10^15 hz c. 1.2 x 10^15 hz d. 4.6 x 10^14 hz e. 3.8 x10^17 hz

c. 1.2×10¹⁵ Hz

Explanation:

Work Function: This is the minimum amount of energy a photon requires to liberate an electron from the surface of a metal.

Mathematically, it can be represented as

E' = hf' Equation 1

Where E' = work function of gold, f' = minimum frequency ( threshold frequency), h = Planck's constant

Making f' the subject of the equation,

f' = E'/h Equation 2

Given: E' = 4.8 ev = 4.8×1.6×10⁻¹⁹ J = 7.68×1.6×10⁻¹⁹ J, h = 6.626×10⁻³⁴Js.

Substituting into equation 2

f' = 7.68×10⁻¹⁹/ 6.626×10⁻³⁴

f' = 1.16×10¹⁵ Hz.

f' ≈ 1.2×10¹⁵ Hz

Thus the minimum frequency =  1.2×10¹⁵ Hz

The right option is c. 1.2×10¹⁵ Hz

1 the 4 is sig

find the sum of the two horizontal vectors and the sum of the two vertical vectors

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