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The threshold frequency of caesium is 5.16×10¹⁴ Hz. Then its work function is ______ eV.

Asked in GUJCET 2020 · Work function and the photoelectric equation

Answer: (2) 2.14

Step-by-step solution

Given: threshold frequency f₀ = 5.16×10¹⁴ Hz; take h = 6.63×10⁻³⁴ J s.

Idea: at threshold the photon energy just equals the work function, W = hf₀.

W = 6.63×10⁻³⁴×5.16×10¹⁴ = 3.42×10⁻¹⁹ J

W = (3.42×10⁻¹⁹)/(1.6×10⁻¹⁹) = 2.14 eV

So the work function of caesium is 2.14 eV.

Why the other options are wrong

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