What is the threshold frequency ν0 of cesium?

You are given a ν0 of cesium and you are asked to find the threshold frequency. First, you have to know the half reaction of cesium which is

Cs(g) → Cs⁺ + e⁻ with a ΔHIP of 375.7 kJ/mol
Then convert kilojoules to joules and then divide this by the Avogadro's constant which is 6.023 x 10²³ atoms/mole to give the energy of photon in joules per photon

E = (375,700 J/mol) / (6.023 x 10²³ photons/ mole)
E = 6.239 x 10⁻¹⁹ J/photon

Use the equation of energy of a photon
E = hv where E is the energy of the photon, h is the planck's constant and v is the threshold frequency
E = hv
6.239 x 10⁻¹⁹ J/photon = 6.626 x 10⁻³⁴ (v)
v = 9.42 x 10¹⁴ s⁻¹ Hz

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