What Is The Threshold Frequency ν0 Of Cesium

What is 0 cut off frequency for cesium? Note that 1 eV (electron volt) = 1.60 × 10 × 19 J.?

I think this is what you are looking for:

Cs (g)  'Cs + + eà  »- HIP = 375.7 mol kJ 1 [1]

Convert to J and divide by Eugadro Const so that E is obtained in J per pton. Get

E = 375700 / 6,022 10 23 = 6,239 × 10 19 J

Plate ratio E = hƒÂ - ½ E to J ½ = frequency (Hz s1)

Planck's constant h = 6,626 × 10 34 J s

6239 × 10 19 = (6,626 × 10 34)

½ = 9.42à- 10 14 s 1 (Hz)

½0 = 9.39ÃÂ - 10 14 Hz

Use E = h * v..place h = Plank const = 6.63E34 E = approx 3E19J = one million, 873eV then frequ = 3E8 / 235E9 = one million, use v = c / L to get 277 E15 / s Then and you = lumbar hv .. resulting in E = 8464E19J .. reducing the mapping function (3E19J) .. leaving 5.474E19J = 3.42eV .. This means that the elasticity that occurs in ptoelectrons after passing through metal r That is E (pton)) = Wo + Ek (electrons).

1.6E19 / 6.63E34 = 2.41E14

2.41E14 * 3.89 = 9.39E14Hz

What Is The Threshold Frequency ν0 Of Cesium