Question

What is the threshold frequency ν0 of cesium?

Answer 1

 I think this is what you're after:

 Cs(g) → Cs^+ + e⁻ ΔHIP = 375.7 kJ mol^-1 [1] 


Convert to J and divide by the Avogadro Const to give E in J per photon 


E = 375700/6.022×10^23 = 6.239×10^-19 J 


Plank relationship E = h×ν E in J ν = frequency (Hz s-1) 


Planck constant h = 6.626×10^-34 J s 


6.239×10^-19 = (6.626×10^-34)ν 


ν = 9.42×10^14 s^-1 (Hz) 


IP are usually given in ev Cs 3.894 eV 


E = 3.894×1.60×10^-19 = 6.230×10^-19 J per photon