The first ionisation energy of potassium is 100kcal mol^-1.Calculate the lowest possible frequency of light that can ionise a potassium atom.

from bohrs theory , E = E_o(1-1/n²)

E₁ (ionisation energy) = 100Kcalmol^-1 = 4.2*10⁵joulesmol^-1 

this E₁ is for 1 mol , for 1 atom E_o will be E₁/N_a                        (N_a = avagadro number)

E_o = E₁/N_a = 7*10^-18 joules

 so , E = E_o(1-1/n²)

 E = hv                          (v is frequency of light emmited)

hv = 7*10^-18 (1-1/n²)

h = 6.6*10^-34 , n should be 2   for minimum frequency ...

hv = 7*10^-18(1-1/4)

 v = 7.95*10¹⁵ hertz