Chem Help!! It requires 183.7 kJ/mol to eject electrons from cesium metal.?

Question

(a) What is the minimum frequency of light necessary to emit electrons from cesium via the photoelectric effect?_________s-1

(b)What is the wavelength of this light? ______nm

(c) If cesium is irradiated with light of 416 nm, what is the maximun possible kinetic energy of the emitted electrons? ______J

Please help..ive been working on this for a while, and cant seem to get the right answer!!

Answer 1

a) An energy of 183.7 kJ/mol corresponds to 183.7 x 10^3 / 6.022 x 10^23 = 30 x 10^-20 Joule per atom.
I divided here by Avogadros constant. This energy is equal to hf where f is the frequency.
f = 30 x 10^-20/6.626 x 10^-34 = 4.6 x 10^14 Hz
b) Wavelength L = c/f = 3x10^8 / 4.6 x 10^14 = 650 nm
c) The wavelength of 416 nm corresponds to an energy of h c / L = 47 x 10^-20 Joule. This is 17 x 10^20 J more than what we used in a)