posted by qwerty123 on .
To eject electrons from the surface of potassium metal, 222kJ/mol of electrons are needed.
a) What minimum frequency of light is needed to supply the required energy?
b) What is the wavelength of this light?
c) If potassium is irradiated with light having wavelength of 350 nm, what is the maximum E(sub K) of an emitted electron?
d) What maximum speed in m/2 will the electrons in part (c) have?
Divide 222 kJ/mol by 6.022E-23 which gives you per photon, then convert to J/photon.
a. E = h*frequency
b. c = frequency*wavelength
c. E = hc/wavelength
d. E = 1/2 mv^2
The number you have from 222,000/6.022E23 (that's not -23 and I've made it Joules by converting kJ to J) is called the work function. For part a,
E = h*frequency-work function. I forgot the work function part.