Answer to Question #156374 in Physics for Jane

Question

Green light of wavelength 550 nm is shone on potassium, which

has a work function of 2.0 eV. Calculate the maximum kinetic energy of the photoelectrons emitted.

Answer 1

We can find the maximum kinetic energy of the photoelectrons emitted from the formula:

$KE_{max}=hf-W,$ $KE_{max}=\dfrac{hc}{\lambda}-W,$ $KE_{max}=\dfrac{6.63\cdot10^{-34}\ J\cdot s\cdot3\cdot10^8\ \dfrac{m}{s}}{550\cdot10^{-9}\ m}-2.0\ eV,$ $KE_{max}=3.62\cdot10^{-19}\ J\cdot\dfrac{1\ eV}{1.6\cdot10^{-19}\ J}-2.0\ eV=0.26\ eV.$

Answer:

$KE_{max}=0.26\ eV.$