Question

The potassium atom has a work function of 2.3 eV. What is the maximum kinetic energy of the photoelectrons if 9.7 x 10¹⁴ Hz radiation is incident upon it?

Answer 1

The equation of photoelectric effect is $h \nu = A + K_{max}$, where $A$ is a work function and $h$ is Planck's constant. Using last equation, maximum kinetic energy is:

$K_{max} = h \nu - A =$

$6.63\cdot 10^{-34} J \cdot Hz^{-1} \cdot 9.7 \cdot 10^{14} Hz - 2.3 \cdot 1.6 \cdot 10^{-19} J \approx 2.75 \cdot 10^{-19} J = 1.72 eV$.