Photons of energy 4.20 eV bombard a surface, which emits photoelectrons with kinetic energies from 0 to 1.80 eV. What is the threshold frequency for this surface?

Question

Quizplus · Accepted Answer

The threshold frequency ($f_0$) can be found using the photoelectric equation $E_{photon} = K_{max} + W$, where $E_{photon} = 4.20 \, \text{eV}$, $K_{max} = 1.80 \, \text{eV}$, and $W$ is the work function. Rearranging for $W$, we get $W = E_{photon} - K_{max} = 4.20 \, \text{eV} - 1.80 \, \text{eV} = 2.40 \, \text{eV}$. To find the threshold frequency, use $f_0 = \frac{W}{h}$, where $h = 4.135667696 \times 10^{-15} \, \text{eV}\cdot\text{s}$. Converting $W$ to Joules if necessary and solving gives a frequency close to the value in choice B.

Photons of Energy 4