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

Question

Quizplus · Accepted Answer

The threshold frequency ($f_0$) can be found using the photoelectric equation $E_{\text{photon}} = \phi + K_{\text{max}}$, where $E_{\text{photon}}$ is the energy of the incoming photon, $\phi$ is the work function (or threshold energy), and $K_{\text{max}}$ is the maximum kinetic energy of the emitted photoelectrons. Given $E_{\text{photon}} = 4.20 \, \text{eV}$ and $K_{\text{max}} = 1.80 \, \text{eV}$, we find $\phi = 4.20 \, \text{eV} - 1.80 \, \text{eV} = 2.40 \, \text{eV}$. To find the threshold frequency, convert the work function from eV to Joules ($1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}$), then use $\phi = h f_0$, where $h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s}$ is Planck's constant. Solving for $f_0$ gives $f_0 = \frac{\phi}{h}$ in Hz. Substituting $\phi = 2.40 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV}$ yields $f_0 \approx 5.80 \times 10^{14} \, \text{Hz}$.

Photons of Energy 4.20eV4.20 \mathrm{eV}4.20eV Bombard a Surface, Which Emits Photoelectrons with Kinetic Energies from Photoelectrons

Photons of Energy $4.20 \mathrm{eV}$ Bombard a Surface, Which Emits Photoelectrons with Kinetic Energies from Photoelectrons