= Solution
Use a representative conduction-electron density $n\sim10^{29}\,\mathrm{m^{-3}}$, roughly one mobile <Electron> per atom in a dense <metal>, and the bare <Electron> <mass> for this estimate. The <plasma frequency> is then
$$
\boxed{\omega_p=\sqrt{\frac{ne^2}{m_e\epsilon_0}}\simeq\sqrt{\frac{(10^{29})(1.602\times10^{-19})^2}{(9.109\times10^{-31})(8.854\times10^{-12})}}\simeq1.8\times10^{16}\,\mathrm{rad\,s^{-1}}.}
$$
This is an order-of-magnitude answer because no particular <metal> or <carrier density> is specified. It corresponds to an ordinary <frequency> $f_p\simeq2.8\times10^{15}\,\mathrm{Hz}$, excitation <energy> $\hbar\omega_p\simeq12\,\mathrm{eV}$, and free-space wavelength near $1.1\times10^{-7}\,\mathrm m$.
The physical <oscillation> is collective motion of the <electron gas> against the positive ionic background. Displacing the <Electrons> relative to the <ions> produces a restoring <electric field>. In the simplest uniform-displacement picture, $\ddot x+\omega_p^2x=0$: the <frequency> increases with the density of mobile <electric charge> and decreases with carrier inertia. Quantizing this longitudinal collective <oscillation> gives a <plasmon>.
The same scale controls the ideal collisionless <Drude model> optical response. Its <cold collisionless plasma dielectric response> is $\epsilon_r(\omega)=1-\omega_p^2/\omega^2$. Below $\omega_p$, the negative <relative permittivity> makes a transverse <electromagnetic wave> evanescent, producing strong <reflection>; above it, the ideal model admits propagation. Real-metal collisions, interband <photon absorption> and background <polarization> modify this simple optical edge. Thus $\omega_p$ is a collective charge-response scale, not a single-electron scattering rate or a sharp transparency threshold for every real <metal>.
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