Answer
Metallic sodium does not show the photoelectric effect for red light.
Work Step by Step
For light of frequency $f$ and wavelength $\lambda$, the photon energy is
$E =hf$
or, $E=\frac{ch}{\lambda}$,
in which $c$ is the speed of light in vacuum, and $h$ is the Planck's constant.
In our case, the wavelength of the incident red light is $\lambda=680\;nm$ $=680\times 10^{-9}$ $m$
Therefore, the energy of a photon of red light
$E=\frac{3\times 10^{8}\times 6.63\times 10^{-34}}{680\times 10^{-9}}$ $J$
or, $E=2.925\times 10^{-19}\;J$
or, $E=\frac{2.925\times 10^{-19}}{1.6\times 10^{-19}}$ $eV$
or, $E\approx 1.828\;eV$
The smallest amount of energy needed to eject an electron from metallic sodium is $2.28\;eV$, that is, the work function $(\phi)$ of the metallic sodium is $\phi=2.28\;eV$
As, $E\lt\phi$, metallic sodium does not show the photoelectric effect for red light.
