Assuming that the incident radiation is capable of ejecting photo electrons from all the given metals, the…

According to the Einstein's theory of photoelectric effect, for the photoelectric effect to occur, the energy of incident radiation should be greater or equal to the threshold energy called work function of the metal. The equation can be illustrated as:

$$\begin{gathered} \mathrm{h\nu } = \mathrm{\phi } + \mathrm{KE} \\ \mathrm{h\nu } - \mathrm{\phi } = \mathrm{KE} \end{gathered}$$

Now, from above relation it can be said that more will be the work function ($\mathrm{\phi }$) of the metal, lesser will be the kinetic energy of the ejected photoelectrons. The order work function for given metals is

$\mathrm{Cu}>\mathrm{Ag}>\mathrm{Li}>\mathrm{Na}$

Therefore, the order of the kinetic energy of ejected photoelectrons will be:

$\mathrm{Cu}<\mathrm{Ag}<\mathrm{Li}<\mathrm{Na}$

Hence, $\mathrm{Cu}$ possess least $\mathrm{K}.\mathrm{E}.$ from the given option.