(a) From $\lambda_{0} = \frac{12375}{W_{0}}$
The maximum wavelength of light required for the photoelectron emission, $\left( \lambda_{0} \right)_{Li} = \frac{12375}{2.3} = 5380\mathring{\mathrm{A}}$
Similarly $\left( \lambda_{0} \right)_{Cu} = \frac{12375}{4} = 3094\ \mathring{\mathrm{A}}$
Since the wavelength $3094\ \mathring{\mathrm{A}}$ does not in the visible region, but it is in the ultraviolet region. Hence to work with visible light, lithium metal will be used for photoelectric cell