Question

Work function of lithium and copper are respectively $2.3eV$ and $4.0eV.$ Which one of the metal will be useful for the photoelectric cell working with visible light ? $\left(h=6.6\times {\left(10\right)}^{-34}J-s,c=3\times {\left(10\right)}^{8}m/s\right)$

• A. Lithium
• B. Copper
• C. both
• D. none of the above

Answer 1

Answer: (A)

From ${\lambda }_0=\frac{12375}{W_0}$
The maximum wavelength of light required for the photoelectron emission, ${\left({\left(\lambda \right)}_0\right)}_{Li}=\frac{12375}{2.3}=5380\stackrel{o}{A}$
Similarly ${\left({\left(\lambda \right)}_0\right)}_{Cu}=\frac{12375}{4}=3094\stackrel{o}{A}$
since the wavelength $3094\stackrel{o}{A}$ 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.