Question

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

• A. Lithium
• B. Copper
• C. Both
• D. None of them can be used

Answer 1

The correct option is A Lithium

From ${\lambda }_0=\frac{12375}{W_0}$
The maximum wavelength of light required for the photoelectron emission, $({\lambda }_0{)}_{Li}=\frac{12375}{2.3}=5380\dot{A}$.
Similarly $({\lambda }_0{)}_{Cu}=\frac{12375}{4}=3094\dot{A}$.
Since the wavelength $3094\dot{A}$ does not lie 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.