Question

The work functions of lithium and copper are $2.3eV$ and $4.0eV$ respectively. Out of these, the one which is suitable for the photoelectric cell that works with the visible light is

• A. lithium
• B. copper
• C. both lithium and copper
• D. neither lithium nor copper.

Answer 1

The correct option is A lithium

The wavelength of light required for the emission of photoelectrons is given by:

${\lambda }_0=\frac{12375}{W_0}$

The maximum wavelength of light required for the photoelectric emission for Lithium, ${\lambda }_{Li}=\frac{12375}{2.3}=5380\dot{A}$

Similarly for Copper, ${\lambda }_{Cu}=\frac{12375}{4}=3094\dot{A}$

Since the wavelength $3094\dot{A}$ does not lie in the visible region, but it is in the ultraviolet region so copper can't be used to work with visible light. Hence to work with visible light, lithium metal will be used for the as photoelectric cell as ${\lambda }_{Li}$ lies in visible region.