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 xx 10^(-34) J - s, c = 3 xx 10^(8) m//s)`

Options

A. Lithium
B. Copper
C. Both
D. None of these

Answer

A

Explanation

From `lambda_(0) = (12375)/(W_(0))`
The maximum wavelength of light required for the photoelectron emission , `(lambda_(0))_(Li) = (12375)/(2.3) = 5380 Å`.
Similarly `(lambda_(0))_(cut) = (12375)/(4) = 3094 Å`. Since the wavelength `3094 Å` 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.