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

Answer

B

Explanation

The work function, `W_(0)=hv_(0)=(hC)/(lambda_(0))` joule
and the work function in `eV=(hC)/(e lambda_(0))=(12375)/(lambda_(0)(Å))`
`:.lambda_(0)=(12375)/(W_(0))`
`:.(lambda_(0))_("Li")=(12375)/(2.3)=5380 Å`
This wavelength lies in the visible region.
For copper `(lambda_(0))_("Cu")=(12375)/(4)=3094Å`
Wavelength 3090 Å lies in the ultraviolet region.
`:.` For the photocell, working with visible light, lithium metal should be used.