A) Lithium
B) Copper
C) Both
D) None
Correct Answer: A
Solution :
Here: work function of lithium \[A{{m}^{2}}\] \[A{{m}^{2}}\] The maximum wavelength of light required for the photoelectron emission from the lithium metal is \[A{{m}^{2}}\] \[A{{m}^{2}}\] \[\sqrt{2}s\] Maximum wave length of light required for photoelectron emission from the copper is given by \[\frac{1}{\sqrt{2}}s\] \[\frac{1}{2}s\] \[\mu F\] Since, the wavelength \[\mu F\]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.
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