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.You need to login to perform this action.
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