A) Lithium
B) Copper
C) Both
D) None of these
Correct Answer: A
Solution :
From \[{{\lambda }_{0}}=\frac{12375}{{{W}_{0}}}\] The maximum wavelength of light required for the photoelectron emission, \[{{({{\lambda }_{0}})}_{Li}}=\frac{12375}{2.3}=5380{\AA}\]. Similarly \[{{({{\lambda }_{0}})}_{Cu}}=\frac{12375}{4}\]= 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.
You need to login to perform this action.
You will be redirected in
3 sec
