Dual Nature Of Light Question 97
Question: 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 ´ 10?34 J-s, c = 3 ´ 108 m/s) [DPMT 2003]
Options:
A) Lithium
B) Copper
C) Both
D) None of these
Show Answer
Answer:
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.
BETA
