Q. 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 $(h = 6.6 \times 10^{- 34}J$-$s,c = 3 \times 10^{8}m/s)$

A) Lithium
B) Copper
C) Both
D) None of these
Explanation

(a) From $\lambda_{0} = \frac{12375}{W_{0}}$
The maximum wavelength of light required for the photoelectron emission, $\left( \lambda_{0} \right)_{Li} = \frac{12375}{2.3} = 5380\mathring{\mathrm{A}}$
Similarly $\left( \lambda_{0} \right)_{Cu} = \frac{12375}{4} = 3094\ \mathring{\mathrm{A}}$
Since the wavelength $3094\ \mathring{\mathrm{A}}$ 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

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.