Q. The work function of a metal is $4.2\ eV,$ its threshold wavelength will be
A) $4000\ \mathring{\mathrm{A}}$
B) $3500\ \mathring{\mathrm{A}}$
C) $2955\ \mathring{\mathrm{A}}$
D) $2500\ \mathring{\mathrm{A}}$
Explanation
(c) $W_{0}(eV) = \frac{12375}{\lambda_{0}} \Rightarrow \lambda_{0} = \frac{12375}{4.2} = 2945\ \mathring{\mathrm{A}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.