Q. The photoelectric threshold wavelength for a metal surface is 6600$\mathring{\mathrm{A}}$. The work function for this metal is
A) 0.87 eV
B) 1.87 eV
C) 18.7 eV
D) 0.18 eV
Explanation
(b) $\phi_{0} = \frac{hc}{e\lambda_{0}}\ (in\ eV) = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{6600 \times 10^{- 10} \times 1.6 \times 10^{- 19}} = 1.87\ eV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.