Q. Light of wavelength 4000 $\mathring{\mathrm{A}}$ incident on a sodium surface for which the threshold wavelength of photoelectrons is 5420 $\mathring{\mathrm{A}}$. The work function of sodium is
A) 0.57 eV
B) 1.14 eV
C) 2.29 eV
D) 4.58 eV
Explanation
(c) $\phi_{0} = hc/\lambda_{0}$ (in$\ eV$)
$= \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{5420 \times 10^{- 10} \times 1.6 \times 10^{- 19}} = 2.29\ eV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.