Q. Light of wavelength $1824\ \mathring{\mathrm{A}},$ incident on the surface of a metal, produces photo-electrons with maximum energy $5.3\ eV.$ When light of wavelength $1216\ \mathring{\mathrm{A}}$ is used, the maximum energy of photoelectrons is $8.7\ eV.$ The work function of the metal surface is
A) $3.5\ eV$
B) $13.6\ eV$
C) $6.8\ eV$
D) $1.5\ eV$
Explanation
(d) $E = W_{0} + K_{\max}.$ From the given data $E$ is $6.78\ eV$ (for $\lambda = 1824\ \mathring{\mathrm{A}}$) or $10.17\ eV$ [for $\lambda = 1216\ \mathring{\mathrm{A}}$]
$\therefore W_{0} = E - K_{\max} = 6.78 - 5.3 = 1.48\ eV$
or
$W_{0} = 10.17 - 8.7 = 1.47\ eV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.