Q. Given that a photon of light of wavelength 10,000 $\mathring{\mathrm{A}}$ has an energy equal to 1.23 eV. When light of wavelength 5000 $\mathring{\mathrm{A}}$ and intensity $I_{0}$ falls on a photoelectric cell, the surface current is $0.40 \times 10^{- 6}$ A and the stopping potential is 1.36 V, then the work function is
A) 0.43 eV
B) 0.55 eV
C) 1.10 eV
D) 1.53 eV
Explanation
(c) $E = \frac{hc}{\lambda}$ or $E \propto \frac{1}{\lambda}\ \therefore\ \frac{E_{2}}{E_{1}} = \frac{\lambda_{1}}{\lambda_{2}}$
or $E_{2} = E_{1} \times \frac{\lambda_{1}}{\lambda_{2}} = 1.23 \times \frac{10,000}{5,000} = 2.46\ eV$
Now, $hv - \phi_{0} = \frac{1}{2}mv_{\max}^{2} = eV_{s}$
or $\phi_{0} = hv_{2} - eV_{s} = E_{2} - eV_{s}$
$= 2.46 - 1.36 = 1.10\ eV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.