Q. For intensity $I$ of a light of wavelength $5000\mathring{\mathrm{A}}$ the photoelectron saturation current is $0.40\ \mu A$ and stopping potential is $1.36\ V,$ the work function of metal is
A) $2.47\ eV$
B) $1.36\ eV$
C) $1.10\ eV$
D) $0.43\ eV$
Explanation
(c) By using $E = W_{0} + K_{\max}$
$E = \frac{12375}{5000} = 2.475\ eV\ $and$\ K_{\max} = eV_{0} = 1.36\ eV$
So $2.475 = W_{0} + 1.36 \Rightarrow W_{0} = 1.1\ eV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.