Q. In photoelectric effect an em-wave is incident on a metal surface and electrons are ejected from the surface. If the work function of the metal is 2.14 eV and stopping potential is 2 V, what is the wavelength of the em-wave ? (Given $hc=1242\,\mathrm{eV\,nm}$ where $h$ is the Planck's constant and $c$ is the speed of light in vaccum.)
A) 300 nm
B) 400 nm
C) 600 nm
D) 200 nm
Explanation
[Solution by Dr. Abubacker] $h\nu=\phi+eV_0=2.14+2=4.14$ eV, so $\lambda=\frac{1242}{4.14}=300$ nm. Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.