Q. Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The de Broglie wavelength of the emitted electron is :
A) $\le 2.8 \times 10^{-12}$ m
B) $< 2.8 \times 10^{-10}$ m
C) $< 2.8 \times 10^{-9}$ m
D) $\ge 2.8 \times 10^{-9}$ m
Explanation
[Solution] Photon energy 2.48 eV, so $K_{max} = 0.20$ eV and $\lambda_{min} = \frac{h}{\sqrt{2mK}} \approx 2.8$ nm; slower electrons have longer wavelengths, so $\lambda \ge 2.8 \times 10^{-9}$ m. Answer (4).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.