Q. The work function of a surface of a photosensitive material is 6.2 eV. The wavelength of the incident radiation for which the stopping potential is 5 V lies in the
A) X-ray region
B) Ultraviolet region
C) Visible region
D) Infrared region
Explanation
[Solution from source] Photon energy = φ + eV₀ = 6.2 + 5 = 11.2 eV, so λ = 12400/11.2 eV·Å ≈ 1100 Å (about 110 nm). This lies in the ultraviolet; visible light is only about 1.8 to 3.1 eV, and X-rays need wavelengths below about 100 Å. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.