When electromagnetic radiation of wavelength 396∘A is incident on a metal sheet that has a work function of 15.6 eV, find the maximum kinetic energy of the photoelectrons emitted. Take h=6.6×10−34Js
A
15.65eV
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
10.48eV
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
9.65eV
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
16.48eV
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A15.65eV Given λ = 396∘A=396×10−10 m From photoelectric effect, hν=W+(KE)max ⇒hcλ=W+(K.E)max where λ = Wavelength of incident radiation W = Work function of the metal sheet (KE)max = Kinetic energy of the emitted photoelectron. So, 6.6×10−34×3×108396×10−10=15.6×1.6×10−19+(KE)max ⇒K.Emax=(0.05×10−16−24.96×10−19)J ⇒K.Emax=25.04×10−19J=15.65eV