1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# The electric field associated with a light wave is E=E0sin(1.57 ×107(x−ct)), where x is in meter and t is in second. If this light is used to produce photoelectric emission, from the surface of a metal has a work function 1.9 eV, then the stopping potential will be ?

A
1.2 V
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
1.5 V
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
1.75 V
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
1.9 V
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is A 1.2 VComparing the the given expression for electric filed with the standard expression, E=E0sin(kx−ωt)we get k=1.57×107m−1 where k is propagation constant which is related to wavelength, λ as follows k=2πλ so putting value of k we get λ=4×10−7m=4000AngstromSo energy of incident photon will be E=hcλ=124004000ev=3.1eVOut of which 1.9eV will be consumed against work function,W=1.9eV so maximum kinetic energy KEmax=3.1eV−1.9eV=1.2eVSo the stopping potential will be V0=KEmaxcharge=1.2eVe=1.2V

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
UV rays
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program