wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A monochromatic light source of intensity 5 mW emits 8 × 1015 photons per second. This light ejects photoelectrons from a metal surface. The stopping potential for this setup is 2.0 V. Calculate the work function of the metal.

Open in App
Solution

Given:
Intensity of light, I = 5 mW
Number of photons emitted per second, n = 8 × 1015
Stopping potential, V0 = 2 V
Energy, E = hv = In = 5×10-38×1015
From Einstein's photoelectric equation, work function,
W0=hv-eV0
Here, h = Planck's constant
e = 1.6×10-19 C
On substituting the respective values, we get:
W0=5×10-38×1015-1.6×10-19×2 =6.25×10-19-3.2×10-19 =3.05×10-19 =3.05×10-191.6×10-15=1.906 eV

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Work Function
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon