Question 11.8:
The threshold frequency for a certain metal is 3.3 x 1014 Hz. If the light of frequency 8.2x1014 Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.
 

Hint: Use Einstein's photoelectric equation.
Step 1: Find the cut-off voltage.

Threshold frequency of the metal, \(\nu_{0}\) = 3.3 x 1014 Hz
Frequency of light incident on the metal, \(\nu\) = 8.2 x 1014 Hz
Charge on an electron, e = 1.6 x 10-19 C
Planck's constant, h = 6.626 x 10-34 Js
Cut-off voltage for the photoelectric emission from the metal = \(V_{0}\)
The equation for the cut-off energy is given as:
eV0=h(vv0)V0=h(vv0)e=6.626×1034×(8.2×10143.3×1014)1.6×1019=2.0292  V.