Question

The threshold frequency for a photo-sensitive metal is \( 3.3 \times 10^{14} \, \text{Hz} \). If light of frequency \( 8.2 \times 10^{14} \, \text{Hz} \) is incident on this metal, the cut-off voltage for the photo-electric emission is nearly

• A. 2V
• B. 3V
• C. 5V
• D. 1V

Hint:

The cut-off voltage can be found from the photoelectric equation \( E = h \nu - W \).

Answer 1

The Correct Option is B

Using the photoelectric equation \( E_{\text{photon}} = h \nu - W \), where \( \nu \) is the frequency, \( h \) is Planck's constant, and \( W \) is the work function of the metal, we can calculate the cut-off voltage. With the given frequency, the cut-off voltage comes out to be approximately 3V.

Step 2: Conclusion.
The cut-off voltage is 3V, corresponding to option (b).