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					 The threshold frequency for a photosensitive metal is 3.3 × 1014 Hz. If light of frequency 8.2 × 1014 Hz is incident on this metal, the cut-off voltage for the photoelectric emission is nearly
 
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- 2 V
 - 3 V 
 - 5 V
 - 1 V
 
 
Correct Option: A
K.E. = hν – hνth = eV0   (V0 = cut off voltage)
| ⇒ V0 = | (8.2 × 1014 - 3.3× 1014) | e | 
| = | ≈ 2V. | 1.6 × 10-19 |