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Question: The threshold frequency of a certain metal is3.3×10<sup>14</sup>Hz. If light of frequency 8.2×10<sup...

The threshold frequency of a certain metal is3.3×1014Hz. If light of frequency 8.2×1024 Hz is incident on the metal, then the cut off voltage for photoelectric emission is (Given h=6.6310-34 j s)

A

2V

B

4V

C

6V

D

8V

Answer

2V

Explanation

Solution

: Given threshold frequency

Frequency of incident light

As eV0=h(v−v0)e V _ { 0 } = h \left( v - v _ { 0 } \right)

Or V0=h(v−v0)eV _ { 0 } = \frac { h \left( v - v _ { 0 } \right) } { e }