When a certain photosensitive surface is illuminated with monochromatic light of frequency \nu , the stopping potential for the photo current is - \frac{V_{0}}{2} . When the surface is illuminated by monochromatic light of frequency \frac{\nu}{2} , the stopping potential is - V_{0} . The threshold frequency for photoelectic emission is :  

Answers (1)

For v frequency stoping potential is \frac{-v_{0}}{2}

For \frac{\nu }{2} frequency stoping potential is -v0

So for \nu frequency

h\nu = w+\frac{v_{0}}{2}e........1

For  \frac{\nu }{2}  frequency

h\frac{\nu }{2} = w+v_{0}e........2

From 1 and 2

We get

w = \frac{3}{2}h\nu

w = hv_{0}= \frac{3}{2}hv

so

v_{0}= \frac{3}{2}v

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