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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 :
Option 1)
2 $\nu$
Option 2)
$\frac{3\nu}{2}$
Option 3)
$\frac{4\nu}{3}$
Option 4)
$\frac{5\nu}{3}$

Answers (1)

best_answer

Stopping Potential /Cut-off Potential -

It is defined as the potential necessary to stop any electron from reaching the other side.

-

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

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

Option 1)

2 $\nu$

Option 2)
$\frac{3\nu}{2}$
Option 3)

$\frac{4\nu}{3}$

Option 4)

$\frac{5\nu}{3}$

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