### JEE Main 12 Jan 2019 Evening Question 22

##### Question: When a certain photosensitive surface is illuminated with monochromatic light of frequency u, the stopping potential for the photo current is $ \frac{-V _0}{2}. $ When the surface is illuminated by monochromatic light of frequency $ \frac{\upsilon }{2}, $ the stopping potential is $ -V _0. $ The threshold frequency for photoelectric emission is [JEE Main Online Paper Held On 12-Jan-2019 Evening]

#### Options:

A) $ \frac{4\upsilon }{3} $

B) $ 2\upsilon $

C) $ \frac{5\upsilon }{3} $

D) $ \frac{3\upsilon }{2} $

## Show Answer

#### Answer:

Correct Answer: D

#### Solution:

Einstein’s photoelectric equation in the two cases is given by $ \frac{eV _0}{2}=h\upsilon -h{\upsilon_0} $ …(i)

and $ eV _0=\frac{h\upsilon }{2}-h{\upsilon_0} $ -(ii)

From eqn. (i) and (ii), $ \frac{1}{2}=\frac{h\upsilon -h{\upsilon_0}}{h\upsilon /2-h{\upsilon_0}} $

$\Rightarrow {\upsilon_0}=\frac{3}{2}\upsilon $