JEE Main 2019 — Physics solution

JEE Main2019Physics

When a certain photosensistive surface is illuminated with monochromatic light of frequency v, the stopping potential for the photo current is $- \frac{V _{0}}{2}$ . When the surface is illuminated by monochromatic light of frequency $\frac{v}{2}$ , the stopping potential is $- V_{0}$ . The threshold frequency for photoeletric emission is:

A
$\frac{3 v}{2}$
B
$\frac{4}{3} v$
C
$\frac{5 v}{3}$
D
$2 v$

Solution

$2 h v = 2 ϕ + e V_{0} \frac{h v}{2} = ϕ + e V_{0} \frac{3 h v}{2} = ϕ v_{0} = \frac{3 v}{2}$

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