When a certain photosensitive surface is illuminated with a monochromatic light of frequency ν, the stopping potential of the photo current is -V_0/2. When the surface is illuminated by monochromatic light of frequency (ν)/2, the stopping potential is -V_0. The threshold frequency for photoelectric emission is

Correct answer: (B) 3v/2. hν=W+V_0/2e (hν)/2=W+V_0e On solving, we get, W=3/2h ν ⇒hν_0=3/2h ν ⇒ν_0=3/2ν

Q63 / 90 · JEE Main 2019 (12 Jan Shift 2) · Physics

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

h ν​ = W + \frac{V_{0}}{2} e

\frac{h ν​}{2} = W + V_{0} e

On solving, we get,

W = \frac{3}{2} h ν​

\Rightarrow h {ν​}_{0} = \frac{3}{2} h ν​

\Rightarrow {ν​}_{0} = \frac{3}{2} ν​

Hence, the correct answer is option: 'B'.