When a certain photosensitive surface is illuminated with a monochromatic light of frequency ν, the stopping potential of the photo current is -(V0/2). When the surface is illuminated by monochromatic light of frequency (ν/2), the stopping potential is -V0. The threshold frequency for photoelectric emission is

Question

When a certain photosensitive surface is illuminated with a monochromatic light of frequency ν, the stopping potential of the photo current is -(V0/2). When the surface is illuminated by monochromatic light of frequency (ν/2), the stopping potential is -V0. The threshold frequency for photoelectric emission is (A) (5 Î½â/3) (B) (3 ν/2) (C) (4/3)ν (D) 2ν

Tardigrade Admin · Accepted Answer

hÎ½â=W+(V0/2)e (h Î½â/2)=W+V0e On solving, we get, W=(3/2)h Î½â ⇒ hÎ½â0=(3/2)h Î½â ⇒ Î½â0=(3/2)Î½â

$\frac{5 ν }{3}$

$\frac{3 ν}{2}$

$\frac{4}{3} ν$

$2 ν$

$h ν = W + \frac{V _{0}}{2} e$
$\frac{h ν }{2} = W + V_{0} e$
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 $

35ν​​

23ν​

34​ν

2ν

hν​=W+2V0​​e 2hν​​=W+V0​e On solving, we get, W=23​hν​ ⇒hν​0​=23​hν​ ⇒ν​0​=23​ν​

$\frac{5 ν }{3}$

$\frac{3 ν}{2}$

$\frac{4}{3} ν$

$2 ν$

$h ν = W + \frac{V _{0}}{2} e$
$\frac{h ν }{2} = W + V_{0} e$
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 $