Question

When a certain photosensitive surface is illuminated with a monochromatic light of frequency $\nu,$ the stopping potential of 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 photoelectric emission is

• A. $\frac{5 ν​}{3}$
• B. $\frac{3 \nu}{2}$
• C. $\frac{4}{3}\nu$
• D. $2\nu$

Answer 1

Answer: (B)

$hν​=W+\frac{V_{0}}{2}e$
$\frac{h ν​}{2}=W+V_{0}e$
On solving, we get, $W=\frac{3}{2}h \, ν​$
$\Rightarrow hν​_{0}=\frac{3}{2}h \, ν​$
$\Rightarrow ν​_{0}=\frac{3}{2}ν​$