Question

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

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

Answer 1

The correct option is D $\frac{3\nu }{2}$

As we know that,

$h\nu =\varphi +eV$

From questions,

$h\nu =\varphi +(-e{V}_0/2)$, and

$h\nu /2=\varphi +(-e{V}_0)$

From both equations, we get

$\varphi =\frac{3}{2}h\nu$

So, the threshold frequency will be $\frac{3}{2}\nu$.

Hence, option $\left(D\right)$ is correct.