When a certain photosensitive surface is illuminated with 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

First, we recall the Einstein photoelectric equation, which links the stopping potential $$V_s$$ with the frequency of the incident radiation:

$$e\,V_s = h\nu - h\nu_0,$$

where $$e$$ is the magnitude of electronic charge, $$h$$ is Planck’s constant, $$\nu$$ is the frequency of the light used, and $$\nu_0$$ is the threshold frequency of the surface.

The problem gives two experimental situations. In the first, the stopping potential is reported as $$-\dfrac{V_0}{2}$$ when the surface is illuminated with monochromatic light of frequency $$\nu$$. Placing this directly into the formula, we write

$$e\left(-\dfrac{V_0}{2}\right)=h\nu - h\nu_0.$$

Simplifying the left‐hand side gives

$$-\dfrac{eV_0}{2}=h\nu-h\nu_0.\qquad(1)$$

In the second experiment the frequency is halved to $$\dfrac{\nu}{2}$$ and the stopping potential becomes $$-V_0$$. Substituting these values into the photoelectric equation, we get

$$e(-V_0)=h\left(\dfrac{\nu}{2}\right)-h\nu_0,$$

or

$$-eV_0=\dfrac{h\nu}{2}-h\nu_0.\qquad(2)$$

Now we possess two simultaneous equations, (1) and (2). Our next task is to eliminate $$\nu_0$$ by subtracting one equation from the other. Subtracting (2) from (1) yields

$$\left(-\dfrac{eV_0}{2}\right)-(-eV_0)=\bigl(h\nu-h\nu_0\bigr)-\bigl(\dfrac{h\nu}{2}-h\nu_0\bigr).$$

The left side simplifies to

$$-\dfrac{eV_0}{2}+eV_0=\dfrac{eV_0}{2},$$

while the right side simplifies to

$$h\nu-h\nu_0-\dfrac{h\nu}{2}+h\nu_0=h\nu-\dfrac{h\nu}{2}=\dfrac{h\nu}{2}.$$

Equating the simplified results gives us

$$\dfrac{eV_0}{2}=\dfrac{h\nu}{2}\quad\Longrightarrow\quad eV_0=h\nu.\qquad(3)$$

This relation directly connects $$V_0$$ with $$\nu$$. We now substitute (3) back into either of our earlier equations to solve for the threshold frequency. Using equation (1):

$$-\dfrac{eV_0}{2}=h\nu-h\nu_0.$$

Replacing $$h\nu$$ by $$eV_0$$ from (3):

$$-\dfrac{eV_0}{2}=eV_0-h\nu_0.$$

Moving terms involving $$eV_0$$ to the left gives

$$h\nu_0=eV_0+\dfrac{eV_0}{2}=\dfrac{3eV_0}{2}.$$

Once more inserting $$eV_0=h\nu$$ from (3):

$$h\nu_0=\dfrac{3}{2}h\nu,$$

and dividing both sides by $$h$$ finally produces

$$\nu_0=\dfrac{3\nu}{2}.$$

Therefore the threshold frequency is $$\dfrac{3\nu}{2}$$, which matches Option B.

Hence, the correct answer is Option B.