1. Use Einstein’s photoelectric equation

For incident frequency $\nu$, the maximum kinetic energy is

$K_{\max}=h(\nu-\nu_0)$

and the stopping potential $V_s$ satisfies

$K_{\max}=e|V_s|$

Here, only the magnitude matters because stopping potential is given as negative.

2. Write equations for the two cases

Case 1: frequency $\nu$

Stopping potential is $-\dfrac{V_0}{2}$, so magnitude is $\dfrac{V_0}{2}$.

Hence,

$e\frac{V_0}{2}=h(\nu-\nu_0) \qquad (1)$

Case 2: frequency $\dfrac{\nu}{2}$

Stopping potential is $-V_0$, so magnitude is $V_0$.

Hence,

$eV_0=h\left(\frac{\nu}{2}-\nu_0\right) \qquad (2)$

3. Solve the two equations

From (1):

$h(\nu-\nu_0)=\frac{eV_0}{2}$

From (2):

$h\left(\frac{\nu}{2}-\nu_0\right)=eV_0$

Multiply (1) by $2$:

$2h(\nu-\nu_0)=eV_0 \qquad (3)$

Now compare (2) and (3):

$2h(\nu-\nu_0)=h\left(\frac{\nu}{2}-\nu_0\right)$

Divide by $h$:

$2(\nu-\nu_0)=\frac{\nu}{2}-\nu_0$

Expand:

$2\nu-2\nu_0=\frac{\nu}{2}-\nu_0$

Bring terms together:

$2\nu-\frac{\nu}{2}=2\nu_0-\nu_0$

$\frac{3\nu}{2}=\nu_0$

So the threshold frequency is

$\boxed{\nu_0=\frac{3\nu}{2}}$

4. Check with options

Option C is

$\frac{3\nu}{2}$

So the correct option is C.

5. Comparison with stored answer

Stored correct answer: C

My derived answer: C

They agree.