A metallic surface is illuminated with radiation of wavelength $$\lambda$$, the stopping potential is $$V_0$$. If the same surface is illuminated with radiation of wavelength $$2\lambda$$, the stopping potential becomes $$\frac{V_0}{4}$$. The threshold wavelength for this metallic surface will be

Using Einstein's photoelectric equation for the two given wavelengths:

$$\frac{hc}{\lambda} = \phi + eV_0 \quad \text{...(i)}$$

$$\frac{hc}{2\lambda} = \phi + \frac{eV_0}{4} \quad \text{...(ii)}$$

From equation (i), we get $$eV_0 = \frac{hc}{\lambda} - \phi$$. Substituting into equation (ii):

$$\frac{hc}{2\lambda} = \phi + \frac{1}{4}\left(\frac{hc}{\lambda} - \phi\right)$$

$$\frac{hc}{2\lambda} = \phi + \frac{hc}{4\lambda} - \frac{\phi}{4}$$

$$\frac{hc}{2\lambda} = \frac{3\phi}{4} + \frac{hc}{4\lambda}$$

Now, solving for $$\phi$$:

$$\frac{hc}{2\lambda} - \frac{hc}{4\lambda} = \frac{3\phi}{4}$$

$$\frac{hc}{4\lambda} = \frac{3\phi}{4}$$

$$\phi = \frac{hc}{3\lambda}$$

Since the work function is also $$\phi = \frac{hc}{\lambda_0}$$ (where $$\lambda_0$$ is the threshold wavelength), we get:

$$\frac{hc}{3\lambda} = \frac{hc}{\lambda_0}$$

$$\lambda_0 = 3\lambda$$

So, the answer is $$3\lambda$$.