A certain metallic surface is illuminated by monochromatic radiation of wavelength $$\lambda$$. The stopping potential for photoelectric current for this radiation is $$3V_0$$. If the same surface is illuminated with a radiation of wavelength $$2\lambda$$, the stopping potential is $$V_0$$. The threshold wavelength of this surface for photoelectric effect is ___ $$\lambda$$.

Using the photoelectric equation $$eV_s = \frac{hc}{\lambda} - \phi$$, where $$\phi = \frac{hc}{\lambda_0}$$ is the work function and $$\lambda_0$$ is the threshold wavelength.

For radiation of wavelength $$\lambda$$ with stopping potential $$3V_0$$:

$$e(3V_0) = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} \quad \cdots (1)$$

For radiation of wavelength $$2\lambda$$ with stopping potential $$V_0$$:

$$e(V_0) = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0} \quad \cdots (2)$$

Subtracting equation (2) from equation (1):

$$2eV_0 = \frac{hc}{\lambda} - \frac{hc}{2\lambda} = \frac{hc}{2\lambda}$$

So $$eV_0 = \frac{hc}{4\lambda}$$.

Substituting back into equation (2):

$$\frac{hc}{4\lambda} = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0}$$

$$\frac{hc}{\lambda_0} = \frac{hc}{2\lambda} - \frac{hc}{4\lambda} = \frac{hc}{4\lambda}$$

$$\lambda_0 = 4\lambda$$

The threshold wavelength is $$4\lambda$$, so the answer is $$4$$.