When a metal surface is illuminated by light of wavelength $$\lambda$$, the stopping potential is $$8$$ V. When the same surface is illuminated by light of wavelength $$3\lambda$$, stopping potential is $$2$$ V. The threshold wavelength for this surface is :

Photoelectric effect: $$eV_s = \frac{hc}{\lambda} - \phi$$.

$$8e = \frac{hc}{\lambda} - \phi$$ ... (1). $$2e = \frac{hc}{3\lambda} - \phi$$ ... (2).

From (1)-(2): $$6e = \frac{hc}{\lambda}(1-1/3) = \frac{2hc}{3\lambda} \Rightarrow \frac{hc}{\lambda} = 9e$$.

From (1): $$\phi = 9e - 8e = e$$. Threshold: $$\frac{hc}{\lambda_0} = \phi = e$$.

$$\lambda_0 = \frac{hc}{e} = 9\lambda \cdot \frac{e}{e} = 9\lambda$$.

The answer is Option (3): $$9\lambda$$.