When radiation of wavelength '$$\lambda$$' is incident on a metallic surface, the stopping potential is 4.8 V. If the surface is illuminated with radiation of double the wavelength then the stopping potential becomes $$1.6 \mathrm{~V}$$. The threshold wavelength for the surface is

The de-Broglie wavelength $$(\lambda)$$ of a particle is related to its kinetic energy (E) as

Dual nature of light is exhibited by

When radiations of wavelength $$\lambda$$ is incident on a metallic surface the stopping potential required is $$4.8 \mathrm{~V}$$. If same surface is illuminated with radiations of double the wavelength, then required stopping potential becomes $$1.6 \mathrm{~V}$$, then the value of threshold wavelength for the surface is