When radiation of wavelength '$\lambda$' is incident on a metallic surface, the stopping potential is $4.8 \text{ V}$. If the surface is illuminated with radiation of double the wavelength then the stopping potential becomes $1.6 \text{ V}$. The threshold wavelength for the surface is
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Use the equation $eV_s = hc(\frac{1}{\lambda} - \frac{1}{\lambda_0})$ for two cases and divide or substitute to eliminate constants like $hc$ and $e$.