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\\
(1) $3\lambda$\\
(2) $4\lambda$\\
(3) $\frac{3}{2}\lambda$\\
(4) $\frac{\lambda}{4}$