When a light of a given wavelength falls on a metallic surface the stopping potential for photoelectrons is 3.2 V . If a second light having wavelength twice of first light is used, the stopping potential drops to 0.7 V . The wavelength of first light is $\_\_\_\_$ m. $$ \left(\mathrm{h}=6.63 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{e}=1.6 \times 10^{-19} \mathrm{C}, \mathrm{c}=3 \times 10^8 \mathrm{~m} / \mathrm{s}\right) $$