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.
$$(h= 6.63\times10^{-34}J.s,e=1.6\times10^{-19}C,c=3\times10^{8}m/s)$$

We need to find the wavelength of the first light given photoelectric stopping potentials. For light of wavelength $$\lambda$$ the stopping potential is $$V_1 = 3.2$$ V, and for light of wavelength $$2\lambda$$ the stopping potential is $$V_2 = 0.7$$ V.

According to Einstein's photoelectric equation, $$eV = \frac{hc}{\lambda} - \phi.$$ Applying this to the first light gives $$eV_1 = \frac{hc}{\lambda} - \phi \quad \cdots (1)$$ and to the second light gives $$eV_2 = \frac{hc}{2\lambda} - \phi \quad \cdots (2).$$

Subtracting equation (2) from (1) yields $$e(V_1 - V_2) = \frac{hc}{\lambda} - \frac{hc}{2\lambda} = \frac{hc}{2\lambda},$$ which leads to $$\lambda = \frac{hc}{2e(V_1 - V_2)}.$$

Substituting the numerical values, we get $$\lambda = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{2 \times 1.6 \times 10^{-19} \times (3.2 - 0.7)}$$ $$= \frac{19.89 \times 10^{-26}}{2 \times 1.6 \times 10^{-19} \times 2.5}$$ $$= \frac{19.89 \times 10^{-26}}{8 \times 10^{-19}} = 2.486 \times 10^{-7} \text{ m} \approx 2.5 \times 10^{-7} \text{ m}.$$

Therefore, the wavelength is Option 2: $$2.5 \times 10^{-7}$$ m.