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 .
$(\text{h} = 6.63\times10^{-34}\ \text{J.s}, \text{e} = 1.6\times10^{-19}\text{C}, \text{c} = 3\times10^8\ \text{m/s})$
Answer: (B) $2.5\times10^{-7}$
$$\frac{hc}{\lambda} = \phi + 3.2\ \text{eV},\qquad \frac{hc}{2\lambda} = \phi + 0.7\ \text{eV}$$
Subtracting: $\dfrac{hc}{2\lambda} = 2.5$ eV, so $\dfrac{hc}{\lambda} = 5$ eV.
$$\lambda = \frac{hc}{5e} = \frac{6.63\times10^{-34}\times3\times10^8}{5\times1.6\times10^{-19}} = 2.49\times10^{-7}\ \text{m}\approx2.5\times10^{-7}\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics