When a light of a given wavelength falls on a metallic surface the stopping potential for photoelect

Question

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 .
$\left(\mathrm{h}=6.63 	imes 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{e}=1.6 	imes 10^{-19} \mathrm{C}ight.$,
$\left.c=3 	imes 10^8 \mathrm{~m} / \mathrm{s}ight)$

Professor PC Thomas Classes · Answer

By Einstein's photoelectric equation, $eV_s = \frac{hc}{\lambda} - \Phi$. For the first light: $3.2 e = \frac{hc}{\lambda} - \Phi$. For the second light: $0.7 e = \frac{hc}{2\lambda} - \Phi$. Subtracting the second equation from the first gives: $3.2 e - 0.7 e = \frac{hc}{\lambda} - \frac{hc}{2\lambda} \implies 2.5 e = \frac{hc}{2\lambda}$. Rearranging to solve for the wavelength $\lambda$ yields $\lambda = \frac{hc}{5e}$. Substituting the given constants: $\lambda = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}{5 	imes 1.6 	imes 10^{-19}} = \frac{19.89 	imes 10^{-26}}{8.0 	imes 10^{-19}} = 2.48625 	imes 10^{-7} 	ext{ m}$, which closely approximates to $2.5 	imes 10^{-7} 	ext{ m}$.