• # question_answer When a metallic surface is illuminated with monochromatic light of wavelength$\lambda$, stopping potential for photoelectric current is $3{{V}_{0}}$, When the same metallic surface is illuminated with a light of wavelength$2\lambda$, the stopping potential is${{V}_{0}}$. The threshold wavelength for the surface is A) $6\lambda$                 B)   $4\lambda$C) $4\lambda /3$              D) $8\lambda$

Einstein?s photoelectric equation, $\frac{hc}{\lambda }=e(3{{V}_{0}})+W\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......(i)$ $\frac{hc}{2\lambda }=e({{V}_{0}})+W\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......(i)$ Solving these equations, we get; $\frac{hc}{\lambda }=\frac{3hc}{2\lambda }+W-3W$ $or\,\,\,\,\,\,2W=\frac{hc}{2\lambda }\,\,\,\,or\,\,\,\,\frac{hc}{{{\lambda }_{0}}}=\frac{hc}{4\lambda }$ So, ${{\lambda }_{0}}\,=\,\,4\lambda$