Question

When one centimeter thick surface is illuminated with light of wavelength $\lambda$, the stopping potential is V. When the same surface is illuminated by light of wavelength 2$\lambda$, the stopping potential is $\frac{V}{3}$ . Threshold wavelength for metallic surface is :

• A. $\frac{4\lambda }{3}$
• B. $4\lambda$
• C. $6\lambda$
• D. $\frac{8\lambda }{3}$

Answer 1

The correct option is B $4\lambda$

${\text{KE}}_{\text{max}}={\text{E}}_{\text{incident}}-\varphi \text{eV}=\frac{\text{hc}}{\lambda }-\varphi \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)\frac{\text{ev}}{3}=\frac{\text{hc}}{2\lambda }-\varphi \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(2\right)\text{Divide}\left(2\right)\text{by}\left(1\right),\text{we get}\frac{1}{3}=\frac{(\frac{\text{hc}}{2\lambda }-\varphi )}{(\frac{\text{hc}}{\lambda }-\varphi )}\Rightarrow \varphi =\frac{\text{hc}}{4\lambda }=\frac{\text{hc}}{{\lambda }_0}\therefore {\lambda }_0=4\lambda$