Question

When a surface is illuminated with light of wavelength , 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. 4/3
• B. 4
• C. 6
• D. 8/3

Answer 1

The correct option is B

4

$eV=hc(\frac{1}{\lambda }-\frac{1}{{\lambda }_0})\cdots \left(1\right)\frac{eV}{3}=hc(\frac{1}{2\lambda }-\frac{1}{{\lambda }_0})\cdots \left(2\right)\text{Dividing eq.(1) by (2), we get;}3=\frac{(\frac{1}{\lambda }-\frac{1}{{\lambda }_0})}{(\frac{1}{2\lambda }-\frac{1}{{\lambda }_0})}Or3=(\frac{1}{2\lambda }-\frac{1}{{\lambda }_0})=\frac{1}{\lambda }-\frac{1}{{\lambda }_0}or={\lambda }_0=4\lambda$