Question

When 1 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

$4\lambda$

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