Question

When 1 cm 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 $4\lambda$

According to the question,

$ev=hc(\frac{1}{\lambda }-\frac{1}{{\lambda }_0})$ ...(i)

$\frac{ev}{3}=hc(\frac{1}{2\lambda }-\frac{1}{{\lambda }_0})$ ... (ii)

Dividing Eq (i) by Eq (ii), we get

$3=\frac{(\frac{1}{\lambda }-\frac{1}{{\lambda }_0})}{(\frac{1}{2\lambda }-\frac{1}{{\lambda }_0})}$

or $3=(\frac{1}{2\lambda }-\frac{1}{{\lambda }_0})=\frac{1}{\lambda }-\frac{1}{{\lambda }_0}$

$\frac{3}{2\lambda }-\frac{1}{\lambda }=\frac{3}{{\lambda }_0}-\frac{1}{{\lambda }_0}$

$\frac{1}{2\lambda }=\frac{2}{{\lambda }_0}$

Threshold wavelength for metallic surface

${\lambda }_0=4\lambda$