Question

When a surface 1 cm thick is illuminated with light of wave length $\lambda$ the stopping potential is $V_0$ ,but when the same surface is illuminated by light of wavelength 3$\lambda$ , the stopping potential is $\frac{V_0}{6}$. The threshold wavelength for metallic surface is

• A. 4$\lambda$
• B. 5$\lambda$
• C. 3$\lambda$
• D. 2$\lambda$

Answer 1

Answer: (B)

5$\lambda$

$V_0=\frac{hc}{e}(\frac{1}{\lambda }-\frac{1}{{\lambda }_0})$ ( l )
$\frac{V_0}{6}=\frac{hc}{e}(\frac{1}{3\lambda }-\frac{1}{{\lambda }_0})$ (ll)
Now equation ( I ) $÷$ equation ( l l )
$6=\frac{3({\lambda }_0-\lambda )}{{\lambda }_0-3\lambda }$
$6{\lambda }_0-18\lambda =3{\lambda }_0-3\lambda$
${\lambda }_0=5\lambda$

So, the answer is option (B).