Question

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

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

Answer 1

Answer: (B)

Let ${\lambda }_0$ be threshold wavelength,
$e{V}_0=hc\left(\frac{1}{\lambda }-\frac{1}{{\left(\lambda \right)}_0}\right)$
Also,
$e\frac{V_0}{6}=hc\left(\frac{1}{3\lambda }-\frac{1}{{\left(\lambda \right)}_0}\right)$
$\therefore \frac{1}{6}\left(\frac{1}{\lambda }-\frac{1}{{\left(\lambda \right)}_0}\right)=\left(\frac{1}{3\lambda }-\frac{1}{{\left(\lambda \right)}_0}\right)$
$\frac{1}{{\left(\lambda \right)}_0}\left(1-\frac{1}{6}\right)=\frac{1}{\lambda }\left(\frac{1}{3}-\frac{1}{6}\right)$
$\frac{5}{6{\lambda }_0}=\frac{1}{6\lambda }$
${\lambda }_0=5\lambda$