Question

When 1 cm thick surface is illuminated with light of wavelength $\lambda$ , the stopping potential obtained as $V_0$ But when the same surface is illuminated by light of wavelength $3\lambda$ , the stopping potential obtained is $\frac{V_0}{6}$ .The threshold wavelength for this metal surface is

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

Answer 1

The correct option is D $5\lambda$

Stopping potential $e{V}_s=\frac{hc}{{\lambda }_{in}}-\frac{hc}{{\lambda }_{th}}$
where ${\lambda }_{in}$ and ${\lambda }_{th}$ are the wavelength of incident photon and threshold wavelength of the surface respectively.
Case 1 : ${\lambda }_{in}=\lambda$ $V_s=V_o$
$\therefore$ $e{V}_o=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_{th}}$ ....(1)
Case 2 : ${\lambda }_{in}=3\lambda$ $V_s=V_o/6$
$\therefore$ $e\frac{V_o}{6}=\frac{hc}{3\lambda }-\frac{hc}{{\lambda }_{th}}$
Or $e{V}_o=\frac{2hc}{\lambda }-\frac{6hc}{{\lambda }_{th}}$ ....(2)
Equating (1) and (2), $\frac{hc}{\lambda }-\frac{hc}{{\lambda }_{th}}$ $=\frac{2hc}{\lambda }-\frac{6hc}{{\lambda }_{th}}$
Or $\frac{5hc}{{\lambda }_{th}}=\frac{hc}{\lambda }$
$⟹{\lambda }_{th}=5\lambda$