Question

When monochromatic light of wavelength '$\lambda$' is incident on a metallic surface, the stopping potential for photoelectric current is '$3V_0$'. When same surface is illuminated with light of wavelength '$2\lambda$', the stopping potentials is '$V_0$. The threshold wavelength for this surface when photoelectric effect takes place is

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

Answer 1

The correct option is D $4\lambda$

Stopping potential $eV=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_{th}}$

where ${\lambda }_{th}$ is the threshold wavelength of the surface.

Incident light of wavelength $\lambda$ leads to the stopping potential $3V_o$.

$\therefore$ $e\left(3V_o\right)=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_{th}}$ ....(1)

Incident light of wavelength $2\lambda$ leads to the stopping potential $V_o$.

$\therefore$ $e\left(V_o\right)=\frac{hc}{2\lambda }-\frac{hc}{{\lambda }_{th}}$ ....(2)

Subtracting (2) from (1) we get $3e{V}_o-e{V}_o=\frac{hc}{\lambda }-\frac{hc}{2\lambda }$

$⟹$ $e{V}_o=\frac{hc}{4\lambda }$

Putting this in equation (2) we get $\frac{hc}{4\lambda }=\frac{hc}{2\lambda }-\frac{hc}{{\lambda }_{th}}$

Or $\frac{1}{{\lambda }_{th}}=\frac{1}{2\lambda }-\frac{1}{4\lambda }$

$⟹$ ${\lambda }_{th}=4\lambda$