Question

When a piece of metal is illuminated by a monochromatic light of wavelength $\lambda$, then stopping potential is $3V$. When same surface is illuminated by light of wavelength $2\lambda$, then stopping potential becomes $V$. The value of threshold wavelength for photoelectric emission will be

• A. $4\lambda$
• B. $8\lambda$
• C. $\frac{4}{3}\lambda$
• D. $6\lambda$

Answer 1

The correct option is A $4\lambda$

Stopping potential is $3V_s$ when a light of wavelength $\lambda$ is incident on the metal surface.

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

where ${\lambda }_{th}$ is the threshold wavelength for photoelectric emission.

Stopping potential is $V_s$ when a light of wavelength $2\lambda$ is incident on the metal surface.

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

From (1) and (2) we get, $3(\frac{hc}{2\lambda }-\frac{hc}{{\lambda }_{th}})=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_{th}}$

Or $3(\frac{1}{2\lambda }-\frac{1}{{\lambda }_{th}})=\frac{1}{\lambda }-\frac{1}{{\lambda }_{th}}$

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

$⟹$ ${\lambda }_{th}=4\lambda$