Question

When light of wavelength '$\lambda$' is incident on photosensitive surface, the stopping potential is ${}^{\prime }{V}^{\prime }$. When light of wavelength '$3\lambda$' is incident on same surface, the stopping potential is $\frac{{}^{\prime }{V}^{\prime }}{6}$. Threshold wavelength for the 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, $eV=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_{th}}$ ........(1)

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

Now a light of wavelength $3\lambda$ is incident on the surface which leads to the stopping potential to be $\frac{V}{6}$.

$\therefore$ $e\frac{V}{6}=\frac{hc}{3\lambda }-\frac{hc}{{\lambda }_{th}}$ .......(2)

Equation (1) - equation (1) gives $eV-e\frac{V}{6}=\frac{hc}{\lambda }-\frac{hc}{3\lambda }$

$⟹$ $eV=\frac{4}{5}\frac{hc}{\lambda }$

Putting this in equation (1) we get $\frac{4}{5}\frac{hc}{\lambda }=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_{th}}$

Or $\frac{1}{{\lambda }_{th}}=\frac{1}{\lambda }-\frac{4}{5\lambda }$

$⟹$ ${\lambda }_{th}=5\lambda$