Question

The stopping potential for a metallic surface illuminated by monochromatic light of wavelength $\lambda$ is $4V_0$ while for another light of wavelength $3\lambda$ it is $V_0$ . The threshold wavelength of the surface for photoelectric emission is

• A. $\lambda$
• B. $3\lambda$
• C. $9\lambda$
• D. $\frac{\lambda }{9}$

Answer 1

Answer: (C)

According to Einsteins photoelectric equation
$K_{max}=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}$
where $\lambda$ is the wavelength of incident light and ${\lambda }_0$ is the threshold wavelength
But $K_{max}=e{V}_0$ where $V_0$ is the stopping potential
$\therefore e{V}_0=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}$
As per question
$e\left(4V_0\right)=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}=hc\left(\frac{1}{\lambda }-\frac{1}{{\lambda }_0}\right)\dots \left(i\right)$
and $e{V}_0=\frac{hc}{3\lambda }-\frac{hc}{{\lambda }_0}=hc\left(\frac{1}{3\lambda }-\frac{1}{{\lambda }_0}\right)\dots \left(ii\right)$
Dividing eqn. $\left(i\right)$ by eqn. $\left(ii\right)$, we get
$4=\frac{\frac{1}{\lambda }-\frac{1}{{\lambda }_0}}{\frac{1}{3\lambda }-\frac{1}{{\lambda }_0}}$
or $4\left(\frac{1}{3\lambda }-\frac{1}{{\lambda }_0}\right)=\left(\frac{1}{\lambda }-\frac{1}{{\lambda }_0}\right)$
or $\frac{4}{3\lambda }-\frac{4}{{\lambda }_0}=\frac{1}{\lambda }-\frac{1}{{\lambda }_0}$
or $\frac{4}{3\lambda }-\frac{1}{\lambda }=\frac{4}{{\lambda }_0}-\frac{1}{{\lambda }_0}=\frac{3}{{\lambda }_0}$
or $\frac{1}{3\lambda }=\frac{3}{{\lambda }_0}$ or ${\lambda }_0=9\lambda$