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 $K_{max}=eV_{0}$ where $V_{0}$ is the stopping potential
$\therefore eV_{0}=\frac{hc}{\lambda}-\frac{hc}{\lambda_{0}}$
As per question
$e \left(4 V_{0}\right)=\frac{hc}{\lambda}-\frac{hc}{\lambda_{0}}=hc \left(\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right) \ldots\left(i\right)$
and $eV_{0}=\frac{hc}{3\lambda}-\frac{hc}{\lambda_{0}}=hc \left(\frac{1}{3\lambda}-\frac{1}{\lambda_{0}}\right)\ldots\left(ii\right)$
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$