A certain metallic surface is illuminated with monochromatic light of wavelength \[\lambda \]. The stopping potential for photo-electric current for this light is \[3{{V}_{0}}\]. If the same surface is illuminated with light of wavelength \[2\lambda \], the stopping potential is \[{{V}_{0}}\]. The threshold wavelength for this surface for the photo-electric effect is:
\[\begin{align}
  & A)\dfrac{\lambda }{4} \\
 & B)\dfrac{\lambda }{6} \\
 & C)4\lambda \\
 & D)6\lambda \\
\end{align}\]

Hint: We are going to use Einstein’s photoelectric equation to get the solution for the question. This equation defines the maximum kinetic energy of a photon ejected from a metal surface when a light beam of threshold wavelength is incident on the surface. We shall compare both the given conditions in question to find the unknown term, which is threshold wavelength.

Formula used:
 \[\begin{align}
  & 1)E=\dfrac{hc}{\lambda } \\
 & 2)E=\dfrac{hc}{{{\lambda }_{0}}} \\
 & 3)E=e{{V}_{s}} \\
\end{align}\]

Complete step by step answer:
The energy of a photon may be given by
\[E=\dfrac{hc}{\lambda }........................(i)\]
where
$h$ is the Planck’s constant
$c$ is the velocity of light
$\lambda $ is the wavelength of incident light
We can write this relation in terms of threshold wavelength also:
\[E=\dfrac{hc}{{{\lambda }_{0}}}.......................(ii)\]
where
${{\lambda }_{0}}$ is the threshold wavelength
We also know that energy, in terms of stopping potential, is given by
\[E=e{{V}_{s}}.......................(iii)\]
where
$E$ is the energy of emitted photon
${{V}_{s}}$ is the stopping potential
Combining all these three equations, we have
\[E=\dfrac{hc}{\lambda }-\dfrac{hc}{{{\lambda }_{0}}}=e{{V}_{s}}\Rightarrow \dfrac{hc}{\lambda }-\dfrac{hc}{{{\lambda }_{0}}}=e{{V}_{s}}......................(iv)\]
Now, from the question, we have two conditions as follows:
Condition 1:
Wavelength \[=\lambda \]
Stopping Potential \[=3{{V}_{0}}\]
Condition 2:
Wavelength \[=2\lambda \]
Stopping Potential \[={{V}_{0}}\]
Putting these values in equation $(iv)$,
Condition 1 yields
\[\dfrac{hc}{\lambda }-\dfrac{hc}{{{\lambda }_{0}}}=e\left( 3{{V}_{0}} \right)......................(v)\]
Condition 2 yields
\[\dfrac{hc}{2\lambda }-\dfrac{hc}{{{\lambda }_{0}}}=e\left( {{V}_{0}} \right)......................(vi)\]
Dividing equation $(v)$ by $(vi)$, we have
\[\dfrac{\dfrac{hc}{\lambda }-\dfrac{hc}{{{\lambda }_{0}}}}{\dfrac{hc}{2\lambda }-\dfrac{hc}{{{\lambda }_{0}}}}=\dfrac{e\left( 3{{V}_{0}} \right)}{e\left( {{V}_{0}} \right)}\Rightarrow \dfrac{\dfrac{1}{\lambda }-\dfrac{1}{{{\lambda }_{0}}}}{\dfrac{1}{2\lambda }-\dfrac{1}{{{\lambda }_{0}}}}=3\Rightarrow \dfrac{1}{\lambda }-\dfrac{1}{{{\lambda }_{0}}}=\dfrac{3}{2\lambda }-\dfrac{3}{{{\lambda }_{0}}}\Rightarrow {{\lambda }_{0}}=4\lambda \]

So, the correct answer is “Option C”.

Additional Information: The phenomenon of photoelectric effect can only be explained with particle properties of light. If we compare equation $(i)$ and $(iii)$, we have
\[E=\dfrac{hc}{\lambda }-e{{V}_{s}}..........................(vii)\]
Equation $(vii)$ is called Einstein’s photoelectric equation.

Note: The minimum energy required by the incident photons to eject a single electron from a metal surface is termed as threshold energy. The corresponding frequency and wavelength of these photons is termed as threshold frequency and threshold wavelength, respectively. Ejection of electrons from a metal surface or the occurrence of photoelectric effect happens only when these threshold values are met.