When radiation of wavelength \[\lambda \] is incident on a metallic surface, the stopping potential is 4.8 V. If the same surface is illuminated with radiation of double the wavelength, then the stopping potential becomes 1.6 V. Then the threshold wavelength for the surface is :
A. \[2\lambda \]
B. \[4\lambda \]
C. \[6\lambda \]
D. \[8\lambda \]

Hint:The stopping potential is the potential which is able to stop the ejected electron. Using the work energy theorem, when a body is acted upon by external force then the change in kinetic energy of the body is equal to the work done on it.

Formula used:
\[K = \dfrac{{hc}}{\lambda } - \phi \]
where K is the kinetic energy of the emitted electron, h is the planck's constant, c is the speed of light, \[\lambda \]is the wavelength of the photon and \[\phi \]is the work function of the metal.

Complete step by step solution:
The threshold wavelength of the metal is the wavelength corresponding to the minimum energy which is sufficient to overcome the attractive force which causes the valence electron to be bound to the shell of the atom of the metal. As the energy of the photon is inversely proportional to the wavelength, so for a minimum value of the energy, the wavelength should be the maximum allowed wavelength.

The work done by the stopping potential of \[{V_0}\] to stop a moving electron must be equal to the kinetic energy of the electron.
\[e{V_0} = K{E_{\max }}\]
From the question,
\[4.8e = \dfrac{{hc}}{{{\lambda _1}}} - \phi \ldots \left( i \right)\]
\[\Rightarrow 1.6e = \dfrac{{hc}}{{{\lambda _2}}} - \phi \ldots \left( {ii} \right)\]
The relation between two wavelengths is given as \[{\lambda _2} = 2{\lambda _1}\]
\[ \Rightarrow {\lambda _1} = \lambda \] and \[{\lambda _2} = 2\lambda \]

On solving both the equations, we get
\[3\left( {\dfrac{{hc}}{{{\lambda _2}}} - \phi } \right) = \left( {\dfrac{{hc}}{{{\lambda _1}}} - \phi } \right)\]
\[\Rightarrow 3\left( {\dfrac{{hc}}{{2\lambda }} - \phi } \right) = \left( {\dfrac{{hc}}{\lambda } - \phi } \right)\]
\[\Rightarrow \dfrac{{3hc}}{{2\lambda }} - \dfrac{{hc}}{\lambda } = \left( {3\phi } \right) - \phi \]
\[\Rightarrow 2\phi = \dfrac{{hc}}{{2\lambda }}\]
If the threshold wavelength is \[{\lambda _0}\]
\[\phi = \dfrac{{hc}}{{{\lambda _0}}} = \dfrac{{hc}}{{4\lambda }}\]
Hence, the threshold wavelength of the given metal is \[4\lambda \].

Therefore, the correct option is B.

Note: The kinetic energy of the released electron is proportional to the incident radiation frequency. If the frequency is higher, the released electrons have the greatest kinetic energy. The number of photoelectrons released is proportional to the amount of incoming light.