If ${\lambda _ \circ }$ and $\lambda $ be the threshold wavelength and the wavelength of the incident light respectively, then the maximum velocity of photo-electros ejected from the metal surface will be:
A. $\sqrt {\dfrac{{2h}}{m}\left( {{\lambda _ \circ } - \lambda } \right)} $
B. $\sqrt {\dfrac{{2hc}}{m}\left( {\dfrac{{{\lambda _ \circ } - \lambda }}{{\lambda {\lambda _ \circ }}}} \right)} $
C. $\sqrt {\dfrac{{2hc}}{m}\left( {{\lambda _ \circ } - \lambda } \right)} $
D. $\dfrac{{2h}}{m}\left[ {\dfrac{1}{{{\lambda _ \circ }}} - \dfrac{1}{\lambda }} \right]$

Hint: We should first know about the threshold wavelength. We will write the relation by Einstein's photoelectric equation. After that we will get Kinetic energy in the equation and we always know the formula of kinetic energy so we would replace the kinetic energy formula in Einstein's photoelectric equation. From that we will get the maximum velocity.

Complete step by step solution:
Step1. Threshold wavelength is the maximum wavelength of the incident light needed to release the electron from a surface in photoelectric effect. If we incident a light more than threshold wavelength then the electron would not be released.
Step2. From Einstein's photoelectric equation we can write
$h\nu = \emptyset + K.E$, here $\nu $ is the incident frequency, $\emptyset $ is the threshold energy and K.E. is kinetic energy.
Step3: $\emptyset = \dfrac{{hc}}{{{\gamma _ \circ }}}$, here the $h$ is the planck's constant, $c$ is the speed of light and ${\lambda _ \circ }$ is given threshold wavelength.
$ \Rightarrow hv = \dfrac{{hc}}{\lambda }$, it is the energy at given wavelength, here the $h$ is the planck's constant, $c$ is the speed of light and $\lambda $ is the wavelength of a given photon.
$ \Rightarrow K.E = \dfrac{1}{2}mV_{\max }^2$, the standard formula of Kinetic energy. $m$ is the mass of a photon. ${V_{\max }}$ is the maximum velocity. K.E is the kinetic energy
Step4: We use the derived relations in Einstein's photoelectric equation.
$ \Rightarrow \dfrac{{hc}}{\lambda } = \dfrac{{hc}}{{{\lambda _ \circ }}} + \dfrac{1}{2}mV_{\max }^2$
$ \Rightarrow hc\left( {\dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _ \circ }}}} \right) = \dfrac{1}{2}m{V^2}_{\max }$
$ \Rightarrow V_{\max }^2 = \dfrac{{2hc}}{m}\left( {\dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _ \circ }}}} \right)$
$ \Rightarrow {V_{\max }} = \sqrt {\dfrac{{2hc}}{m}\left( {\dfrac{{{\gamma _ \circ } - \gamma }}{{\gamma {\gamma _ \circ }}}} \right)} $


Hence, the correct option is B.

Note: From Einstein's photoelectric experiment, Einstein gave a theory that when a photon falls on the surface of the metal, the energy of the photon is entirely transferred to the electron. One part is consumed to remove the electron from the metal and another part provides velocity to the escaped electron. This experiment stabilised the fact that a light ray also showed the particle behaviour.