Write Einstein’s photoelectric equation and mention which important features in photoelectric effect can be explained with the help of this equation. The maximum kinetic energy of the photoelectrons gets doubled when the wavelength of light incident on the surface changes from ${{\lambda }_{1}}\,to\,{{\lambda }_{2}}$. Derive the expressions for the threshold wavelength ${{\lambda }_{0}}$ and work function for the metal surface.

Hint: Einstein’s photoelectric equation is generally expressed in the terms of frequency we can express frequency in terms of wavelength and speed of light.

Formula used: $h\nu =h{{\nu }_{0}}+K.E$,
Where, c is speed of light,
h is planck’s constant,
K.E is kinetic energy,
\[\nu \] and \[{{\nu }_{0}}\] are incident frequency and threshold frequency respectively.

Complete step-by-step solution:
Einstein’s photoelectric equation is $h\nu =h{{\nu }_{0}}+K.E$
This equation explains that photoelectric emission will only take place when $\nu \ge {{\nu }_{0}}$ because if \[\nu \]<${{\nu }_{0}}$ then K.E will come out to be negative which is not possible. We can also see kinetic energy is directly proportional to frequency which explains the fact that kinetic energy only depends on the frequency of incident radiation.
Also, we can write $\nu $as $\dfrac{c}{\lambda }$.
Now, it is given that kinetic energy gets doubled when wavelength changes from ${{\lambda }_{1}}\,to\,{{\lambda }_{2}}$.
Putting in the above formula we get two equations.
$\dfrac{hc}{{{\lambda }_{1}}}=\dfrac{hc}{{{\lambda }_{0}}}+K.E\,$ …..(i)
$\dfrac{hc}{{{\lambda }_{2}}}=\dfrac{hc}{{{\lambda }_{0}}}+2K.E$ …..(ii)
Now, subtracting (ii) from twice of (i), we get
$\dfrac{2hc}{{{\lambda }_{1}}}-\dfrac{hc}{{{\lambda }_{2}}}=\dfrac{2hc}{{{\lambda }_{0}}}-\dfrac{hc}{{{\lambda }_{0}}}+2K.E-2K.E$
$\begin{align}
  & \dfrac{2hc}{{{\lambda }_{1}}}-\dfrac{hc}{{{\lambda }_{2}}}=\dfrac{hc}{{{\lambda }_{0}}} \\
 & \\
\end{align}$
$\begin{align}
  & \dfrac{2}{{{\lambda }_{1}}}-\dfrac{1}{{{\lambda }_{2}}}=\dfrac{1}{{{\lambda }_{0}}} \\
 & \\
\end{align}$
$\begin{align}
  & \dfrac{2hc}{{{\lambda }_{1}}}-\dfrac{hc}{{{\lambda }_{2}}}=\dfrac{hc}{{{\lambda }_{0}}} \\
 & \dfrac{2}{{{\lambda }_{1}}}-\dfrac{1}{{{\lambda }_{2}}}=\dfrac{1}{{{\lambda }_{0}}} \\
 & \dfrac{2{{\lambda }_{2}}-{{\lambda }_{1}}}{{{\lambda }_{1}}{{\lambda }_{2}}}=\dfrac{1}{{{\lambda }_{0}}} \\
 & {{\lambda }_{0}}=\dfrac{{{\lambda }_{1}}{{\lambda }_{2}}}{2{{\lambda }_{2}}-{{\lambda }_{1}}} \\
\end{align}$
Now, work function is given by the formula \[({{W}_{0}})=\dfrac{hc}{{{\lambda }_{0}}}\]
Putting the value of ${{\lambda }_{0}}$ from above we get,
${{W}_{0}}=hc(\dfrac{2{{\lambda }_{2}}-{{\lambda }_{1}}}{{{\lambda }_{1}}{{\lambda }_{2}}})$
Therefore, threshold wavelength ${{\lambda }_{0}}$ is $\dfrac{{{\lambda }_{1}}{{\lambda }_{2}}}{2{{\lambda }_{2}}-{{\lambda }_{1}}}$ and work function ${{W}_{0}}=hc(\dfrac{2{{\lambda }_{2}}-{{\lambda }_{1}}}{{{\lambda }_{1}}{{\lambda }_{2}}})$ is ${{W}_{0}}=hc(\dfrac{2{{\lambda }_{2}}-{{\lambda }_{1}}}{{{\lambda }_{1}}{{\lambda }_{2}}})$.

Note: If a similar question comes with numerical values and your work function or threshold wavelength comes out to be negative then you must know that they cannot be negative and you may have interchanged the value of wavelengths.