The kinetic energy of emitted electron is E when the light incident...

Using Einstein's photoelectric equation, kinetic energy of an electron is $E$ when the incident wavelength is $\lambda$ is $E = \dfrac{hc}{\lambda} - ϕ . . . \left(1\right)$

When kinetic energy doubles, then, $2E = \dfrac{hc}{{\lambda}^{'}} - ϕ . . . \left(2\right)$

Subtracting \left(2\right) - \left(1\right)$, we have </p><p>$E = hc\left(\dfrac{1}{{\lambda}^{'}} - \dfrac{1}{\lambda}\right)$</p><p>$⇒ {\lambda}^{'} = \dfrac{hc\lambda}{E\lambda + hc}