The kinetic energy of emitted electron is E when the light incident on the metal has wavelength $$\lambda$$. To double the kinetic energy, the incident light must have wavelength:

We use Einstein's photoelectric equation: $$KE = \frac{hc}{\lambda} - \phi$$, where $$\phi$$ is the work function of the metal.

For the first case, the kinetic energy of the emitted electron is $$E$$ when the incident wavelength is $$\lambda$$: $$E = \frac{hc}{\lambda} - \phi$$. This gives us the work function $$\phi = \frac{hc}{\lambda} - E$$.

For the second case, we need the kinetic energy to be $$2E$$ with a new wavelength $$\lambda'$$: $$2E = \frac{hc}{\lambda'} - \phi$$.

Substituting the expression for $$\phi$$: $$2E = \frac{hc}{\lambda'} - \left(\frac{hc}{\lambda} - E\right) = \frac{hc}{\lambda'} - \frac{hc}{\lambda} + E$$.

Simplifying: $$2E - E = \frac{hc}{\lambda'} - \frac{hc}{\lambda}$$, so $$E = \frac{hc}{\lambda'} - \frac{hc}{\lambda}$$.

Solving for $$\frac{1}{\lambda'}$$: $$\frac{hc}{\lambda'} = E + \frac{hc}{\lambda} = \frac{E\lambda + hc}{\lambda}$$.

Therefore $$\lambda' = \frac{hc\lambda}{E\lambda + hc}$$.

Hence, the correct answer is Option B.