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:
\( \dfrac{hc}{E\lambda - hc} \)
\( \dfrac{hc\lambda}{E\lambda + hc} \)
\( \dfrac{h\lambda}{E\lambda + hc} \)
\( \dfrac{hc\lambda}{E\lambda - hc} \)