PhysicsEasy145×since 2002Q6653The kinetic energy of emitted electron is E when the light incident on the metal has wavelength λ\lambdaλ. To double the kinetic energy, the incident light must have wavelength:AhcEλ−hc\frac{\mathrm{hc}}{\mathrm{E} \lambda-\mathrm{hc}}Eλ−hchcBhcλEλ+hc\frac{\mathrm{hc} \lambda}{\mathrm{E} \lambda+\mathrm{hc}}Eλ+hchcλChλEλ+hc\frac{\mathrm{h} \lambda}{\mathrm{E} \lambda+\mathrm{hc}}Eλ+hchλD hc λEλ−hc\frac{\text { hc } \lambda}{\mathrm{E} \lambda-\mathrm{hc}}Eλ−hc hc λCheck answerSkip