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:
- A$$\frac{\mathrm{hc}}{\mathrm{E} \lambda-\mathrm{hc}}$$
- B$$\frac{\mathrm{hc} \lambda}{\mathrm{E} \lambda+\mathrm{hc}}$$
- C$$\frac{\mathrm{h} \lambda}{\mathrm{E} \lambda+\mathrm{hc}}$$
- D$$\frac{\text { hc } \lambda}{\mathrm{E} \lambda-\mathrm{hc}}$$
✓ Correct answer: B
Want the full step-by-step reasoning?