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:

Question

$\frac{\mathrm{hc}}{\mathrm{E} \lambda-\mathrm{hc}}$

$\frac{\mathrm{hc} \lambda}{\mathrm{E} \lambda+\mathrm{hc}}$

$\frac{\mathrm{h} \lambda}{\mathrm{E} \lambda+\mathrm{hc}}$

$\frac{\text { hc } \lambda}{\mathrm{E} \lambda-\mathrm{hc}}$

Solution

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