JEE MAIN - Physics (2022 - 29th July Morning Shift - No. 16)
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:
$$\frac{\mathrm{hc}}{\mathrm{E} \lambda-\mathrm{hc}}$$
$$\frac{\mathrm{hc} \lambda}{\mathrm{E} \lambda+\mathrm{hc}}$$
$$\frac{\mathrm{h} \lambda}{\mathrm{E} \lambda+\mathrm{hc}}$$
$$\frac{\text { hc } \lambda}{\mathrm{E} \lambda-\mathrm{hc}}$$
Explanation
$$k = {{hc} \over \lambda } - \phi = E$$
and, $$2k = {{hc} \over {{\lambda _2}}} - \phi = 2E$$
$$ \Rightarrow {{hc} \over \lambda } - E = {{hc} \over {{\lambda _2}}} - 2E$$
$$ \Rightarrow {{hc} \over {{\lambda _2}}} = {{hc} \over \lambda } + E$$
$$ \Rightarrow {\lambda _2} = {{hc\lambda } \over {hc + \lambda E}}$$
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