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:
Solution
<p>$k = {{hc} \over \lambda } - \phi = E$</p>
<p>and, $2k = {{hc} \over {{\lambda _2}}} - \phi = 2E$</p>
<p>$\Rightarrow {{hc} \over \lambda } - E = {{hc} \over {{\lambda _2}}} - 2E$</p>
<p>$\Rightarrow {{hc} \over {{\lambda _2}}} = {{hc} \over \lambda } + E$</p>
<p>$\Rightarrow {\lambda _2} = {{hc\lambda } \over {hc + \lambda E}}$</p>
About this question
Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: Photoelectric Effect
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