Answer:
According to Einstein?s photoelectric equation, for negligible work function \[\frac{hc}{\lambda }=K{{E}_{\max }}+0\] (\[\phi \]is negligible) \[\frac{hc}{\lambda }=\frac{{{p}^{2}}}{2m}\] \[\left( KE=\frac{{{p}^{2}}}{2m} \right)\] where, p = momentum of electron, m = mass of electron. \[\Rightarrow \] \[p=\sqrt{\frac{2mhc}{\lambda }}\] ? (i) \[\therefore \] de-Broglie wavelength\[{{\lambda }_{1}}=\frac{h}{p}\] \[{{\lambda }_{1}}=\frac{h}{\sqrt{\frac{2mhc}{\lambda }}}\] [from Eq. (i)] \[{{\lambda }_{1}}=\frac{h\lambda }{2mc}\] Squaring both sides, we get \[\lambda _{1}^{2}=\frac{h\lambda }{2mc}\Rightarrow \lambda =\left( \frac{2mc}{h} \right)\lambda _{1}^{2}\]
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