question_answer

An electromagnetic wave of wavelength \[\lambda \] is incident on a photosensitive surface of negligible work function. If the photo-electrons emitted from this surface have the de-Broglie wavelength  \[{{\lambda }_{1}},\] prove that \[\lambda =\left( \frac{2mc}{h} \right)\lambda _{1.}^{2}\]

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                         \[\lambda _{1}^{2}=\frac{h\lambda }{2mc}\Rightarrow \lambda =\left( \frac{2mc}{h} \right)\lambda _{1}^{2}\]