• # question_answer Why are de-Broglie waves associated with a moving football not visible? The wavelength, $(\lambda )$ of a photon and the de-Broglie wavelength of an electron have the same value. Show that the energy of the photon is $2\,\,\lambda mc/h$ times the kinetic energy of the electron, where ${{m}_{e}}=m,$ C and h have their usual meanings.

The magnitude of the wavelength of de-Broglie waves associated with a moving football is extremely small $\left( \lambda =\frac{h}{mv}<{{10}^{-34}}m \right),$ which is much less than that of visible region and therefore they are not visible. Wavelength of photon $=\lambda$ $\therefore$ Energy of photon, ${{E}_{p}}=hc/\lambda$                        ?(i) Kinetic energy of electron, ${{E}_{e}}=\frac{1}{2}m{{v}^{2}}$ or $m{{v}^{2}}=2{{E}_{e}}$ or         $v=\sqrt{\frac{2{{E}_{e}}}{m}}$                     $(\because {{m}_{e}}=m)$ de-Broglie wavelength of electron             $\lambda =\frac{h}{mv}$             $=\frac{h}{\sqrt{2{{E}_{e}}m}}$ or         ${{E}_{e}}=\frac{{{h}^{2}}}{2{{\lambda }^{2}}m}$                                   ?(ii) On dividing Eq. (i) by Eq. (ii), we get             $\frac{{{E}_{p}}}{{{E}_{e}}}=\frac{hc}{\lambda }\cdot \frac{2{{\lambda }^{2}}m}{{{h}^{2}}}=\frac{2\lambda mc}{h}$ or         ${{E}_{p}}=\frac{2\lambda mc}{h}\cdot {{E}_{e}}$
You will be redirected in 3 sec 