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.

Answer:

            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         \[{{E}_{p}}=\frac{2\lambda mc}{h}\cdot {{E}_{e}}\]