question_answer

An electron and a photon have same de- Broglie wavelength (say\[\overset{\mathbf{o}}{\mathop{\mathbf{A}}}\,\]). Which one possesses more kinetic energy?                     Or Compare the energy of an electron of de-Broglie wavelength \[1\overset{\mathbf{o}}{\mathop{\mathbf{A}}}\,\] with that of an X-ray photon of the same wavelength.                                 

Answer:

                de-Broglie wavelength of electron, \[\lambda =\frac{h}{p}\] \[\therefore \] Momentum of an electron, \[p=\frac{h}{\lambda }\] K.E. of an electron \[=\frac{{{p}^{2}}}{2m}=\frac{{{h}^{2}}}{2m{{\lambda }^{2}}}\] The energy of X-ray photon is totally kinetic. Energy of X-ray photon \[=hv=\frac{hc}{\lambda }\] \[\therefore \]  \[\frac{\text{Electron energy}}{\text{Energy of X-ray photon}}\] \[=\frac{{{h}^{2}}}{2m{{\lambda }^{2}}}\times \frac{\lambda }{hc}=\frac{h}{2m\lambda c}\] \[=\frac{6.6\times {{10}^{-34}}}{2\times 9.11\times {{10}^{-31}}\times 1\times {{10}^{-10}}\times 3\times {{10}^{8}}}\] \[=\frac{11}{911}<1\] Thus the K.E. of a photon is greater than that of an electron of same wavelength.