10. Show the ratio of the Compton wavelength to the de Broglie wavelength for a relativistic electron with Energy E is given by: frac{lambda_c}{lambda} = sqrt{frac{E^2}{E_0^2} - 1} 11. Non-relativistic electrons and protons are accelerated from rest though the same potential difference Delta V. Show the ratio of their de Broglie wavelengths is given by: frac{lambda_e}{lambda_p} = sqrt{frac{m_p}{m_e}}
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The Compton wavelength of an electron is given by the formula: $\lambda_C = \frac{h}{m_e c}$ where $h$ is the Planck's constant, $m_e$ is the mass of the electron, and $c$ is the speed of light. Show more…
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Show that the ratio of the Compton wavelength $\lambda_{\mathrm{C}}$ to the de Broglie wavelength $\lambda=h / p$ for a relativistic electron is $$\frac{\lambda_{\mathrm{C}}}{\lambda}=\left[\left(\frac{E}{m_{e} c^{2}}\right)^{2}-1\right]^{1 / 2}$$ where $E$ is the total energy of the electron and $m_{e}$ is its mass.
Show that the ratio of the Compton wavelength $\lambda_{\mathrm{C}}$ to the de Broglie wavelength $\lambda=h / p$ for a relativistic electron is $$ \frac{\lambda_{\mathrm{C}}}{\lambda}=\left[\left(\frac{E}{m_{c} c^{2}}\right)^{2}-1\right]^{1 / 2} $$ $$ \text { where } E \text { is the total energy of the electron and } m_{e} \text { is its mass. } $$
Maitreya E.
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