Question

Show that the ratio of the Compton wavelength $\lambda_{\mathrm{C}}$ to the de Broglie wavelength $\lambda=h / p$ for a relativistic electron is $$ \frac{\lambda_{\mathrm{C}}}{\lambda}=\left[\left(\frac{E}{m_{c} c^{2}}\right)^{2}-1\right]^{1 / 2} $$ $$ \text { where } E \text { is the total energy of the electron and } m_{e} \text { is its mass. } $$

          Show that the ratio of the Compton wavelength $\lambda_{\mathrm{C}}$ to the de Broglie wavelength $\lambda=h / p$ for a relativistic electron is $$
\frac{\lambda_{\mathrm{C}}}{\lambda}=\left[\left(\frac{E}{m_{c} c^{2}}\right)^{2}-1\right]^{1 / 2}
$$ $$
\text { where } E \text { is the total energy of the electron and } m_{e} \text { is its mass. }
$$

Added by Tracey M.

University Physics with Modern Physics

Hugh D. Young 14th Edition

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Solved by Expert Maitreya E

05/04/2022

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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 $$ \frac{\lambda_{\mathrm{C}}}{\lambda}=\left[\left(\frac{E}{m_{c} c^{2}}\right)^{2}-1\right]^{1 / 2} $$ $$ \text { where } E \text { is the total energy of the electron and } m_{e} \text { is its mass. } $$