Question

10. Show the ratio of the Compton wavelength to the de Broglie wavelength for a relativistic electron with Energy E is given by 소 11. Non-relativistic electrons and protons are accelerated from rest though the same potential difference Δ V, show the ratio of their de Broglie wavelengths is given by: 弖

Answer 1

10) The Compton wavelength is

$\lambda _c=\frac{h}{m_ec}$

The de Broglie wavelength is $\lambda =\frac{h}{p}$ . By taking the ratio of the Compton wavelength to the de Broglie wavelength and square it

$\left ( \frac{\lambda _c}{\lambda } \right )^2=\frac{p^2}{(m_ec)^2}$

the momentum for a slowly-moving or rapidly moving object is described by

$p^2=\frac{E^2-m_e^2c^4}{c^2}$

Substituting and simplifying

$\left ( \frac{\lambda_c}{\lambda } \right )^2=\frac{(E^2-m_e^2c^4)}{(m_ec^2)^2}=\left (\frac{E^2}{m_ec^2} \right )^2-1$

$\frac{\lambda_c}{\lambda }=\sqrt{\left (\frac{E^2}{E_o^2} \right )^2-1}$