Answer to Question #132350 in Mechanics | Relativity for Viccy

Question

What is ratio of De-broglie wavelegth of proton and electron at same kinetic energy.

Answer 1

De Broglie wavelength is

$\lambda = \frac{h}{p}$

$E = \frac{p^2}{2m} \Rightarrow p = \sqrt{2mE}$

$\frac{\lambda_p}{\lambda_e} = \frac{p_e}{p_p} = \frac{\sqrt{2m_eE_e}}{\sqrt{2m_pE_p}}$

$E_e = E_p$ (same kinetic energy), so

$\frac{\lambda_p}{\lambda_e} = \frac{\sqrt{m_e}}{\sqrt{m_p}} = \sqrt{\frac{9.1 \cdot 10^{-31}}{1.67 \cdot 10^{-27}}} = \sqrt{5.449 \cdot 10^{-4}} = 2.334 \cdot 10^{-2} = 0.023$