Question

Find the ratio of the de Broglie wavelengths of an electron and a proton when both are moving with the 

(a) same speed, 

(b) same energy and 

(c) same momentum? 

State which of the two will have the longer wavelength in each case?

Answer 1

Data: m_p = 1836 m_e

(a) The de Broglie wavelength, \(\lambda\) = \(\cfrac{h}p\) = \(\cfrac{h}{mv}\)

(b) \(\lambda\) = \(\cfrac{h}p\) = \(\cfrac{h}{\sqrt{2mK'}}\) where K denotes the kinetic energy \((\cfrac{1}{2}mv^2)\)

∴ \(\cfrac{\lambda_e}{\lambda_p}\) = \(\sqrt{\cfrac{m_pK_p}{m_eK_e}}\) = \(\sqrt{\cfrac{m_p}{m_e}}\) = \(\sqrt{1836}\) = 42.85

as K_p = K_e

Thus, \(\lambda\)_e > \(\lambda\)_p.

(c) \(\lambda\) = \(\cfrac{h}p\) 

∴  \(\cfrac{\lambda_e}{\lambda_p}\) = \(\cfrac{\lambda_p}{\lambda_e}\) = 1 as p_p = p_e