Question

An electron and proton are accelerated through the same potential. Which one of the two has a greater value of de-Broglie wavelength associated with it and less momentum?

Answer 1

$\frac{{\lambda }_e}{{\lambda }_p}=\sqrt{\frac{m_p}{m_e}}$
The mass of an electron is less than that of the proton, hence the de-Broglie wavelength of the electron will be greater than that of the proton.
The mass of an electron is less than that of the proton, hence its momentum (mv) will be less than that of the proton.