Arrange correct order of De broglie wavelength of electron,proton and alpha particle all having the same kinetic energy

Calculate the de Broglie wavelength of an electron, proton and uranium atom, all having the same kinetic energy 100eV .

lambda_(e),lambda_(p) and lambda_(alpha) are the de-Broglie wavelength of electron, proton and alpha particle. If all the accelerated by same potential, then

Find the ratio of de Broglie wavelength of proton and alpha -perticle of same energy

The ratio of the de Broglie wavelength of a proton and alpha particles will be 1:2 if their

What will be the ratio of de - Broglie wavelengths of proton and alpha - particle of same energy ?

The magnitude of the de-Broglie wavelength (lambda) of an electron (e) ,proton (p) ,neutron (n) and alpha particle (a) all having the same energy of Mev , in the incresing order will follow the sequence:

Find the ratio of de Broglie wavelength of a proton and as alpha -particle which have been accelerated through same potential difference.

What should be the order of arrangement of de-Broglie wavelength of electron (lamda_(e)) , an alpha- particle (lamda_(alpha)) and proton (lamda_(p)) given that all have the same kinetic energy ?

For a given kinetic energy, which of the following has the smallest de Broglie wavelength : electron, proton and alpha-particle ?