Question

A proton, an electron, and a helium nucleus all move at speed v. Rank their de Broglie wavelengths from largest to smallest. Write any relevant formulas you use below. a) helium nucleus, proton, electron
b) helium nucleus, electron, proton
c) electron, proton, helium nucleus
d) proton, electron, helium nucleus
e) proton, helium nucleus, electron

Answer 1

Answer:

The correct option is 'c':electron,proton,helium nucleus

Explanation:

The De-Broglie's wavelength of particle is given by

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

Thus we can see that wavelength is inversely related to mass of the particle since 'h' (Plank's constant) and velocity is same for all the particles  

Thus we conclude that the the lightest particle will have the most wavelength

Electron being the lightest of the 3 particles will have the largest wavelength thus the correct option is 'c'. Since electron has the largest wavelength followed by proton and the least wavelength among the 3 is of helium.