Which of these particles (having the same kinetic energy) has the shortest de-Broglie wave length?

Question

Which of these particles (having the same kinetic energy) has the shortest de-Broglie wave length? (A) Electron (B) Alpha particle (C) Proton (D) Neutron

Tardigrade Admin · Accepted Answer

de-Broglie relation is given by λ =(h/mv) This can also be written as λ =(h/√2mv) So, λ propto (1/√m) Because, an alpha particle having the highest mass among the others, so alpha particle have the least wavelength.

Q. Which of these particles (having the same kinetic energy) has the shortest de-Broglie wave length?

Electron

Alpha particle

Proton

Neutron

de-Broglie relation is given by $λ = \frac{h}{m v}$
This can also be written as $λ = \frac{h}{2 m v}$
So, $λ \propto \frac{1}{m}$
Because, an alpha particle having the highest mass among the others,
so alpha particle have the least wavelength.

Q. Which of these particles (having the same kinetic energy) has the shortest de-Broglie wave length?

Electron

Alpha particle

Proton

Neutron

de-Broglie relation is given by λ=mvh​ This can also be written as λ=2mv​h​ So, λ∝m​1​ Because, an alpha particle having the highest mass among the others, so alpha particle have the least wavelength.

de-Broglie relation is given by $λ = \frac{h}{m v}$
This can also be written as $λ = \frac{h}{2 m v}$
So, $λ \propto \frac{1}{m}$
Because, an alpha particle having the highest mass among the others,
so alpha particle have the least wavelength.