An electron, an alpha particle, and a proton have the same kinetic energy.Which of these particles has the shortest, de-Broglie wavelength?

kinetic energy is given by E= (1/2)mv^2.
Same kinetic energy means electron has highest vellocity( to make for its small mass)
Assume mass of electron as 1 unit.
mass of proton aprrox 1836 and mass of alpha particle is 4 times ie is 7344.

now calculate velocity in each case
For electron v= squareroot(2E)
F0r proton v = squareroot(2E/1836)
for alpha particle v=squareroot(2E/7344)

Also de-Broglie wavelength is given by Lambda= h/mv

Now calculate momentum in easch case
For electron mv= squareroot(2E)*1
For proton v = squareroot(2E/1836)*1836
For alpha particle v=squareroot(2E/7344)*7344

Clearly electron has least momentum. Hnece it has largest de-Broglie wavelength. And Alpha particle has least wavelength.