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

21 Points
13 years ago

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.

Sushmitha Naidu
18 Points
8 years ago
among alpha particle,electron,neutron and proton whch has highest wavelenght?
Anmol
13 Points
8 years ago
alpha particle has the highest wavelength
Mihir
69 Points
8 years ago
Kinetic Energy is realted to momentum as
K= P2/2m, where m is the mass of the particle.
Now, P= sqrt(2mK), which means the linear momentum is proportional to square root of mass. Hence, de Broglie wavelength is inversely prorprtional to the square root of mass. Therefore, the electron having the least mass shall possess maximum de Broglie wavelength.
Shubham Singh
199 Points
8 years ago
K= P2/2m
the more the mass the more the momentum
the more the momentum , the less the wavelength
so mass is inversely  proportional to wavelength