 # An electron, an alpha particle, and a proton have the same kinetic energy.Which of these particles has the shortest, de-Broglie wavelength? AskIITiansExpert Kartik-IITMadras
21 Points
12 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.

7 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.
7 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
4 years ago
The De brollies formula is as follows: Wavelength =h/p. WhereP is momentum. h is planks constant. Hence wavelength is inversely proportional to momentum. Also momentum is directly proportional to mass so alpha particle has more mass than electron according to relation mass is inversely proportional to wavelength Therefore electron has large wavelength than alpha particle