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Shashank Y Grade: 12
        

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

7 years ago

Answers : (5)

AskIITiansExpert Kartik-IITMadras
21 Points
										

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
Sushmitha Naidu
18 Points
										
among alpha particle,electron,neutron and proton whch has highest wavelenght???
 
2 years ago
Anmol
13 Points
										
alpha particle has the highest wavelength
 
2 years ago
Mihir
69 Points
										
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.
2 years ago
Shubham Singh
199 Points
										
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
2 years ago
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