calculate de brogile wavelength(in cm) of an α -particle emitted from radium having an energy of 4.8MeV. Mass of α -particle=6.6x10^-24g

							
Dear Prince
Solve using the formula:
The derivation of De Brogiles wave length: 
We know that the energy of light E=mc^2 
Again if the frequency of light is f 
Then E=h*f [where h is planck constant] 
So, 
mc^2=h*f 
or, mc^2=h*(c/l) [where c is the velocity of light,l is the wavelength ,& c=l*f] 
or,l=(h*c)/mc^2 
or,l=h/m*c 
or,l=h/p.[ where p is the momentum=m*c] 
this is the De Broglies equation. 

That derivation is completely wrong. The momentum of a photon is NOT given by m*c, as c=constant, m=0. Neither is the energy of light mc^2, as that also goes to zero. The derivation works by a miracle. 

The proper derivation is this: 
From a relativistic invariant we know that 

E^2-p^2c^2=(mc^2)^2 
For a photon, the mass = 0, so 
E/c=p 

For a photon, 
E=hc/lambda 

Therefore, 
h/lambda=p 


The first step works as a result of contra/covariant vectors, but can be checked using algebra. The relativistic Energy = m*c^2/sqrt(1-v^2/c^2) , momentum p = m*v/sqrt(1-v^2/c^2). 
 
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