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# 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

9 years ago

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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