Question

Question 6.30 Consider the decay of a free neutron at rest: n→p+ e–

Show that the two-body decay of this type must necessarily give an electron of fixed energy and, therefore, cannot account for the observed continuous energy distribution in the β-decay of a neutron or a nucleus (Fig. 6.19).


[Note: The simple result of this exercise was one among the several arguments advanced by W. Pauli to predict the existence of a third particle in the decay products of β-decay. This particle is known as neutrino. We now know that it is a particle of intrinsic spin ½ (like e–, p or n), but is neutral, and either massless or having an extremely small mass (compared to the mass of electron) and which interacts very weakly with matter. The correct decay process of neutron is: n→p + e–+ ν]

Chapter Work, Energy And Power Page 138

Answer 1

Hey there !!!!!!!!

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The decay process for a free neutron which is at rest is 

               n+p⇒e⁻

Using Einstein's energy and mass relation 

               E =mc²

"m" gives mass defect and c="velocity of light"

As m and c remain constant as a result Energy remains constant 

So continuous 2 body decay cannot be observed and 

Correct decay which can be observed is n+p⇒e⁻+ν

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Hope this helped you................

Answer 2

WE know that the decay process of free neutron at rest is given as :-
n→p + e^- 

from Einstein's energy mass relation 

E= mc^2
 
energy of electron is given by delta m c^2 
where delta m is mass defect and c is speed of light 

we know delta m = mass of neutron -(mass of proton + mass of electron )

now as delta m and c are constants this , so the given two bodies can not explain the continuous energy distribution in beta decay of a neutron or nucleus . 

Thus  Correct decay which can be observed is n+p⇒e⁻+ν