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Question
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
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Solution 1
The decay process of free neutron at rest is given as:
n → p + e–
From Einstein’s mass-energy relation, we have the energy of electron as Δmc2
Where,
Δm = Mass defect = Mass of neutron – (Mass of proton + Mass of electron)
c = Speed of light
Δm and c are constants. Hence, the given two-body decay is unable to explain the continuous energy distribution in the β-decay of a neutron or a nucleus. The presence of neutrino νon the LHS of the decay correctly explains the continuous energy distribution.
Solution 2
Let the masses of the electron and proton be m and M respectively. Let v and V be the velocities of electron and proton respectively. Using law of conservation of momentum. Momentum of electron + momentum of proton = momentum of neutron
`:. mv + MV = 0 => V = - m/M v`
Clearly the electron and the proton move in opposite directions. If mass `trianglem` has been conveted into energy in the reaction then
`1/2mv^2 + 1/2 MV^2 = trianglem xx c^2`
or `1/2mv^2 + 1/2M[-m/M]^2v^2 = trianglemc^2`
or `1/2mv^2[1+m/M] = trianglemc^2`
or `v^2 = (2Mtrianglemc^2)/(m(M+m))`
Thus, it is proved that the value of v2 is fixed since all the quantities in right hand side are constant. It establishes that the emitted electron must have a fixed energy and thus we cannot account for the continuous energy distribution in the β-decay of a neutron.
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