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Mass Energy Equivalence

Consider the ...

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 b-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 βdecay. 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 – + υ ]

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Solution

The decay process of free neutron is,

n→p+ e −

Here, n is the neutron, p is the proton and e − is the electron.

The energy of electron E by Einstein’s mass-energy relation is,

E=Δm c 2

Here, Δm is the mass defect, which is mass of neutron-(mass of proton + mass of electron) and c is the speed of light.

As Δm and c are constants; therefore, energy distribution in β decay of neutron or nucleus is not being understood by the given two body decay.

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Similar questions

β

-decay process, discovered around 1900, is basically the decay of a neutron

(n)

. In the laboratory, a proton

(p)

and an electron

(e^{-})

are observed as the decay products of the neutron. Therefore, considering the decay of a neutron as a two-body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant. But experimentally, it was observed that the electron kinetic energy has a continuous spectrum. Considering a three-body decay process, i.e.

n \to p + e^{-} + {¯ ¯ ¯ v}_{e}

, around 1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino

({¯ ¯ ¯ v}_{e})

to be massless and possessing negligible energy, and the neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is

0.8 \times 10^{6} e V

. The kinetic energy carried by the proton is only the recoil energy.
What is the maximum energy of the anti-neutrino?

Q. When subatomic particles undergo reactions, energy is conserved, but mass is not necessarily conserved. However, a particle’s mass “contributes” to its total energy, in accordance with Einstein’s famous equations,

E = m c^{2}

In this equation, E denotes the energy a particle carries because of its mass. The particle can also have additional energy due to its motion and its interactions with other particles.
Consider a neutron at rest, and well separated from other particles. It decays into a proton, an electron and an undetected third particle:
Neutron

\to

proton + electron + ???
Table 1 summarizes some data from a single neutron decay. An MeV (mega electron volt) is a unit of energy.
Table 1
Data from a single neutron decay
Column-1 shows the rest mass of the particle times the speed of light squared and
Column-2 shows its kinetic energy.

\begin{matrix} Column-1 & Column-2 particle & mass \times c^{2} (M e V) & Kinetic energy (MeV) Neutron & 940.97 & 0.00 Proton & 939.67 & 0.01 Electron & 0.51 & 0.39 \end{matrix}

Given table 1, which properties of the undetected third particle can we calculate?

Q. When subatomic particles undergo reactions, energy is conserved, but mass is not necessarily conserved. However, a particle’s mass “contributes” to its total energy, in accordance with Einstein’s famous equations,

E = m c^{2}

In this equation, E denotes the energy a particle carries because of its mass. The particle can also have additional energy due to its motion and its interactions with other particles.
Consider a neutron at rest, and well separated from other particles. It decays into a proton, an electron and an undetected third particle:
Neutron

\to

proton + electron + ???
Table 1 summarizes some data from a single neutron decay. An MeV (mega electron volt) is a unit of energy.
Table 1
Data from a single neutron decay
Column-1 shows the rest mass of the particle times the speed of light squared and
Column-2 shows its kinetic energy.

\begin{matrix} Column-1 & Column-2 particle & mass \times c^{2} (M e V) & Kinetic energy (MeV) Neutron & 940.97 & 0.00 Proton & 939.67 & 0.01 Electron & 0.51 & 0.39 \end{matrix}

Assuming table 1 contains no major errors, what can we conclude about the mass

\times c^{2}

of the undetected third particle?

Q. A plot of the number of neutrons (N) against the number of protons (P) of stable nuclei exhibits upward deviation from linearity for atomic number, Z > 20. For an unstable nucleus having

\frac{N}{P}

ratio less than 1, the possible mode(s) of decay is(are)

Q. A free neutron decays to a proton but a free proton does not decay to a neutron. This is because
(a) neutron is a composite particle made of a proton and an electron whereas proton is a fundamental particle
(b) neutron is an uncharged particle whereas proton is a charged particle
(c) neutron has large rest mass than the proton
(d) weak forces can operate in a neutron but not in a proton.

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