DoAssignment.ca

C3.5 · Explain how conservation laws supported the neutrino prediction

Learn to explain how conservation laws supported the neutrino prediction through clear examples and targeted practice.

Ontario Grade 12 Physics

Energy and Momentum

How conservation laws made sense of beta decay

A conservation law says that a quantity stays constant for a chosen physical system, provided the system is treated consistently. In beta decay, physicists observed an electron leaving a nucleus. The electron did not always carry the same amount of energy. If the electron and the recoiling nucleus were the only products, energy and momentum did not appear to balance in the expected way. Physicists proposed an additional particle: the neutrino. Its energy and momentum could be carried away without being detected in the early observations. The prediction was not a guess that replaced conservation laws. It was a way to preserve them.

What you will learn

1. The system, frame, and beta-decay evidence

A physical system is the object or group of objects being studied. Here, choose the decaying nucleus and all the particles produced in its decay as the system. Use a laboratory reference frame: the frame in which the apparatus and observer are at rest. For a simple one-dimensional sketch, choose right as the positive direction. A negative velocity or momentum then points left.
Beta decay is a change in a nucleus that produces an electron and a new nucleus. In the process considered here, a neutron in the nucleus changes into a proton. The proton remains in the new nucleus. Early observations focused on the outgoing electron and the recoiling nucleus. The proposed neutrino is a very difficult-to-detect particle with no electric charge.
A scalar has magnitude only. Energy is a scalar. A vector has magnitude and direction. Momentum is a vector, so its direction matters. Conservation means that the total amount of a conserved quantity before the change equals the total amount after it. For momentum, this is a vector comparison; for energy, it is a scalar comparison.
The important evidence was that beta-decay electrons were observed with a range of energies, rather than one fixed energy for each decay. If only the electron and recoiling nucleus shared the available energy, conservation appeared not to account for what was observed. The neutrino proposal supplied another product that could carry energy and momentum. The varying electron energy could then reflect different shares of the available energy among the products.
total before=total after\text{total before} = \text{total after}

2. Conservation laws and the neutrino proposal

For a decay, energy is conserved when the energy of the initial system equals the combined energy of all final products. The available energy is shared among the outgoing electron, the recoiling nucleus, and the proposed neutrino. If the electron receives less energy in one decay, the other products can receive more. The neutrino does not create extra energy; it carries part of the energy already available.
Momentum conservation adds a direction test. If the nucleus begins at rest, its initial momentum is zero. The vector sum of the momenta of the electron, recoiling nucleus, and neutrino must also be zero. An electron moving in one direction therefore requires the other products to balance its momentum. An unseen neutrino could carry some of that balancing momentum.
A further conservation rule concerns angular momentum, the rotational counterpart of motion-related quantities. In this context, the key idea is that the total angular momentum before and after a decay must agree. The early beta-decay accounting appeared to leave a mismatch if the known products alone were included. A proposed particle with its own angular momentum could help balance the total. At this level, the important point is the conservation requirement, not a detailed model of the particle.
The neutrino was therefore proposed to make the full set of conservation accounts balance. It was not directly seen in the early beta-decay observations described here. The proposal connected the observed electron and nucleus to an additional particle that could carry away energy, momentum, and angular momentum.
Einitial=Enucleus+Eelectron+EneutrinoE_{\text{initial}} = E_{\text{nucleus}} + E_{\text{electron}} + E_{\text{neutrino}}

3. Reading the evidence as conservation bookkeeping

A conservation account is not the same as a direct observation of every product. The early evidence was the behavior of the observed decay products, especially the range of electron energies and the apparent imbalance when only known products were counted. The neutrino was proposed to explain the imbalance while keeping conservation laws intact.
In a simple model, label the available energy as a fixed amount. For each decay, the electron, recoil, and neutrino share that amount. The exact share going to the electron can vary, while the total remains fixed. This model explains why a range of electron energies does not require energy to disappear.
Momentum gives a second check. Draw each momentum as an arrow in the chosen frame. The arrows must add to the initial momentum. For an initially resting system, the arrows must balance to a zero vector. The direction of the proposed neutrino cannot be chosen independently of the other products; it must fit the momentum account.
These examples use deliberately simple hypothetical values to practise bookkeeping. They are not measurements from a laboratory or claims about a particular historical decay. The historical reasoning was qualitative: conservation laws indicated what an unobserved particle would need to account for.
p⃗initial=p⃗nucleus+p⃗electron+p⃗neutrino\vec{p}_{\text{initial}} = \vec{p}_{\text{nucleus}} + \vec{p}_{\text{electron}} + \vec{p}_{\text{neutrino}}

Worked example

Finding the energy assigned to the proposed particle

In an illustrative energy account, a system has an available energy of 4.0×10−14 J4.0\times10^{-14}\ \mathrm{J}. The electron carries 2.4×10−14 J2.4\times10^{-14}\ \mathrm{J}, and the recoiling nucleus carries 0.10×10−14 J0.10\times10^{-14}\ \mathrm{J}. What energy remains for the neutrino?
  1. Set the system and identify the unknown
    Use the decaying system and all its products in the laboratory frame. Energy is scalar, so no direction is needed. The unknown is the energy carried by the neutrino.
  2. Apply conservation of energy
    The available energy equals the sum of the energies assigned to the nucleus, electron, and neutrino. Rearrange to find the remaining share.
    Eneutrino=Eavailable−Eelectron−EnucleusE_{\text{neutrino}}=E_{\text{available}}-E_{\text{electron}}-E_{\text{nucleus}}
  3. Substitute with units
    All terms are in joules, so they can be subtracted directly.
    Eneutrino=(4.0−2.4−0.10)×10−14 JE_{\text{neutrino}}=(4.0-2.4-0.10)\times10^{-14}\ \mathrm{J}
  4. Evaluate and round
    The result is positive and has two significant figures, consistent with the least precise given value.
    Eneutrino=1.5×10−14 JE_{\text{neutrino}}=1.5\times10^{-14}\ \mathrm{J}
Answer: In this illustrative account, the neutrino carries 1.5×10−14 J1.5\times10^{-14}\ \mathrm{J}.
Check: The units remain joules. Adding the three product energies gives 2.4×10−14+0.10×10−14+1.5×10−14=4.0×10−14 J2.4\times10^{-14}+0.10\times10^{-14}+1.5\times10^{-14}=4.0\times10^{-14}\ \mathrm{J}, so the account balances. The values are hypothetical, not measured decay data.

Worked example

Using momentum direction to find a missing contribution

An initially resting system decays in one dimension. The electron has momentum +3.0×10−22 kg m/s+3.0\times10^{-22}\ \mathrm{kg\,m/s}, and the recoiling nucleus has momentum −1.0×10−22 kg m/s-1.0\times10^{-22}\ \mathrm{kg\,m/s}. Find the momentum assigned to the proposed neutrino. Right is positive.
  1. Set the system and direction
    Include the nucleus, electron, and neutrino in the system. The initial system is at rest, so its total momentum is zero. The signs show direction: positive is right and negative is left.
  2. Apply vector momentum conservation
    The signed momenta of all products must add to the initial momentum. In one dimension, the signs keep track of the vector directions.
    0=pnucleus+pelectron+pneutrino0=p_{\text{nucleus}}+p_{\text{electron}}+p_{\text{neutrino}}
  3. Substitute and solve
    Insert the two given signed momenta and isolate the neutrino momentum.
    pneutrino=−(+3.0×10−22−1.0×10−22) kg m/sp_{\text{neutrino}}=-(+3.0\times10^{-22}-1.0\times10^{-22})\ \mathrm{kg\,m/s}
  4. State the vector result
    The negative sign means the neutrino momentum points left, opposite the chosen positive direction.
    pneutrino=−2.0×10−22 kg m/sp_{\text{neutrino}}=-2.0\times10^{-22}\ \mathrm{kg\,m/s}
Answer: The neutrino momentum is 2.0×10−22 kg m/s2.0\times10^{-22}\ \mathrm{kg\,m/s} to the left.
Check: The unit is momentum, and the signed sum is −1.0×10−22+3.0×10−22−2.0×10−22=0 kg m/s-1.0\times10^{-22}+3.0\times10^{-22}-2.0\times10^{-22}=0\ \mathrm{kg\,m/s}. The direction is required to balance the other two momenta.

Worked example

Explaining different electron energies

Two hypothetical decays have the same available energy, 5.0×10−14 J5.0\times10^{-14}\ \mathrm{J}. In both, the nucleus receives 0.20×10−14 J0.20\times10^{-14}\ \mathrm{J}. In the first decay, the electron receives 3.0×10−14 J3.0\times10^{-14}\ \mathrm{J}. In the second, it receives 1.8×10−14 J1.8\times10^{-14}\ \mathrm{J}. Find the neutrino energy in each case and explain what the comparison shows.
  1. Identify the system and fixed quantity
    For each decay, include the nucleus and all products. The available energy is the same, so the total energy must be the same in both accounts.
  2. Write the energy balance
    The neutrino receives whatever energy remains after accounting for the electron and nuclear recoil.
    Eneutrino=Eavailable−Eelectron−EnucleusE_{\text{neutrino}}=E_{\text{available}}-E_{\text{electron}}-E_{\text{nucleus}}
  3. Calculate the first decay
    Substitute the first electron energy and the shared recoil energy.
    Eneutrino,1=(5.0−3.0−0.20)×10−14 J=1.8×10−14 JE_{\text{neutrino,1}}=(5.0-3.0-0.20)\times10^{-14}\ \mathrm{J}=1.8\times10^{-14}\ \mathrm{J}
  4. Calculate the second decay
    Use the same available and recoil energies with the second electron energy.
    Eneutrino,2=(5.0−1.8−0.20)×10−14 J=3.0×10−14 JE_{\text{neutrino,2}}=(5.0-1.8-0.20)\times10^{-14}\ \mathrm{J}=3.0\times10^{-14}\ \mathrm{J}
Answer: The neutrino energies are 1.8×10−14 J1.8\times10^{-14}\ \mathrm{J} and 3.0×10−14 J3.0\times10^{-14}\ \mathrm{J}. The lower-energy electron corresponds to a higher neutrino energy in these accounts.
Check: Each account sums to 5.0×10−14 J5.0\times10^{-14}\ \mathrm{J}. The joule units are consistent, and both remaining energies are positive. This illustrates how variable electron energy can fit energy conservation when another product carries a changing share.

Common mistakes and how to avoid them

Saying that the neutrino was directly observed in the original energy-balance reasoning.
Correction: Distinguish the observed electron-energy pattern from the proposed neutrino used to account for the missing contributions.
Treating momentum as a scalar and adding only momentum magnitudes.
Correction: Momentum has direction. Choose a positive direction and include signs, or add vector arrows.
Claiming that the neutrino supplies extra energy to the decay.
Correction: The neutrino carries part of the energy already available. The total energy of all products still equals the initial energy.
Balancing energy but ignoring momentum and angular momentum.
Correction: The neutrino proposal addressed a broader conservation-accounting problem. Energy, momentum, and angular momentum all had to be considered.

Lesson summary

Check your understanding

Question 1

A decay system is initially at rest. Which statement best describes the momentum account?
  1. The magnitudes of the product momenta must add to zero.
  2. The vector sum of all product momenta must be zero.
  3. Only the electron momentum must be zero.
  4. Momentum need not be considered if energy is conserved.
Show answer and explanation
The vector sum of all product momenta must be zero.
Momentum is a vector. For an initially resting system, the vector sum of all final momenta must equal zero.

Question 2

Why did the neutrino proposal help explain a range of electron energies?
  1. It allowed the total available energy to change randomly.
  2. It carried a varying share of the available energy, so the total could remain conserved.
  3. It made the electron energy irrelevant to conservation.
  4. It removed the need to include the recoiling nucleus.
Show answer and explanation
It carried a varying share of the available energy, so the total could remain conserved.
The products can share a fixed total energy in different ways. A changing neutrino share can accompany a changing electron share.

Question 3

Which statement correctly distinguishes evidence from the proposal?
  1. The observed range of electron energies was evidence; the neutrino was proposed to help account for it.
  2. The neutrino was directly seen whenever an electron was emitted.
  3. The electron-energy range proved that energy was not conserved.
  4. The recoiling nucleus was proposed, while the electron was never observed.
Show answer and explanation
The observed range of electron energies was evidence; the neutrino was proposed to help account for it.
The electron-energy pattern was observed. The neutrino was proposed as an additional product to make the conservation account consistent.

Key terms

Conservation law
A rule that a specified physical quantity has the same total before and after a process.
System
The object or group of objects chosen for study.
Reference frame
The viewpoint used to describe positions, motion, and directions.
Scalar
A quantity with magnitude but no direction, such as energy.
Vector
A quantity with both magnitude and direction, such as momentum.
Momentum
A vector quantity that describes an object's motion; its SI unit is kilogram metre per second.
Angular momentum
A quantity describing rotational motion that must also be included in the decay conservation account.

Continue through SPH4U

View the complete SPH4U Ontario Grade 12 Physics curriculum and lessons

About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 12 Physics (SPH4U), expectation C3.5. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

Official curriculum reference

Report a correction or ask a question