I was doing this question:

Reactions of radioactive decay, as mentioned in the text, do not obey Lavoisier's law (conservation of mass), once there is no conservation of mass in the process, but rather energy.

The answer says the item is True, but i think it's false.

The text mentions this decay:

$$\rm ^{60}Co \to {}^{60}Ni + \beta + γ $$

Here are my questions:

  1. Since mass is energy, there's no conservation, right?

  2. The Law of conservation of mass by Lavoisier is wrong? Since there's a slightly change in the mass of the final product.

  3. Using the equation $E= mc^2$, do i have to use the momentum in this case?

  4. I'm very confused by this also: If the number of protons in the 60Co was 27, and after the decay is 28 (60Ni), does the electron emitted in the decay make the 28 protons of 60Ni 27 protons? So the charge in the end is the same? How is this possible? I'm confused: the energy is conserved? The mass is conserved?

  1. Total energy is conserved, not total mass.
  2. Yes, mass is not conserved exactly.
  3. The correct equation would be $E = \sqrt{m^2 c^4 + p^2 c^2}$. For example the photon ($\gamma$) has energy $E = pc$ since it is massless.
  4. Charge is conserved, yes. One of the neutrons is converted into a proton and releases an electron (and a anti-electron neutrino), see the Beta Decay Wikipedia article. Thus the overall charge is the same.
  • $\begingroup$ 1.But in the process, there's some energy that is not conserved? $\endgroup$ – Rhafael Apr 23 '18 at 21:06
  • $\begingroup$ No energy is absolutely conserved. If the Co atom is initially at rest, in the final state all three resulting particles are moving away from each ither. In such a way that the total (relativistic) energy is conserved and overall momentum is conserved $\endgroup$ – Kai Apr 23 '18 at 21:50
  1. The Law of conservation of mass by Lavoisier is wrong? Since there's a slightly change in the mass of the final product.

Physics uses mathematical models to fit data and observations and , more important, then uses them as tools for further predictions, this is what makes the models useful.

The models depend on the framework, the dimensions in space and time, the dimensions in energy and momentum.

Broadly two categories exist: Classical physics, which are the models used to describe observations before the twentieth century, and the quantum regime, which had to be invoked to explain data in small dimensions.

In addition, there is special relativity and General relativity used when momenta are large or gravitational masses ( as classically defined) are large.

Each mathematical model is successful in its frame of reference.

Thermodynamics was first modeled with its laws ( laws are extra axioms needed to define physical solutions for the differential equations used), and then it was discovered that it emerged from statistical mechanics in a continuous, provable manner.

Similarly classical mechanics can be shown to emerge from the underlying level of quantum mechanics, classical electrodynamics from quantum electrodynamics .

The law of conservation of mass fitted macroscopic observations until the twentieth century, when it was discovered that for large energies and momenta one had to use special relativity to describe the data, and in special relativity the only laws are conservation of energy and momentum, in a four vector . This makes mass the "length" of the four vector, given in formula 3. in the answer of Kai.

When adding three dimensional vectors, the resultant vector has the length given by the vectorial addition, not the sum of the lengths of each vector. Similarly the addition of two four vectors does not conserve the mass. The rest mass is invariant under Lorentz transformations for each object under consideration. In the framework of small dimensions and/or large energies where special relativity has to be used mass is not conserved.

Physics frameworks can be shown in overlap regions to be mathematically consistent, within the experimental errors in each framework.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.