Changes

Accurate measurements and detailed calculations allowed for verifying the theoretical values with an accuracy of ±0.5%. This was the first time a nucleus was artificially split, and thereby the first transmutation of elements using accelerated particles:

Accurate measurements and detailed calculations allowed for verifying the theoretical values with an accuracy of ±0.5%. This was the first time a nucleus was artificially split, and thereby the first transmutation of elements using accelerated particles:

Probably the best empirical verification of '''E=mc²''' was done in 2005 by Simon Rainville et al., as published in ''Nature''<ref>[http://www.nature.com/nature/journal/v438/n7071/full/4381096a.html Simon Rainville, James K. Thompson, et. al ''World Year of Physics: A direct test of E=mc&sup2;''] Nature 438, 1096-1097 (22 December 2005)] doi:10.1038/4381096a; Published online 21 December 2005</ref>, as the authors write in their abstract:

Some claim that the best empirical verification of '''E=mc²''' was done in 2005 by Simon Rainville et al., as published in ''Nature'' (which is not a leading physics journal).<ref>[http://www.nature.com/nature/journal/v438/n7071/full/4381096a.html Nature 438, 1096-1097 (22 December 2005)] doi:10.1038/4381096a; Published online 21 December 2005</ref> The authors state in their article in ''Nature'' magazine that &quot;Einstein's relationship is separately confirmed in two tests, which yield a combined result of 1-?mc²/E=(-1.4±4.4)×10^-7, indicating that it holds to a level of at least 0.00004%. To our knowledge, this is the most precise direct test of the famous equation yet described.&quot;

{{cquote|Einstein's relationship is separately confirmed in two tests, which yield a combined result of 1−Δmc²/E&#061;(−1.4±4.4)×10⁻⁷, indicating that it holds to a level of at least 0.00004%. To our knowledge, this is the most precise direct test of the famous equation yet described.}}

==A Famous Example -- Nuclear Fission of Uranium==

==A Famous Example -- Nuclear Fission of Uranium==

The [[Theory of Relativity]] played no role in this work, but later tried to retrofit the theory to the data in order to explain the explain the observed mass changes.  Here is the most famous example of the mass change.

The [[Theory of Relativity]] played no role in this work, but later tried to retrofit the theory to the data in order to explain the explain the observed mass changes.  Here is the most famous example of the mass change.

Nuclear fission, which is the basis for nuclear energy, was discovered in experiments by [[Otto Hahn]] and [[Fritz Strassman]], and analyzed by [[Lise Meitner]], in 1938.

Nuclear fission, which is the basis for nuclear energy, was discovered in experiments by [[Otto Hahn]] and [[Fritz Strassmann]], and analyzed by [[Lise Meitner]], in 1938.

The decay path of [[Uranium]] that figured in the ''Hahn-Strassmann experiment'' may have been this:

The decay path of [[Uranium]] that figured in the ''Hahn-Strassmann experiment'' may have been this:

:²³⁵U  &rarr;  ¹⁴⁰Xe  +  ⁹¹Sr  +  4n

:²³⁵U  &rarr;  ¹⁴⁰Xe  +  ⁹¹Sr  +  4n

(The [[Xenon]] decayed within about a minute to ¹⁴⁰Ba.  There are a large number of fission paths and fission products, but they were searching for the chemical signature of [[Barium]].)

(The [[Xenon]] decayed within about a minute to ¹⁴⁰Ba.  There are a large number of fission paths and fission products, but Hahn and Strassmann were searching for the chemical signature of [[Barium]].)

 

The masses of the particles are:

The masses of the particles are:

A well-accepted physical law of classical physics was the law of conservation of mass.  This was not easy to deduce.  It required careful analysis of such phenomena as combustion, in the 1700's, to eliminate the various confounding sub-phenomena that made the law difficult to see.  But, by 1900, the law was well established:

A well-accepted physical law of classical physics was the law of conservation of mass.  This was not easy to deduce.  It required careful analysis of such phenomena as combustion, in the 1700's, to eliminate the various confounding sub-phenomena that made the law difficult to see.  But, by 1900, the law was well established:

:::*'''In all interactions, mass is precisely conserved.'''

::*'''In all interactions, mass is precisely conserved.'''

For example, the mass of a TNT molecule is 227.1311 Daltons, or 227.1311 g/mol, which is, for all practical purposes, the same as the mass of its constituent Carbon, Hydrogen, Nitrogen, and Oxygen atoms.  It is essentially impossible to measure the difference.  The principle of conservation of mass is upheld.

For example, the mass of a TNT molecule is 227.1311 Daltons, or 227.1311 g/mol, which is, for all practical purposes, the same as the mass of its constituent Carbon, Hydrogen, Nitrogen, and Oxygen atoms.  It is essentially impossible to measure the difference.  The principle of conservation of mass is upheld.

Radium-226 decays into Radon-222 by emission of an alpha particle with an energy of 4.78 MeV.

Radium-226 decays into Radon-222 by emission of an alpha particle with an energy of 4.78 MeV.

1 kg of Radium-226 = <math>\frac{.622 \times 10^{27}}{226.0254}</math> atoms.  (The numerator is Avogadro's number, and the denominator is the atomic weight of Radium-226.)  This is 2.6643647 * 10²⁴ atoms.

1 kg of Radium-226 = <math>\frac{.60221413 \times 10^{27}}{226.0254}</math> atoms.  (The numerator is Avogadro's number, and the denominator is the atomic weight of Radium-226.)  This is 2.6643648 * 10²⁴ atoms.

That number of Radon-222 atoms has mass .98226836 kg.  That number of alpha particles has mass .01770863 kg.

That number of Radon-222 atoms has mass .98226836 kg.  That number of alpha particles has mass .01770864 kg.

The mass lost is .00002301 kg.

The mass lost is .00002299 kg.

Each emitted alpha particle has energy of 4.78 MeV, or 4.78 * .1602 * 10^-18 Joules.  The total alpha energy from the decay of 1 kg of radium is 2.04 * 10¹² Joules.

Each emitted alpha particle has energy of 4.78 MeV, or 4.78 * .16021765 * 10^-18 Joules.  The total alpha energy from the decay of 1 kg of radium is 2.040 * 10¹² Joules.

Also, Radon-222 decays into Polonium-218 by emission of an alpha particle with an energy of 5.49 MeV.

Also, Radon-222 decays into Polonium-218 by emission of an alpha particle with an energy of 5.49 MeV.

1 kg of Radon-222 = <math>\frac{.622 \times 10^{27}}{222.0176}</math> atoms.  This is 2.7124611 * 10²⁴ atoms.

1 kg of Radon-222 = <math>\frac{.60221413 \times 10^{27}}{222.0176}</math> atoms.  This is 2.7124612 * 10²⁴ atoms.

 

 

 

 

 

That number of Polonium-218 atoms has mass .98194467 kg.  That number of alpha particles has mass .01802830 kg.

For the Cockcroft-Walton experiment, we us Avogadro's number of particles.  The mass of that many of the various atoms, in kilograms, is just their atomic mass.  Avogadro's number of Lithium-7 atoms weighs 7.01600455 kg.  The Hydrogen atoms weigh 1.007825032 kg, and the Helium atoms weigh 4.00260325 kg each.  The atoms weigh 8.023829582 kg before, and 8.0052065 kg after.  The mass lost is .018623082 kg.

The mass lost is .00002703 kg.

The energy released is 17.3 MeV per reaction, or 1669 * 10^-12 Joules per Avogadro's number.

It looks as thought we have to rewrite the law of conservation of mass:

It looks as though we have to rewrite the law of conservation of mass:

:::*'''In all "ordinary" interactions, mass is precisely conserved.'''

::*'''In all "ordinary" interactions, mass is precisely conserved.'''

:::*'''In nuclear interactions, there is a small but measurable loss of mass.'''

::*'''In nuclear interactions, there is a small but measurable loss of mass.'''

:By the way, we can clearly see that atomic weights of pure isotopes are not integers, and that it has something to do with the energy released by nuclear disintegration.  In retrospect, the formula E=mc&sup2; explains the non-integer character of atomic weights.

:By the way, we can clearly see that atomic weights of pure isotopes are not integers, and that it has something to do with the energy released by nuclear disintegration.  In retrospect, the formula E=mc&sup2; explains the non-integer character of atomic weights.

Making special cases like this is unsatisfactory, of course.

Making special cases out of nuclear interaction versus non-nuclear ones is unsatisfactory, of course.

We do this for a few other interactions, including the explosion of TNT.  This would include the Lithium-plus-Hydrogen and Uranium fission phenomena described above.  We won't bother with the details. As observational scientists, we look for patterns in the behavior of nature.  We make a table:

We do this for a few other interactions, including the explosion of TNT.  This would include many other radioactive decays, and the Uranium fission phenomena described above.  We won't bother with the details.

 

As observational scientists, we look for patterns in the behavior of nature.  We make a table:

{| class="wikitable" border="1" cellpadding="8" cellspacing="0"  

{| class="wikitable" border="1" cellpadding="8" cellspacing="0"  

|-

|-

| alpha decay of Ra-226

| alpha decay of Ra-226

| 2.04 * 10¹²

| 2.040 * 10¹²

| .00002301 kg

| .00002299 kg

|-

|-

| alpha decay of Rn-222

| alpha decay of Rn-222

| 2.39 * 10¹²

| 2.386 * 10¹²

| .00002703 kg

| .00002713 kg

| .018623082 kg

|}

|}

We plot these, and a few others, not shown, on graph paper, and find to our amazement that the relationship is linear.

We plot these, and a few others, not shown, on graph paper, and find to our amazement that the relationship is linear.

For Ra, m/E = .112794118 E-16

For Radium decay, m/E = .1126 * 10^-16

For Po, m/E = .113096234 E-16

For Polonium decay, m/E = .1137 * 10^-16

 

For the Cockcroft-Walton experiment, m/E = .1116 * 10^-16

If this is linear, the mass defect for TNT would have been .47 * 10^-10.  We couldn't possibly have measured this.

As a linear relationship, the mass defect for TNT would have been .47 * 10^-10.  We couldn't possibly have measured this.

So we can rewrite the rule for conservation of mass in a more satisfactory way:

So we can rewrite the rule for conservation of mass in a more satisfactory way:

:::*'''In all interactions, there is a loss of mass, equal to about .113 * 10^-16 kg per Joule of energy released.'''

::*'''In all interactions, there is a loss of mass, equal to about .112 * 10^-16 kg per Joule of energy released.'''

What we thought was exact conservation is just very nearly exact, and we hadn't been able to measure it before.

What we thought was exact conservation is just very nearly exact, and we hadn't been able to measure it before.

But maybe there's more.  This constant has dimensions of kilograms per Joule.  From high-school physics, we know that that is seconds squared divided by meters squared.  That is, it is the reciprocal of the square of a velocity.  We calculate that velocity.  It is about 2.97 * 10⁸ meters per second.  Very close to the speed of light!  Very interesting!  (The calculations above were not extremely precise.  The formula has been verified with great precision, but not here.)

But maybe there's more.  This constant has dimensions of kilograms per Joule.  From high-school physics, we know that that is seconds squared divided by meters squared.  That is, it is the reciprocal of the square of a velocity.  We calculate that velocity.  It is about 2.97 * 10⁸ meters per second.  Very close to the speed of light!  Very interesting!  (The calculations above were not extremely precise.  The formula has been verified with great precision, but not here.)

We don't understand why (that will have to wait for the invention of relativity), but we can formulate a hypothesis:

Since we are just making empirical observations, we don't understand why this is so (that will have to wait for the invention of relativity), but we can formulate a hypothesis:

:::*'''In all interactions, there is a loss of mass, equal to <math>\frac{1}{c^2}</math> times the amount of energy released.'''

::*'''In all interactions, there is a loss of mass, equal to <math>\frac{1}{c^2}</math> times the amount of energy released.'''

We don't have to give the units any more, since everything is now dimensionally correct.

We don't have to give the units any more, since everything is now dimensionally correct.

This is a common way that scientific discoveries are made.  But, as stated above, this is not how E=mc&sup2; was discovered.  But it could have been. But in the event, it wasn't.

This sort of inductive approach is a common way that scientific discoveries are made.  For example, Newton's law of motion, F=ma, is empirical and backed up by enormous amounts of confirming data.

 

The formula E=mc&sup2; '''could have''' been discovered this way in our alternative universe, but, as stated above, that is not historically how it happened. It was formulated on theoretical grounds in order to obtain conservation of momentum in relativistic interactions, and then confirmed by experiment.

==See also==

==See also==