http://physics.stackexchange.com/questi ... -reactionsSo that's why we say that mass is converted to energy in nuclear reactions: the "mass" that is being converted is really just binding energy, but there's enough of this energy that when you look at the nucleus as a particle, you need to factor in the binding energy to get the right mass.
However, the overall mass of the products is actually less, in real terms...Tero wrote:The old e=mc2 equation.
I've looked for the answer. In fusion, actual particles disappear: electron and positron.
In regular earthly nuclear reactions it's complicated:http://physics.stackexchange.com/questi ... -reactionsSo that's why we say that mass is converted to energy in nuclear reactions: the "mass" that is being converted is really just binding energy, but there's enough of this energy that when you look at the nucleus as a particle, you need to factor in the binding energy to get the right mass.
E = mc2 applies to any system that absorbs or emits energy. Even chemical reactions. When you burn hydrogen in oxygen to make water, heat energy is given off and there is a slight difference between the mass of the hydrogen and oxygen and the mass of the water. However, since the amount of energy produced is relatively small, the lost mass is almost impossible to measure.JimC wrote:However, the overall mass of the products is actually less, in real terms...Tero wrote:The old e=mc2 equation.
I've looked for the answer. In fusion, actual particles disappear: electron and positron.
In regular earthly nuclear reactions it's complicated:http://physics.stackexchange.com/questi ... -reactionsSo that's why we say that mass is converted to energy in nuclear reactions: the "mass" that is being converted is really just binding energy, but there's enough of this energy that when you look at the nucleus as a particle, you need to factor in the binding energy to get the right mass.
The binding energy showing as mass effect is real, as shown by the fact that you can't simply add up the mass of the various nucleons in a given isotope and get an exact answer.
Yes - Tero's original post shouldn't have conflated fusion (where two light nuclei combine to create a larger nuclei with a little less overall mass) with the annihilation reaction that occurs when positron and electron meet; in that case the 2 gamma rays emitted have all the energy provided by the original mass of the particles.Xamonas Chegwé wrote:E = mc2 applies to any system that absorbs or emits energy. Even chemical reactions. When you burn hydrogen in oxygen to make water, heat energy is given off and there is a slight difference between the mass of the hydrogen and oxygen and the mass of the water. However, since the amount of energy produced is relatively small, the lost mass is almost impossible to measure.JimC wrote:However, the overall mass of the products is actually less, in real terms...Tero wrote:The old e=mc2 equation.
I've looked for the answer. In fusion, actual particles disappear: electron and positron.
In regular earthly nuclear reactions it's complicated:http://physics.stackexchange.com/questi ... -reactionsSo that's why we say that mass is converted to energy in nuclear reactions: the "mass" that is being converted is really just binding energy, but there's enough of this energy that when you look at the nucleus as a particle, you need to factor in the binding energy to get the right mass.
The binding energy showing as mass effect is real, as shown by the fact that you can't simply add up the mass of the various nucleons in a given isotope and get an exact answer.
Rearranging Einstein's equation, Mass (in kilograms) = Energy (in joules) / 89,875,517,873,681,760 (c squared)
So, even if a few million joules are emitted in a reaction, the change in mass would still be measured in nanograms! It is only in nuclear reactions that the mass loss becomes experimentally measurable. However, even in the most efficient fusion reactions, the energy emitted is still only a tiny fraction of that which is locked up in matter.
...plus their kinetic energy.JimC wrote:Yes - Tero's original post shouldn't have conflated fusion (where two light nuclei combine to create a larger nuclei with a little less overall mass) with the annihilation reaction that occurs when positron and electron meet; in that case the 2 gamma rays emitted have all the energy provided by the original mass of the particles.Xamonas Chegwé wrote:E = mc2 applies to any system that absorbs or emits energy. Even chemical reactions. When you burn hydrogen in oxygen to make water, heat energy is given off and there is a slight difference between the mass of the hydrogen and oxygen and the mass of the water. However, since the amount of energy produced is relatively small, the lost mass is almost impossible to measure.JimC wrote:However, the overall mass of the products is actually less, in real terms...Tero wrote:The old e=mc2 equation.
I've looked for the answer. In fusion, actual particles disappear: electron and positron.
In regular earthly nuclear reactions it's complicated:http://physics.stackexchange.com/questi ... -reactionsSo that's why we say that mass is converted to energy in nuclear reactions: the "mass" that is being converted is really just binding energy, but there's enough of this energy that when you look at the nucleus as a particle, you need to factor in the binding energy to get the right mass.
The binding energy showing as mass effect is real, as shown by the fact that you can't simply add up the mass of the various nucleons in a given isotope and get an exact answer.
Rearranging Einstein's equation, Mass (in kilograms) = Energy (in joules) / 89,875,517,873,681,760 (c squared)
So, even if a few million joules are emitted in a reaction, the change in mass would still be measured in nanograms! It is only in nuclear reactions that the mass loss becomes experimentally measurable. However, even in the most efficient fusion reactions, the energy emitted is still only a tiny fraction of that which is locked up in matter.
Most of that weight loss is not actual mass conversion to energy, but neutrons which have escaped the rods, and become absorbed in the control rods or the material of the containment vessel.Tero wrote:They weigh the rods in the Finnish reactors. The table is in Finnish, but I calculated about 3% weight loss in the year that they added and removed the same number of rods, 2005. It does not happen every year, that the number matches.
Good point...Tero wrote:Plus there is also some gas product, Kr and Xe. The fuel absorbs some gas, not all.
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