2006-07-30 04:28:10 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Gamma rays of sufficient energy often convert into electron-positron pairs when they have a glancing collision with a heavy nucleus. And particel accelerators
2006-07-30 04:34:00 · answer #1 · answered by yankovicfan6 3 · 0⤊ 0⤋
During a fission reaction, or the splitting of an atomic nucleus, the resulting parts of the split do not have total mass equalling the original atom. Where did the mass go? It was converted into energy. The rate of conversion is c^2. As, is the Energy [E]=mass[m]*the speed of light in a vacuum[c] squared. This also works for a fusion reaction. The mass of the original atoms that end up being fused together is a little more than the resulting mass. Some of the mass has been converted to energy.
2006-07-30 12:25:36 · answer #2 · answered by quntmphys238 6 · 0⤊ 0⤋
When a neutron combines with a Uranium atom during fission, the products of this reaction are Barium, Krypton, 2.4 neutrons (this is an average of many fission events), and energy.
n(1) + U(235) = Ba(144) + Kr(89) + n(2.4) + energy
If you add up the masses on both sides of the reaction (convervation of mass) the mass on the right side is a little low. The missing mass has been converted into that energy. And the amount of energy can be found with the famous equation e=mc^2
2006-07-30 11:55:48 · answer #3 · answered by Steve S 4 · 0⤊ 0⤋
Einstein suggested this theory to prove the atomic model formed by rutherford.If the electrons revolvearound the nucleus,then they get the energy by the conversion of its mass to energy
2006-07-30 11:36:05 · answer #4 · answered by the great man of lake mauvia 2 · 0⤊ 0⤋
Another example of an energy to mass conversion occurs in supernovae, where iron nucleii fuse to form heavier elements. Often, the elements that form have slightly more mass than the nucleii that fused to form the element.
2006-07-30 17:28:03 · answer #5 · answered by Mad Scientist Matt 5 · 0⤊ 0⤋
The relationship is through the famous equation, e = mc^2
The gamma ray explanation that's offered by that other guy is really the only way I can recall of a visible method where this is shown.
2006-07-30 11:40:45 · answer #6 · answered by ddr_machine 2 · 0⤊ 0⤋
Arc welding. Soldering copper pipe. Ice and snow.
2006-07-30 11:35:48 · answer #7 · answered by Anonymous · 0⤊ 0⤋
i wish all such stuff would start taking place in our daily lives.
2006-07-30 11:41:36 · answer #8 · answered by smarts 1 · 0⤊ 0⤋