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Publicado: 16 de septiembre de 2014
CENTRO DE BACHILLERATO TECNOLOGICO INDUSTRIAL Y DE SERVICIOS #92
NUCLEAR ENERGY
NOMBRE: Julio Enrique Ozuna Hernández.
PROFESOR: Francisco Jadón Lara.
ESPECIALIDAD:Mecánica Industrial
4to SEMESTRE GRUPO: “E”
NUMERO DE LISTA: 30
FECHA: 23 DE MAYO DEL 2014
.SAN CRISTOBAL DE LAS CASAS CHIAPAS.
We consider stability of nuclei whit respect toconversion of mass to energy. That such a conversion can exist is seen by considering the helium nuclus. A helium nucleus is believed to contain two neutrons and two protons. Since the mass of a neutronis 1.00866 A.M.U. We might expec that the mass of a helium nucleus woud be:
2(1.00866) +2(1.00732) = 4.03196 A.M.U.However, the experimentally observed mass of the helium nucleus is 4.002 A.M.U. ¿ what has happened to the missing mass of 0.030 A.M.U.? At present it is believed that, in forming a helium nucleusfrom protons and neutron, mass is converted to energy. The amount of energy equivalent to the lost mass can be calculated from the Einstein relation E= mc^2 where E is the energy equivalent in ergs(1 ergs is 2.39 x 10^-8 cal) of the mass M in grams and C is the velocity of light, 3.00x10^10 cm per sec. For the helium nucleus, the loss of 0.030 A.M.U. of mass corresponds to the liberation of 650million kcal per mole of helium. Conversely, the same amount of energy is required to break up 1 mole of helium nuclei into protons and neutrons, and so gives a measure of the binding energy of thenucleus.
TRANSPARENTES SIMILARES
A:A CALCULATED:CALCULARCORRESPONDS:CORRESPONDE
CONVERSION CONSIDER:CONSIDERAR CONVESELY:A LA INVERSA
NEUTRON CONVERTED:CONVERTIR ENERGY:ENERGIA
EQUIVALET:EQUIVALENTE EXIST:EXISTE...
NUCLEAR ENERGY
NOMBRE: Julio Enrique Ozuna Hernández.
PROFESOR: Francisco Jadón Lara.
ESPECIALIDAD:Mecánica Industrial
4to SEMESTRE GRUPO: “E”
NUMERO DE LISTA: 30
FECHA: 23 DE MAYO DEL 2014
.SAN CRISTOBAL DE LAS CASAS CHIAPAS.
We consider stability of nuclei whit respect toconversion of mass to energy. That such a conversion can exist is seen by considering the helium nuclus. A helium nucleus is believed to contain two neutrons and two protons. Since the mass of a neutronis 1.00866 A.M.U. We might expec that the mass of a helium nucleus woud be:
2(1.00866) +2(1.00732) = 4.03196 A.M.U.However, the experimentally observed mass of the helium nucleus is 4.002 A.M.U. ¿ what has happened to the missing mass of 0.030 A.M.U.? At present it is believed that, in forming a helium nucleusfrom protons and neutron, mass is converted to energy. The amount of energy equivalent to the lost mass can be calculated from the Einstein relation E= mc^2 where E is the energy equivalent in ergs(1 ergs is 2.39 x 10^-8 cal) of the mass M in grams and C is the velocity of light, 3.00x10^10 cm per sec. For the helium nucleus, the loss of 0.030 A.M.U. of mass corresponds to the liberation of 650million kcal per mole of helium. Conversely, the same amount of energy is required to break up 1 mole of helium nuclei into protons and neutrons, and so gives a measure of the binding energy of thenucleus.
TRANSPARENTES SIMILARES
A:A CALCULATED:CALCULARCORRESPONDS:CORRESPONDE
CONVERSION CONSIDER:CONSIDERAR CONVESELY:A LA INVERSA
NEUTRON CONVERTED:CONVERTIR ENERGY:ENERGIA
EQUIVALET:EQUIVALENTE EXIST:EXISTE...
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