Physics, asked by simransodhi013, 9 months ago

how is energy released in the fission reaction determined give an example​

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Answered by gavitanmol74
1

Answer:

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Answered by shwetayadav1701011
1

Answer:

Nuclear fission is the splitting of the nucleus of an atom into nuclei of lighter atoms, accompanied by the release of energy, brought on by a neutron bombardment. The original concept of this nuclei splitting was discovered by Enrico Femi in 1934—who believed transuranium elements might be produced by bombarding uranium with neutrons, because the loss of Beta particles would increase the atomic number. However, the products that formed did not correlate with the properties of elements with higher atomic numbers than uranium (Ra, Ac, Th, and Pa). Instead, they were radioisotopes of much lighter elements such as Sr and Ba. The amount of mass lost in the fission process is equivalent to an energy of  3.20×10−11J .Consider the neutron bonbardment

U92235+n10→U92236→fission products(6)

which releases 3.20×10−11J per U235 atom.

How much energy would be released if 1.00g of U235 were to undergo fission?

SOLUTION

(1.00gU235)×(1molU235235gU235)×(6.022×1023atomsU2351molU235)×(3.20×10−11J1atomU235)=8.20×1010J(7)

Clearly, the fission of a small amount of atoms can produce an enormous amount of energy, in the form of warmth and radiation (gamma waves). When an atom splits, each of the two new particles contains roughly half the neutrons and protons of the original nucleus, and in some cases a 2:3 ratio.Consider the neutron bonbardment

U92235+n10→U92236→fission products(6)

which releases 3.20×10−11J per U235 atom.

How much energy would be released if 1.00g of U235 were to undergo fission?

SOLUTION

(1.00gU235)×(1molU235235gU235)×(6.022×1023atomsU2351molU235)×(3.20×10−11J1atomU235)=8.20×1010J(7)

Clearly, the fission of a small amount of atoms can produce an enormous amount of energy, in the form of warmth and radiation (gamma waves). When an atom splits, each of the two new particles contains roughly half the neutrons and protons of the original nucleus, and in some cases a 2:3 ratio.

Explanation:

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