   # How much energy is released in the fission of 1 kg of 2 3 5 U according to the equation below? Ellie experimentally determined masses of the products are 136.92532 u and 96.91095 u, respectively. (The masses of 2 3 5 U and the neutron are given in Example Problem 14.6.) U 92 235 + n 0 1 → T 52 137 e + Z 40 97 r + 2 n 0 1 EXAMPLE PROBLEM 14.6 Calculate the energy released by a nucleus of uranium-235 if it splits into a barium-141 nucleus and a krypton-92 nucleus according to the equation above. Strategy Einstein's equation ( E = m c 2 ) relates the energy released to the difference in mass between the fissile uranium nuclide and the resulting fission products. The fission reaction is described by the equation above, so we just need to account for I the masses of all participating particles. ### Chemistry for Engineering Students

4th Edition
Lawrence S. Brown + 1 other
Publisher: Cengage Learning
ISBN: 9781337398909 ### Chemistry for Engineering Students

4th Edition
Lawrence S. Brown + 1 other
Publisher: Cengage Learning
ISBN: 9781337398909
Chapter 14, Problem 14.53PAE
Textbook Problem
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## How much energy is released in the fission of 1 kg of 235U according to the equation below? Ellie experimentally determined masses of the products are 136.92532 u and 96.91095 u, respectively. (The masses of 235U and the neutron are given in Example Problem 14.6.)   U 92 235 + n 0 1 → T 52 137 e +   Z 40 97 r + 2   n 0 1 EXAMPLE PROBLEM 14.6Calculate the energy released by a nucleus of uranium-235 if it splits into a barium-141 nucleus and a krypton-92 nucleus according to the equation above.Strategy Einstein's equation ( E = m c 2 ) relates the energy released to the difference in mass between the fissile uranium nuclide and the resulting fission products. The fission reaction is described by the equation above, so we just need to account for I the masses of all participating particles.

Interpretation Introduction

To determine: The amount of energy released in the fusion of 1kg of 235U according to the equation

92235U+01n52137Te+4097Zr+201n

### Explanation of Solution

92235U+01n52137Te+4097Zr+201n

Apply Einstein’s equation

E=Δmc2

Δm= Mass of product mass of reactant

C: speed of light.

Mass of 1 atom of U=235.0439231u(given)

Mass of 1 neutron =1.0086649u(given)

Mass of 1 atom of Te=136.92532u(given)

Mass of 1 atom of Zr=96.91095u(given)

Mass of reactant

=mU+mn=235.0439231u+1.0086649u=236.0525880u

Mass of product

=mTe+mZr+2(mn)=136.92532u+96.91095u+(2×1.0086649)u=235.8535989u

Δm=mass of productmass of reactants=235.85359984u236.0525880u=0.1989882u

Energy released in the fusion of the U-atom =(0.1989882u)×C2

1u=1.66053886×1027kgC=3×108m/s

=(Δm)C2=(0

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