Question

Nuclear fission and electrical energy

Electrical energy currently produced by nuclear reactors is derived from the decay of 235 92U, through a process known as nuclear fission, whereby a slow-moving neutron is captured by the 235 92U nucleus, creating 236 92U. The uranium nucleus 236 92U is very unstable, and it will immediately decay into two lighter elements and release neutrons. Some of the neutrons will go on to destabilize other nuclei, creating a chain reaction. During fission reactions the total mass of the products is less than the total mass of reactants. The difference between these two totals is called the mass defect, Δm. The energy equivalent of this mass, ΔE, can be calculated when Einstein's equation is written as

ΔE=Δmc2

where c = 2.998×10⁸ m/s is the speed of light.

Part A

When 235 92U captures a neutron and decays, one possible set of fission products is rubidium-90, 9037Rb, and cesium-144, 144 55Cs. The reaction also produces two neutrons as shown in the reaction

235 92U+10n→9037Rb+144 55Cs+210n

Using the mass data given in the table determine how much energy, in joules, is released when one mole of uranium-235 decays.

Particle	Exact mass (amu)
10n	1.0086649
144 55Cs	143.9320773
9037Rb	89.9148027
235 92U	235.0439222

Express your answer to four significant figures and include the appropriate energy per mol units.

Hints

ΔE =

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Nuclear binding energy

When the actual measured mass of an atom is compared to the sum of the component subatomic particles (electrons, protons, and neutrons), the mass of the isotope is always smaller. This mass loss or defect has been converted to energy in the formation of the atom. The energy equivalent of the missing mass is known as the total nuclear binding energy and can be calculated from the mass defect. Binding energies are often expressed in electronvolts (eV). The mass defect can be related to this energy term as

1 amu=9.31×108 eV

where a mass change of 1amu corresponds to an energy release of 9.31×108 eV. Often in such calculations the total mass of an electron (0.0005485 amu) and a proton (1.0072765 amu) is expressed in terms of the mass of a 11H atom, or 1.0078250 amu.

Part B

The predominant isotope of gold, 197 79Au, has an experimentally determined exact mass of 196.967 amu. What is the total nuclear binding energy of gold in electronvolts per atom?

Express your answer to three significant figures and include the appropriate units.

Hints

ΔE =

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Part C

You have been chosen as the recipient of a $5 billion grant to investigate the rather unexplored field of low mass fission. The government wants you to create a reactor that will split scandium-45, 4521Sc, by means of the following reaction:

4521Sc→2010Ne+2311Na+210n

The experimentally determined masses of the isotopes involved are given in the table.

Particle	Exact mass (amu)
4521Sc	44.95591
2311Na	22.98977
2010Ne	19.99244
10n	1.00866

Consider the thermodynamic feasibility of this fission reaction as an alternative to the uranium fission reaction described above. Which of the following would be the correct assessment of this reaction based on thermodynamic calculations?

Hints

You have been chosen as the recipient of a $5 billion grant to investigate the rather unexplored field of low mass fission. The government wants you to create a reactor that will split scandium-45, , by means of the following reaction:

The experimentally determined masses of the isotopes involved are given in the table.

Particle	Exact mass ()
	44.95591
	22.98977
	19.99244
	1.00866

Consider the thermodynamic feasibility of this fission reaction as an alternative to the uranium fission reaction described above. Which of the following would be the correct assessment of this reaction based on thermodynamic calculations?

This reaction will give us energy because the products weigh more than the reactants.

This reaction will not give us energy because the products weigh more than the reactants.

This reaction will give us energy because the products weigh less than the reactants.

This reaction will not give us energy because the products weigh less than the reactants.

This reaction will not give us energy because the products weigh the same as the reactants.

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Answer 1

Part a

First, add up the change in masses.
Start: U-235 + n = 235.0439222 + 1.0086649 = 236.0525871
Finish: Rb-90 + Cs-144 + 2n = 89.9148027 + 143.9320773 + 2*1.0086649 = 235.8642098
The difference, per U-235 going in, is:

236.0525871 - 235.8642098 = 0.1883773 amu
Convert to kg:
1 amu = 1.66e-27 kg

0.1883773 amu * 1.66e-27 kg/amu = 0.312706318e-27 kg

Convert to energy equivalent: E = mc^2
0.312706318e-27 kg * 9e16 m^2 / s^2 = 2.814356862e-11 J per U-235 atom.

Multiply by Avogadro's number to get Joules/mole
2.814356862e-11 J/atom * 6.02e23 atom/mole = 16.94e12 J = 16.94e9 KJ

Part b

Use either "1 amu = 9.31x10^8 eV" or

"E= m*c2

but not both. They're equivalent.

1amu = 9.31x10^8 eV (or...)

E = 1.558amu * c^2
E = 1.66x10^-27kg * (3.00x10^8m/s)^2
E = 1.49x10^-10 J = 9.31x10^8 eV

So when you did this part:
deltam=1.674 amu(9.31x10^8 eV/1 amu) = 1.558x10^9 eV
then assuming 1.674amu is the correct delta m, you were done.

You can go the long method as well:
E = deltam * c^2
E = 1.674amu * c^2
E = 2.780x10^-27kg * (3.00x10^-8m/s)^2
E = 2.498x10^-10 J
E = 1.559x10^9 eV