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Tritium is naturally rare, but can be produced by the reactionn+2Hโ†’3H+ฮณ. How much energy in MeV is released in this neutron capture?

Short Answer

Expert verified

The amount of energy that is released in this neutron capture isE=6.26MeV.

Step by step solution

01

Concept Introduction

The power kept in the centre of an atom is known as nuclear energy. All stuff in the cosmos is made up of tiny particles called atoms. The centre of an atom's nucleus is typically where most of its mass is concentrated. Neutrons and protons are the two subatomic particles that make up the nucleus. Atomic bonds that hold them together carry a lot of energy.

02

Information Provided

  • The reaction for Tritium is: n+2Hโ†’3H+ฮณ.
03

Analysing the reactions

Calculate the energy release for the following reaction โ€“

n+2Hโ†’3H+ฮณ

The energy release is happening due to the different total mass of nuclei before and after the nuclear reaction. Energydifference can be calculated by the free mass-energy formula โ€“

E=ฮ”mc2

To get these mass differences take parent nuclei and subtract from them daughter nuclei and masses of all otherby-products of the nuclear reaction. Since all atomic masses will be expressed with the unified atomic unit, remember how to convert it to eV โ€“

lu=931.5MeVc2

Now analyse the reactions โ€“

1H+1Hโ†’2H+e++ve

Looking in the Appendix for the atomic masses the following is obtainedโ€“

3H=3.016050u2H=2.014102uneutron=1.008665u

04

Energy Calculation

So, for the mass difference, subtract atomic masses of parent and daughter nuclei, deuterium, and neutron from tritium โ€“

ฮ”m=1.008665u+2.014102uโˆ’3.016050u

And the result is obtained as โ€“

ฮ”m=0.006717u

And from the energy relation it is obtained that โ€“

E=ฮ”mc2=0.00045084โ‹…931.5MeVc2โ‹…c2

And the result is obtained as โ€“

E=6.26MeV

Therefore, the neutron capture releases more than 6 million electron-volts of energy.

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Most popular questions from this chapter

(a) Calculate the energy released in the neutron-induced fission reactionn+235Uโ†’92Kr+142Ba+2n,

Given m(92Kr)=91.926269u and

m(142Ba)=141.916361u.

(b) Confirm that the total number of nucleons and total charge are conserved in this reaction.

The naturally occurring radioactive isotope 232Th does not make good fission fuel, because it has an even number of neutrons; however, it can be bred into a suitable fuel (much as 238U is bred into239P).

(a) What are Z and N for232Th?

(b) Write the reaction equation for neutron captured by 232Th and identify the nuclide AXproduced inn+232Thโ†’AX+ฮณ.

(c) The product nucleus ฮฒโˆ’decays, as does its daughter. Write the decay equations for each, and identify the final nucleus.

(d) Confirm that the final nucleus has an odd number of neutrons, making it a better fission fuel.

(e) Look up the half-life of the final nucleus to see if it lives long enough to be a useful fuel.

The energy produced by the fusion of a 1.00โˆ’kg mixture of deuterium and tritium was found in Example Calculating Energy and Power from Fusion. Approximately how many kilograms would be required to supply the annual energy use in the United States?

Fallout from nuclear weapons tests in the atmosphere is mainly 90Sr and 137Cs , which have 28.6 - and 32.2y half-lives, respectively. Atmospheric tests were terminated in most countries in 1963, although China only did so in 1980. It has been found that environmental activities of these two isotopes are decreasing faster than their half-lives. Why might this be?

Verify that the total number of nucleons, total charge, and electron family number are conserved for each of the fusion reactions in the carbon cycle given in the above problem. (List the value of each of the conserved quantities before and after each of the reactions.)

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