Question

Tritium is naturally rare, but can be produced by the reaction\(n{ + ^2}H{ \to ^3}H + \gamma \). How much energy in \(MeV\) is released in this neutron capture?

Short answer

The amount of energy that is released in this neutron capture is\(E = 6.26{\rm{ }}MeV\).

Step-by-step solution

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.

Information Provided

• The reaction for Tritium is: \(n{ + ^2}H{ \to ^3}H + \gamma \).

Analysing the reactions

Calculate the energy release for the following reaction –

\(n{ + ^2}H{ \to ^3}H + \gamma \)

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 –

\(E = \Delta m{c^2}\)

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 \(eV\) –

\(lu = \frac{{931.5{\rm{ }}MeV}}{{{c^2}}}\)

Now analyse the reactions –

\(^1H{ + ^1}H{ \to ^2}H + {e^ + } + {v_e}\)

Looking in the Appendix for the atomic masses the following is obtained–

\(\begin{array}{c}^3H = 3.016050{\rm{ }}u\\^2H = 2.014102{\rm{ }}u{\rm{ }}\\neutron{\rm{ }} = 1.008665{\rm{ }}u\end{array}\)

Energy Calculation

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

\(\Delta m = 1.008665{\rm{ }}u + 2.014102{\rm{ }}u - 3.016050{\rm{ }}u\)

And the result is obtained as –

\(\Delta m = 0.006717{\rm{ }}u\)

And from the energy relation it is obtained that –

\(\begin{array}{c}E = \Delta m{c^2}\\ = 0.00045084 \cdot \frac{{931.5MeV}}{{{c^2}}} \cdot {c^2}\end{array}\)

And the result is obtained as –

\(E = 6.26{\rm{ }}MeV\)

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