Question

(a) Calculate the energy released in the neutron-induced fission reaction\(n{ + ^{239}}Pu{ \to ^{96}}Sr{ + ^{140}}Ba + 4n\), given \(m{(^{96}}Sr) = 95.921750{\rm{ }}u\)

And

\(m{(^{140}}Ba) = 139.910581{\rm{ }}u\).

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

Short answer

(a) The energy released in the reaction\(n{ + ^{239}}Pu{ \to ^{96}}Sr{ + ^{140}}Ba + 4n\)is\(E = 180.55MeV\).

(b) It is confirmed that the total number of nucleons and total charge are conserved in \(n{ + ^{239}}Pu{ \to ^{96}}Sr{ + ^{140}}Ba + 4n\) reaction.

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.

Energy calculation

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 reaction –

\(n{ + ^{239}}Pu{ \to ^{96}}Sr{ + ^{140}}Ba + 4n\)

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

\(\begin{align}{}^{96}Sr & = 95.921750{\rm{ }}u\\^{140}Ba & = 139.910581{\rm{ }}u\\^{239}Pu & = 239.052157{\rm{ }}u\\n & = 1.008665{\rm{ }}u\end{align}\)

Expression for mass difference will be –

\(\Delta m = ({m_n} + m\left( {^{239}Pu} \right) - \{ m\left( {^{96}Sr} \right) + m\left( {^{140}Ba} \right) + 4{m_n})\} \)

Plugging in the values –

\(\begin{align}{}\Delta m & = 1.008665{\rm{ }}u + 239.052157{\rm{ }}u - \{ 95.921750{\rm{ }}u + 139.910581{\rm{ }}u + 4 \times 1.008665{\rm{ }}u\} \\\Delta m & = 0.193831{\rm{ }}u\end{align}\)

So, from the energy relation it is obtained that –

\(\begin{align}{}E & = \Delta m{c^2}\\ & = 0.193831 \times \frac{{931.5MeV}}{{{c^2}}} \times {c^2}\\E & = 180.55MeV\end{align}\)

Therefore, the energy value for the reaction is obtained as \(E = 180.55MeV\).

Step 3:Number of nucleons and charge

To check that everything is conserved we carefully analyse nuclear reaction –

\(_0^1{n_1} + { - ^{239}}Pul \to _{38}^{96}Sr + _{56}^{140}56 + 4_0^1{n_1}\)

For the nucleon number, we got: \(1 + 239\) on the left side, and \(96 + 140 + 4\) on the right side, both ofthem gives \(240\), which confirms that it is conserved.

For the charge number neutron have zero charges, so on the left, we got \(94\), and on the right, wehave \(56\) and \(38\), which also gives \(94\), which confirms charge conservation.

Therefore, all the quantities are conserved.