Question

(a) Calculate the energy released in the neutron-induced fission (similar to the spontaneous fission in Example\(32.3\)) \(n{ + ^{238}}U{ \to ^{96}}Sr{ + ^{140}}Xe + 3n\), given \(m{(^{96}}Sr) = 95.921750{\rm{ }}u\) and \(m{(^{140}}Xe) = 139.92164{\rm{ }}u\).

(b) This result is about \(6{\rm{ }}MeV\) greater than the result for spontaneous fission. Why?

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

Short answer

(a) The energy released from the reaction\(n + {}^{238}U \to {}^{96}Sr + {}^{140}Xe + 3n\)is\(E = 177.08{\rm{ }}MeV\).

(b) In difference to the spontaneous fission we have by-products here which yields more energy.

(c) It is confirmed that the total number of nucleons and total charge are conserved in\(n + {}^{238}U \to {}^{96}Sr + {}^{140}Xe + 3n\)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 –

\(n + {}^{238}U \to {}^{96}Sr + {}^{140}Xe + 3n\)

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

\(\begin{array}{c}neutron = 1.008665{\rm{ }}u\\^{238}U = 238.05078{\rm{ }}u\\^{96}Sr = 95.92175{\rm{ }}u{\rm{ }}\\^{140}Xe = 139.92164{\rm{ }}u\end{array}\)

Expression for mass difference will be –

\(\Delta m = m(n) + m{(^{238}}U) - \{ m{(^{96}}Sr) + m{(^{140}}Xe) - 3m(n)\} \)

Plugging in the values –

\(\begin{array}{c}\Delta m = 1.008665{\rm{ }}u + 238.05078{\rm{ }}u - 95.92175{\rm{ }}u{\rm{ }} - 139.92164{\rm{ }}u - 31.008645{\rm{ }}u\\\Delta m = 0.1901{\rm{ }}u\end{array}\)

So, from the energy relation it is obtained that –

\(\begin{array}{c}E &= \Delta m{c^2}\\ &= 0.1901 \cdot \frac{{931.5MeV}}{{{c^2}}} \cdot {c^2}\\E &= 177.08{\rm{ }}MeV\end{array}\)

Therefore, the energy value for the reaction is obtained as \(E = 177.08{\rm{ }}MeV\).

Greater Result

In the spontaneous fission there are notby-products from stable to unstable uranium, meaning\(^{238}U\)goes directly to the\(^{239}U\). Because of the by-products and elements in

between this process is more energy-rich.

Therefore, because of by-products the result obtained is greater.

Number of nucleons and charge

Regarding the conservation of nucleons and charge, check the reaction –

\(n + {}^{238}U \to {}^{96}Sr + {}^{140}Xe + 3n\)

To be conserved we need a total charge on the left and right sides to be the same, as the total number of nucleons on the left and rightsides to be the same.

For the left side –

	Nucleons	Charges
\(n\)	\(1\)	\(0\)
\({}^{238}U\)	\(238\)	\(92\)

And for the right side –

	Nucleons	Charges
\(3n\)	\(3\)	\(0\)
\({}^{96}Sr\)	\(96\)	\(38\)
\({}^{140}Xe\)	\(140\)	\(54\)

Regarding the number of nucleons on the left side we have\(1 + 238 = 239\), and on the right side, there is\(3 + 96 + 140 = 239\), which isconserved.

When it comes to the conservation of the charge, on the left we got\(92\)from uranium itself, and on the right side, we got\(38 + 54 = 92\)from\(Sr\)and\(Xe\). The number is equal so the charge is also conserved.

Therefore, all the quantities are conserved.