Nuclear Engineering

Fundamentals of Nuclear Reactions, Energy and Model

Energetic of nuclear reaction — Q value

The first and foremost point of nuclear reaction is to understand whether a nuclear reaction yields energy or just has a reverse effect. Therefore, it is important to calculate energy gain or loss in a nuclear reaction in terms of its Q-value. Let us first calculate it from the following consideration:


When a target nucleus is bombarded with an energetic projectile a different nucleus forms along with another particle. In this process either energy gets released or absorbed depending upon how masses of the reactants change, and is known as nuclear reaction. Nuclear physicists understood this process on the basis of formation of a ‘compound nucleus’ when a projectile interacts with a target nucleus. The compound nucleus can decay in different channels forming various product nuclei and other particles and does not remember the way it was formed. The application of nuclear reaction is immense. The two most important nuclear reactions, i.e. fission and fusion are the most efficient ways of electrical power production and hence are part and parcel of human civilization.

Let us first discuss what the Q-value of a nuclear reaction is and how to calculate it.

In general, any nuclear reaction can be represented as

a + A = B + b —————— (m3.1)

where, A = target nucleus with mass M_A B = Product nucleus of mass M_B a = incident particle or projectile of mass m_a b = product particle of mass m_b Let us take k_A, k_B, k_a, k_b be the kinetic energies of A, B, a and b respectively. From conservation of energy principle, we can write (K.E. + Rest mass energy)_L.H.S. = (K.E. + Rest mass energy)_R.H.S ——————– (m3.2) or, M_Ac² + (k_a + m_ac²) = (k_B +M_Bc²) + (k_b +m_bc²) —————– (m3.3) Considering target nucleus is at rest i.e., k_A = 0. or, k_B + k_b –k_a = (M_A +m_a –M_B –m_a)c² —————– (m3.4) where, c is the velocity of light in vacuum. The difference in K.E. of reaction products and projectile is known as the energy balance or Q-value of the reaction. ∴ Q = k_B + k_b –k_a = (M_A +m_a –M_B –m_a)c² ——————-(m3.5) Depending on whether Q-value is positive or negative, the nuclear reactions can be classified into two categories; namely, exoergic and endoergic. In exoergic reaction, (M_A +m_a) > (M_B +m_b), i.e. (k_B +k_b) > k_a and hence Q > 0. When the input mass is greater than the output mass, some mass is lost in the form of energy at the expense of lost mass. On the other hand, in endoergic reaction Q < 0, i.e. the output mass being larger than the input mass, some mass has been created at the expense or ‘loss’ of the output kinetic energy. Let us see the following example to understand how to calculate Q-value of a typical nuclear reaction.

Example m3.1: The masses of the different nuclei taking part in reaction in amu are as follows: = 7.01822, = 1.00814, = 1.00898 and mass of the product nucleus = 7.01915. Calculate the Q-value of this reaction in MeV. Is it exoergic or endoergeic? Solution: Masses of the reactants = 7.01822 + 1.00814 = 8.02636 amu Masses of the products = 7.01915 + 1.0898 = 8.02813 amu Q-value of the reaction = 8.02636-8.02813 = -0.00117 amu = -0.00117 X 931 MeV = -1.089 MeV The negative sign of the Q-value indicates that the reaction is endoergic.


Now we will see the physical quantities those remain conserved during a nuclear reaction. You know that when two billiard balls hit each other, their motion after collision is strictly decided by the conservation of energy and momentum. Similarly, in all physical processes some quantities must be conserved. In nuclear reaction the following quantities are conserved:

(a) Conservation of energy (including mass-energy) (b) Conservation of linear momentum (c) Conservation of angular momentum (spin and orbital) (d) Conservation of electric charge (e) Conservation of statistics and (f) Conservation of parity (except in weak interaction).


Let me just explain the conservation of last two quantities, i.e. statistics and parity. It is known that Fermi-Dirac (F-D) statistics is applied to a particle of half-integral spin (e.g. a nucleon) or a group of nucleons of odd number. When the number of nucleons is even, Bose-Einstein (B-E) statistics is applied, which is otherwise applicable to particles of integral spin. Let us denote F-D statistics by -1 and B-E statistics by +1¹. The simplest example of conservation of statistics can be shown in electron-positron annihilation . The statistics in the left hand side of the above reaction is (-1) X (-1) = +1. Annihilation process gives rise two gamma photons, which are bosons and hence the statistics in the right hand side is (+1) X (+1) = +1. Parity is a purely mathematical concept, which describes the kind of spatial symmetry of physical phenomena. If the sign of the wave function ψ may or may not change in sign with reflection symmetry (i.e. ). First case it is known as even parity and the second case the odd parity. Therefore,
		Even parity
		Odd parity


The product of the parities of individual particle gives the parity of a system of particles. All the elementary particle and nuclear energy state have either even or odd parity. Apart from nuclear reactions involving weak interaction, parity is conserved in all other nuclear reactions. However, in weak interaction a quantity combining charge and parity remains We know that nucleons being spin ½ particles, satisfy Fermi-Dirac statistics.

Let us now address the question: How nuclear reaction takes place? —Compound nucleus formation For relatively lower projectile energies, reactions may not involve a direct mechanism, but may proceed via an unstable intermediate state known as compound nucleus state. In this case, the incident particle loses so much energy that it cannot escape the struck nucleus. The compound nucleus thus formed has its mass number equal to the mass numbers of the target nucleus and that of the projectile. Compound nucleus is in highly excited state and therefore breaks into a product nucleus and other particles. Classically, compound nucleus can be understood by this qualitative manner. According to liquid drop model, nucleus can be considered to be a conglomeration of particles held together by attractive forces acting between them. An energetic nucleon approaching this nucleus enters into the target nucleus under an attractive potential and probably collides with one of the nucleons, and the energy is shared between both of them. Each nucleon again has the probability of colliding with other nucleons and thereby sharing the energy and ultimately shared between many nucleons. In this situation, it is called a compound nucleus. A typical example of nuclear reaction through compound nuclear formation is



One important characteristics of compound nucleus formation theory is that, compound nucleus may be formed in a number of different ways; however, the properties of compound nucleus are independent of the mode of formation. One important concept in the compound nucleus picture of nuclear reaction is the ‘level width’. The compound nucleus, on an average, remains in a given excited state for a certain time before decaying either through particle or γ-ray emission. The reciprocal of this mean life time, τ, is called the decay constant which represents the probability per unit time of the emission of a particle or γ-ray. The level width, Γ is defined as


where, λ is the decay constant. From energy-time uncertainty principle,



considering Δt to be the time uncertainty corresponding to the energy uncertainty ΔE, to be the mean life time of an excited state, the level width Γ is then identified with ΔE. In nuclear reaction, different decay modes posses different level widths. The total width may be written as the sum of the partial widths



where etc. are the partial level widths for the emission of deuteron, α-particles, neutrons, γ-rays etc. The concept of level width is useful in that it can be obtained from the measurement of resonances. By knowing the total level width, the mean life time can be calculated from the relation



Following example illustrates mean life of a compound nucleus.


Example m3.2: The reaction shows among others a resonance for all excitation energy of the compound nucleus of 13.23 MeV. The width of this level as found experimentally is 130 keV. Calculate the mean life of the nucleus for this excitation.¹ Solution: The reaction is The mean lifetime is given by

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