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Q1. (a) Calculate the density of nuclear matter with nucleon mass = 1.67 x 10 ^-24 g and R ₀ = 1.24 fm. (b) How many grams of ¹² C can produce 8.2 x 10 ⁴ MJ energy, if 4 MeV of energy is produced when a ¹² C atom is burnt ? Express your result in tons.

Ans. (a) Calculation of Nuclear Matter Density The density of nuclear matter (ρ) is approximately constant, independent of the mass number (A), because the nuclear radius is given by R = R ₀ A ^1/3 , where R ₀ is a constant. The mass of a nucleus ≈ A m _n , where A is the mass number and m _n is the mass of a single nucleon. The volume of a nucleus V = (4/3)πR ³ = (4/3)π(R ₀ A ^1/3 ) ³ = (4/3)πR ₀ ³ A. Therefore, the density (ρ) can be calculated as: ρ = Mass / Volume = (A m _n ) / [(4/3)πR ₀ ³ A] = m _n / [(4/3)πR ₀ ³ ] The given values are: Nucleon mass (m _n ) = 1.67 x 10 ^-24 g R ₀ = 1.24 fm = 1.24 x 10 ^-13 cm (since 1 fm = 10 ^-13 cm) Substituting the values into the formula: ρ = (1.67 x 10 ^-24 g) / [(4/3)π (1.24 x 10 ^-13 cm) ³ ] ρ = (1.67 x 10 ^-24 g) / [(4/3) 3.14159 1.906624 x 10 ^-39 cm ³ ] ρ = (1.67 x 10 ^-24 g) / [7.986 x 10 ^-39 cm ³ ] ρ ≈ 2.09 x 10 ¹⁴ g/cm ³ Thus, the density of nuclear matter is approximately 2.09 x 10 ¹⁴ g/cm ³ . (b) Calculation of Mass of ¹² C Required First, we convert all energy values to the SI unit of Joules (J). Energy produced per ¹² C atom burnt = 4 MeV = 4 x 10 ⁶ eV = 4 x 10 ⁶ (1.602 x 10 ^-19 J/eV) = 6.408 x 10 ^-13 J Total energy required = 8.2 x 10 ⁴ MJ = 8.2 x 10 ⁴ x 10 ⁶ J = 8.2 x 10 ¹⁰ J Next, we calculate the number of ¹² C atoms needed: Number of atoms = Total Energy / Energy per atom = (8.2 x 10 ¹⁰ J) / (6.408 x 10 ^-13 J/atom) ≈ 1.279 x 10 ²³ atoms Now, we find the number of moles of ¹² C: Number of moles = Number of atoms / Avogadro’s Number (N _A ) = (1.279 x 10 ²³ ) / (6.022 x 10 ²³ mol ^-1 ) ≈ 0.2124 moles The molar mass of ¹² C is 12 g/mol. Therefore, the required mass is: Mass = Number of moles Molar mass = 0.2124 mol 12 g/mol ≈ 2.549 g Finally, we express this mass in tons (1 ton = 1000 kg = 10 ⁶ g): Mass in tons = 2.549 g / (10 ⁶ g/ton) = 2.549 x 10 ^-6 tons Thus, the required mass of ¹² C is 2.549 grams or 2.549 x 10 ^-6 tons .

Q2. Establish the four factor formula : K _∞ = ηεpf for an infinite reactor. How this relation would be modified if the reactor is made only of ²³⁵ U or is made finite.

Ans. Establishment of the Four Factor Formula The four-factor formula, K _∞ = ηεpf, describes the neutron multiplication cycle in a reactor of infinite size. It defines the ratio of the number of neutrons in one generation to the number in the preceding generation. Let’s establish it by following the life cycle of one generation, starting with N thermal neutrons being absorbed in the fuel. 1. Reproduction Factor (η): The average number of fission neutrons produced per thermal neutron absorbed by the fuel. If Σ _f ^fuel and Σ _a ^fuel are the macroscopic cross sections for fission and absorption for thermal neutrons in the fuel, and ν neutrons are produced per fission, then: η = ν (Σ _f ^fuel / Σ _a ^fuel ) After the absorption of N thermal neutrons, the number of fast neutrons produced is = Nη. 2. Fast Fission Factor (ε): This factor accounts for the fission caused by fast neutrons (energy > 1 MeV) in fissile isotopes like ²³⁸ U. It is defined as: ε = (Total number of fast neutrons from thermal + fast fission) / (Number of fast neutrons from thermal fission) Since fast fission produces extra neutrons, the value of ε is slightly greater than 1 (typically 1.02 to 1.04). The total number of neutrons after fast fission = Nηε. 3. Resonance Escape Probability (p): As the fast neutrons slow down, they pass through a resonance energy region where they have a high probability of being captured by non-fissile materials like ²³⁸ U. ‘p’ is the probability that a neutron will escape this resonance capture and slow down to thermal energies. The number of neutrons reaching thermal energy = Nηεp. 4. Thermal Utilization Factor (f): This is the ratio of thermal neutrons absorbed in the fuel to the total thermal neutrons absorbed in the reactor (fuel + moderator + structural materials). f = (Thermal neutrons absorbed in fuel) / (Total thermal neutrons absorbed in reactor) = Σ _a ^fuel / Σ _a ^total The number of thermal neutrons that will be absorbed in the fuel for the next generation = Nηεp f. The infinite multiplication factor, K _∞ , is defined as the ratio of neutrons produced in one generation to the neutrons absorbed in the preceding generation: K _∞ = (Thermal neutrons absorbed in next generation) / (Thermal neutrons absorbed in initial generation) K _∞ = (Nηεpf) / N K _∞ = ηεpf Modification for a Reactor made only of ²³⁵ U: If the reactor is made of pure ²³⁵ U only, then:

• Fast Fission Factor (ε) ≈ 1: In the absence of ²³⁸ U, there is no significant contribution from fast fission.
• Resonance Escape Probability (p) ≈ 1: ²³⁵ U has very small resonance absorption peaks compared to ²³⁸ U. Therefore, nearly all neutrons escape resonance capture during slowdown.

In this case, the formula simplifies to:

K

_∞

≈ ηf

. Since there is no moderator, f=1 (all absorptions are in fuel), so

K

_∞

≈ η

.

Modification for a Finite Reactor:

In a finite reactor, neutrons can leak out of the system’s boundaries. The four-factor formula only describes the neutron cycle in an infinite system. For a finite reactor, we must account for neutron leakage. This is done using the

effective multiplication factor (K

_eff

)

.

K

_eff

= K

_∞

* P

_NL

Where P

_NL

is the

non-leakage probability

. This can be split into two parts:

• P _F : The non-leakage probability for a fast neutron (while it is slowing down).
• P _T : The non-leakage probability for a thermal neutron (before it is absorbed).

So, P

_NL

= P

_F

P

_T

.

K

_eff

= ηεpf

P

_F

* P

_T

For a reactor to be critical, K

_eff

must equal 1.

Q3. (a) A 2.6 MeV neutron collides with hydrogen. Calculate the probability that the energy of neutron is within the energy range 0.63 and 0.75 MeV after collision. If a neutron loses 0.75 MeV in LAB system, what is the scattering angle in the CM system ? (b) On an average, the neutron loses half of its energy per collision with a target nucleus. How many elastic collisions are needed for a 2 MeV neutron to become a thermal neutron having an energy 0.4 eV?

Ans. (a) Neutron Collision with Hydrogen Part 1: Probability in Energy Range When a neutron collides elastically with a target nucleus, its energy E’ after the collision lies in a range relative to the initial energy E: αE ≤ E’ ≤ E, where α = ((A-1)/(A+1)) ² and A is the mass number of the target nucleus. For hydrogen, A = 1, so α = ((1-1)/(1+1)) ² = 0. This means the neutron’s energy E’ after collision is uniformly distributed between 0 and the initial energy E. The probability density function p(E’) is: p(E’) = 1/E, for 0 ≤ E’ ≤ E p(E’) = 0, otherwise Given: Initial neutron energy (E) = 2.6 MeV Energy range = 0.63 MeV to 0.75 MeV The probability of finding the neutron within this energy range is calculated by integration: P(0.63 ≤ E’ ≤ 0.75) = ∫ _0.63 ^0.75 p(E’) dE’ = ∫ _0.63 ^0.75 (1/E) dE’ = (1/E) [E’] _0.63 ^0.75 = (1/2.6) (0.75 – 0.63) = 0.12 / 2.6 ≈ 0.04615 Thus, the probability that the neutron’s energy is between 0.63 and 0.75 MeV after collision is approximately 0.0462 or 4.62% . Part 2: Scattering Angle in CM system If a neutron loses 0.75 MeV in the LAB system, its final energy (E’) after the collision is: E’ = E – energy loss = 2.6 MeV – 0.75 MeV = 1.85 MeV The relationship between the LAB energy (E’) and the CM (Center of Mass) scattering angle (θ _c ) is given by: E’ = (E/2) [(1 + α) + (1 – α)cos(θ _c )] For hydrogen (A=1, α=0), this simplifies to: E’ = (E/2) [1 + cos(θ _c )] Substituting the values: 1.85 = (2.6 / 2) [1 + cos(θ _c )] 1.85 = 1.3 (1 + cos(θ _c )) 1 + cos(θ _c ) = 1.85 / 1.3 ≈ 1.423 cos(θ _c ) = 1.423 – 1 = 0.423 θ _c = arccos(0.423) θ _c ≈ 64.97° Therefore, the scattering angle in the CM system is approximately 65° . (b) Number of Elastic Collisions Required When a neutron loses, on average, half of its energy per collision, its energy (E _n ) after n collisions will be: E _n = E ₀ (1/2) ⁿ where E ₀ is the initial energy. Given: Initial energy (E ₀ ) = 2 MeV = 2 x 10 ⁶ eV Final thermal energy (E _n ) = 0.4 eV We want to find n: 0.4 = (2 x 10 ⁶ ) (1/2) ⁿ (1/2) ⁿ = 0.4 / (2 x 10 ⁶ ) = 0.2 x 10 ^-6 = 2 x 10 ^-7 Taking the logarithm of both sides: n log(1/2) = log(2 x 10 ^-7 ) -n log(2) = log(2) + log(10 ^-7 ) -n log(2) = log(2) – 7 log(10) Using base-10 log (log(2) ≈ 0.301): -n * (0.301) = 0.301 – 7 -0.301n = -6.699 n = 6.699 / 0.301 ≈ 22.25 Since the number of collisions must be an integer, we must round up to the next higher integer to bring the neutron to or below the desired energy level. Therefore, 23 elastic collisions are needed for a 2 MeV neutron to become a 0.4 eV thermal neutron.

Q4. (a) Derive an expression for the slowing down time for neutrons in a scattering medium. If 2 MeV neutrons are slowed down to 0.025 eV in graphite, obtain the corresponding time for slowing down. Take Σ _s =0.385cm ^-1 , ξ=0.158 and m=1.67×10 ^-27 kg. (b) A neutron beam is incident on a stationary target of ¹⁹ F atoms. If the Q-value of the reaction ¹⁹ F(n, p) ¹⁹ O is -3.9 MeV, calculate the threshold energy of the reaction.

Ans. (a) Slowing Down Time for Neutrons Derivation of the Expression: The slowing down time (t _s ) is the average time it takes for a neutron to slow from an initial high energy E ₀ to a final low energy E _f . The average time between two successive collisions is dt = λ _s / v, where λ _s = 1/Σ _s is the scattering mean free path and v is the neutron’s speed. The average logarithmic energy decrement per collision is ξ = -dE/E, so we can write dE = -ξE per collision. The time taken for the energy to change by dE is dt. The rate of time change with respect to energy is: dt/dE = (time per collision) / (energy loss per collision) = (λ _s /v) / (ξE) The total slowing down time t _s is obtained by integrating from the initial energy E ₀ down to the final energy E _f : t _s = ∫ _{E f} ^{E ₀} (λ _s / (ξEv)) dE We know E = (1/2)mv ² , so v = √(2E/m). Substituting this into the formula: t _s = ∫ _{E f} ^{E ₀} [λ _s / (ξE √(2E/m))] dE t _s = (λ _s / ξ) √(m/2) ∫ _{E f} ^{E ₀} E ^-3/2 dE Performing the integration: ∫ E ^-3/2 dE = -2E ^-1/2 t _s = (λ _s / ξ) √(m/2) [-2E ^-1/2 ] _{E f} ^{E ₀} t _s = (2λ _s / ξ) √(m/2) [E _f ^-1/2 – E ₀ ^-1/2 ] Since E ₀ >> E _f usually, the term E ₀ ^-1/2 becomes negligible. t _s ≈ (2λ _s / ξ) √(m/2) (1/√E _f ) As v _f = √(2E _f /m), this can also be written as: t _s ≈ 2λ _s / (ξv _f ) Calculation of Slowing Down Time: Given values: E ₀ = 2 MeV = 3.204 x 10 ^-13 J E _f = 0.025 eV = 4.005 x 10 ^-21 J Σ _s = 0.385 cm ^-1 ⇒ λ _s = 1/Σ _s = 1/0.385 cm ≈ 2.597 cm = 0.02597 m ξ = 0.158 m = 1.67 x 10 ^-27 kg (This is the neutron mass) We will use the more exact formula: t _s = (2 0.02597 m / 0.158) √(1.67×10 ^-27 kg / 2) [1/√(4.005×10 ^-21 J) – 1/√(3.204×10 ^-13 J)] t _s ≈ 0.3287 √(8.35×10 ^-28 ) [1/(6.328×10 ^-11 ) – 1/(5.66×10 ^-7 )] t _s ≈ 0.3287 (2.89×10 ^-14 ) [1.58×10 ¹⁰ – 1.766×10 ⁶ ] Since 1.58×10 ¹⁰ >> 1.766×10 ⁶ , the approximation holds: t _s ≈ 0.3287 (2.89×10 ^-14 ) (1.58×10 ¹⁰ ) t _s ≈ 1.50 x 10 ^-4 s Thus, the slowing down time in graphite is approximately 1.50 x 10 ^-4 seconds or 0.15 milliseconds . (b) Threshold Energy of the Reaction The reaction ¹⁹ F(n, p) ¹⁹ O is an endoergic reaction because its Q-value is negative. For such a reaction to occur, the incident particle (neutron) must have a minimum kinetic energy, called the threshold energy (E _th ), to make the reaction possible. The threshold energy is calculated by the formula: E _th = -Q (1 + m _i / M _T ) Where: Q = Q-value of the reaction = -3.9 MeV m _i = mass of incident particle (neutron, m _n ≈ 1 amu) M _T = mass of target nucleus ( ¹⁹ F, M _F ≈ 19 amu) Plugging the values into the formula: E _th = -(-3.9 MeV) (1 + 1 / 19) E _th = 3.9 MeV (20 / 19) E _th = 3.9 MeV 1.05263 E _th ≈ 4.105 MeV Therefore, the threshold energy for the reaction is approximately 4.105 MeV . This means the incident neutron must have at least 4.105 MeV of kinetic energy for this reaction to take place.

Q5. (a) Using the method of spherical harmonics for a non-multiplying 1-D plane geometry, write the exact infinite coupled equations. Hence, for a large absorbing system, obtain equations under P ₁ approximation. (b) Discuss briefly the different kinds of initial and boundary conditions which help to solve the transport equation.

Ans. (a) Spherical Harmonics Method and P ₁ Approximation Exact Infinite Coupled Equations: For a non-multiplying medium in 1-D plane geometry, the steady-state Boltzmann transport equation (assuming isotropic scattering and source) is: μ (∂ψ(x, μ)/∂x) + Σ _t ψ(x, μ) = (Σ _s /2)∫ _-1 ¹ ψ(x, μ’)dμ’ + S(x)/2 where ψ(x, μ) is the angular neutron flux, μ = cos(θ), Σ _t is the total macroscopic cross section, Σ _s is the scattering cross section, and S(x) is the isotropic source. In the spherical harmonics method, we expand the angular flux in a series of Legendre polynomials P _l (μ): ψ(x, μ) = Σ _l=0 ^∞ [(2l+1)/2] ψ _l (x) P _l (μ) where ψ _l (x) are the l-th moments of the flux. Substituting this expansion into the transport equation, using the recurrence relation for Legendre polynomials: (2l+1)μP _l (μ) = (l+1)P _l+1 (μ) + lP _l-1 (μ) and multiplying by P _k (μ) and integrating over μ, we obtain an exact infinite set of coupled equations for the moments ψ _l (x): (l+1) (d/dx)ψ _l+1 (x) + l (d/dx)ψ _l-1 (x) + (2l+1)Σ _t ψ _l (x) = (2l+1)Σ _s ψ ₀ (x)δ _l0 + (2l+1)S _l (x) This is an infinite hierarchy of equations, valid for l = 0, 1, 2, … . δ _l0 is the Kronecker delta. Equations under P ₁ Approximation: In the P ₁ approximation, we truncate the series at l=1, assuming that ψ _l = 0 for l ≥ 2. This approximation is suitable for a large, weakly absorbing medium where the flux is nearly isotropic. Setting up the equations for l = 0: (1)(d/dx)ψ ₁ (x) + 0 + Σ _t ψ ₀ (x) = Σ _s ψ ₀ (x) + S ₀ (x) ⇒ (d/dx)ψ ₁ (x) + (Σ _t – Σ _s )ψ ₀ (x) = S ₀ (x) Since Σ _a = Σ _t – Σ _s , we get: 1. (d/dx)ψ ₁ (x) + Σ _a ψ ₀ (x) = S ₀ (x) Setting up the equations for l = 1: (2)(d/dx)ψ ₂ (x) + (1)(d/dx)ψ ₀ (x) + 3Σ _t ψ ₁ (x) = 3S ₁ (x) Using the P ₁ approximation (ψ ₂ = 0): 2. (d/dx)ψ ₀ (x) + 3Σ _t ψ ₁ (x) = 3S ₁ (x) These two equations represent the P ₁ approximation. ψ ₀ (x) is the scalar flux Φ(x), and ψ ₁ (x) is the neutron current J(x). These equations can be combined to yield the famous diffusion equation. (b) Initial and Boundary Conditions To solve the transport equation, specifying initial and boundary conditions is necessary to obtain a unique solution. Initial Condition: This defines the neutron distribution in the system at time t=0. Mathematically, it is written as: ψ( r , E, Ω , 0) = f( r , E, Ω ) where f is a known function describing the initial neutron flux. Boundary Conditions: These conditions specify the behavior of the neutron flux at the physical boundaries of the system. Some common types are:

• Vacuum Boundary Condition: This states that no neutrons enter the system from a vacuum. If n is the outward normal vector to the surface, then: ψ( r _s , E, Ω ) = 0, for all Ω such that Ω · n < 0 where r _s is a point on the surface. This is an exact condition.
• Interface Condition: At the interface between two different media, the angular flux must be continuous for all directions. ψ ₁ ( r _s , E, Ω ) = ψ ₂ ( r _s , E, Ω )
• Reflective Boundary Condition: If the system has a plane of symmetry, the flux of neutrons leaving in one direction is equal to the flux entering from the symmetric point in the opposite direction. ψ( r _s , E, Ω ) = ψ( r _s , E, Ω’ ), where Ω’ is the reflection of Ω .
• Extrapolated Boundary Condition: This is an approximate condition used in diffusion theory. It assumes that the scalar flux goes to zero at an “extrapolated distance” ‘d’ beyond the physical boundary. Φ(R + d) = 0, where R is the physical boundary.

Q6. (a) Write down the transport equation for a non-multiplying system. Obtain 1-D equation for a homogeneous moderating infinite slab. (b) State Fick’s law. Write its mathematical expression. Discuss its importance in neutron transport.

Ans. (a) The Transport Equation Transport Equation for a Non-multiplying System: The Boltzmann transport equation is a balance equation for neutrons in a given phase-space volume element. For a non-multiplying system, where there is no fission, the time-dependent equation can be written in terms of the angular neutron flux ψ( r , E, Ω , t): (1/v) (∂ψ/∂t) + Ω·∇ψ + Σ _t (r, E)ψ = ∫ ₀ ^∞ dE’ ∫ _4π dΩ’ Σ _s (r, E’→E, Ω’→Ω)ψ(r, E’, Ω’, t) + S(r, E, Ω, t) Each term in this equation has a physical meaning:

• (1/v) (∂ψ/∂t): The rate of change of the angular flux in time.
• Ω·∇ψ: The net leakage of neutrons out of the volume element due to streaming in direction Ω .
• Σ _t (r, E)ψ: The loss of neutrons from the flux due to collisions (absorption or scattering).
• ∫∫ Σ _s (…)ψ(…) dE’ dΩ’: The gain of neutrons into energy E and direction Ω via scattering from other energies E’ and directions Ω’.
• S(r, E, Ω, t): The rate of production of neutrons from external sources.

1-D Equation for a Homogeneous Moderating Infinite Slab:

We make the following simplifications:

1.

Steady State:

No change with time, so (∂ψ/∂t) = 0.

2.

Infinite Slab:

The flux varies only along one direction, say the x-axis. Thus, ∇ψ → (∂ψ/∂x)

i

.

3.

1-D Geometry:

Ω·∇ψ

→ μ(∂ψ/∂x), where μ = cos(θ) and θ is the angle with the x-axis.

4.

Homogeneous:

The cross sections (Σ

_t

, Σ

_s

) do not depend on position.

5.

Moderating:

We retain the scattering term.

Under these conditions, the equation becomes:

μ (∂ψ(x, E, μ)/∂x) + Σ

_t

(E)ψ(x, E, μ) = ∫

₀

^∞

dE’ ∫

_-1

¹

dμ’ Σ

_s

(E’→E, μ’→μ)ψ(x, E’, μ’) + S(x, E, μ)

If we further assume that scattering is isotropic in the lab frame (a common approximation), then Σ

_s

(E’→E, μ’→μ) = Σ

_s

(E’→E) / 2. The equation simplifies to:

μ (∂ψ(x, E, μ)/∂x) + Σ

_t

(E)ψ(x, E, μ) = (1/2) ∫

₀

^∞

dE’ Σ

_s

(E’→E) ∫

_-1

¹

dμ’ ψ(x, E’, μ’) + S(x, E, μ)

This is the 1-D transport equation for a homogeneous moderating infinite slab.

(b) Fick’s Law

Statement:

Fick’s law states that the net diffusion of particles (e.g., neutrons) across a surface is proportional to the concentration gradient across that surface. In simple terms, particles move from a region of higher concentration to a region of lower concentration.

Mathematical Expression:

In the context of neutron transport, Fick’s law is written as:

J(r) = -D ∇Φ(r)

Where:

• J(r): Is the net neutron current density vector (neutrons/cm²-s), representing the direction and magnitude of the net flow of neutrons.
• Φ(r): Is the scalar neutron flux (neutrons/cm²-s), which is a measure of the neutron “concentration”.
• ∇Φ(r): Is the gradient of the scalar flux, pointing in the direction in which the flux increases most rapidly.
• D: Is the diffusion coefficient (in cm), a property of the medium that describes how easily neutrons spread through it.
• The negative sign indicates that the net flow is down the gradient, i.e., from high flux to low flux.

Importance in Neutron Transport:

Fick’s law is the foundation of

diffusion theory

, which is a widely used approximation to the more complex transport theory. Its importance is as follows:

1.

Simplification:

It greatly simplifies the neutron balance equation. The leakage term ∇·

J

in the continuity equation becomes ∇·(-D∇Φ) = -D∇

²

Φ (for constant D).

2.

Ease of Solution:

It transforms the complex integro-differential transport equation into a much simpler partial differential equation—the diffusion equation. This equation is far easier to solve mathematically.

3.

Practical Application:

Although it is an approximation (working best in weakly absorbing media, far from boundaries and sources), it provides surprisingly accurate results for predicting the overall neutron balance, criticality calculations, and flux distribution in large reactors. It is an indispensable tool for design and analysis in reactor physics.

Q7. Consider an infinite homogeneous moderating medium with a monoenergetic point source emitting S neutrons per second. Show that L² is equal to one-sixth the average of the square of the vector (crow-flight) distance that a neutron travels from the point where it is emitted to the point where it is finally absorbed.

Ans. We need to prove that L ² = 2>/6 , where L is the thermal diffusion length and 2> is the mean square of the straight-line (crow-flight) distance a neutron travels from its point of emission to its point of absorption. We start with the steady-state thermal neutron diffusion equation for a monoenergetic point source (S neutrons/sec) located at the origin (r=0) in an infinite, homogeneous medium: D∇ ² Φ( r ) – Σ _a Φ( r ) + Sδ( r ) = 0 where D is the diffusion coefficient, Σ _a is the macroscopic absorption cross section, Φ( r ) is the scalar flux, and δ( r ) is the Dirac delta function localizing the source at the origin. The mean square distance 2> is defined as: 2> = [∫ _V r ² (absorption rate per unit volume) dV] / (Total absorption rate) The absorption rate per unit volume is Σ _a Φ( r ). In an infinite medium, all neutrons emitted from the source are eventually absorbed, so the total absorption rate is equal to the source strength S. Thus, 2> = (1/S) ∫ _V r ² Σ _a Φ( r ) dV Now, we need to solve the diffusion equation for Φ( r ). Due to spherical symmetry, ∇ ² Φ = (1/r ² ) d/dr (r ² dΦ/dr). For r > 0, the equation is: D(1/r ² ) d/dr (r ² dΦ/dr) – Σ _a Φ(r) = 0 To solve this, we substitute u(r) = rΦ(r). The equation transforms to d ² u/dr ² – (Σ _a /D)u = 0. Defining the diffusion length L ² = D/Σ _a , we have: d ² u/dr ² – (1/L ² )u = 0 The solution to this is u(r) = Ae ^-r/L + Be ^r/L . Since the flux must be finite as r→∞, B must be 0. Therefore, Φ(r) = (A/r)e ^-r/L . The constant A is determined from the source condition. Integrating the original diffusion equation over a small sphere around the origin and using Gauss’s theorem, we find: A = S / (4πD). So, the flux distribution is: Φ(r) = [S / (4πDr)] e ^-r/L We now use this flux to calculate 2>. In spherical coordinates, dV = 4πr ² dr. 2> = (Σ _a /S) ∫ ₀ ^∞ r ² [S / (4πDr)] e ^-r/L (4πr ² dr) 2> = (Σ _a /D) ∫ ₀ ^∞ r ³ e ^-r/L dr This is a standard gamma function integral, ∫ ₀ ^∞ x ⁿ e ^-ax dx = n!/a ⁿ⁺¹ . Here, n=3 and a=1/L. ∫ ₀ ^∞ r ³ e ^-r/L dr = 3! / (1/L) ⁴ = 6L ⁴ Substituting this result into the expression for 2>: 2> = (Σ _a /D) (6L ⁴ ) Since L ² = D/Σ _a , we have Σ _a /D = 1/L ² . 2> = (1/L ² ) * (6L ⁴ ) = 6L ² Therefore, we find: L ² = 2>/6 This shows that the square of the diffusion length is equal to one-sixth of the mean square crow-flight distance a neutron travels from its birth to its absorption.

Q8. Derive two-group diffusion equations for a critical homogeneous reactor for steady state case. Ignore the presence of delayed neutrons.

Ans. The two-group diffusion equations describe neutron behavior by dividing the neutron energy spectrum into two energy groups: Group 1 (fast neutrons) and Group 2 (thermal neutrons). We will consider the steady-state case for a critical, homogeneous reactor, and ignore the presence of delayed neutrons. The criticality condition means the rate of neutron production equals the rate of neutron loss (absorption and leakage), leading to a steady neutron population (∂Φ/∂t = 0). The neutron balance equation for each group can be written as: [Production] + [In-scattering] = [Absorption] + [Leakage] + [Out-scattering] Equation for Group 1 (Fast Neutrons):

• Production: All fission neutrons are born in the fast group. The production rate is from both fast and thermal fissions. Since delayed neutrons are ignored, the production rate is ν(Σ _f1 Φ ₁ + Σ _f2 Φ ₂ ), where ν is neutrons per fission, Σ _f1 and Σ _f2 are fission cross sections in groups 1 and 2, and Φ ₁ , Φ ₂ are the group fluxes.
• Losses:
  • Leakage: From Fick’s Law (J ₁ = -D ₁ ∇Φ ₁ ), the leakage rate is ∇·J ₁ = -D ₁ ∇ ² Φ ₁ .
  • Absorption: The absorption rate in the fast group is Σ _a1 Φ ₁ .
  • Out-scattering (Slowing Down): Fast neutrons are removed from Group 1 by slowing down into Group 2. This removal rate is represented as Σ _R1 Φ ₁ , where Σ _R1 is the removal cross section from Group 1.

The balance equation for the fast group is:

ν(Σ

_f1

Φ

₁

+ Σ

_f2

Φ

₂

) = Σ

_a1

Φ

₁

+ Σ

_R1

Φ

₁

– D

₁

∇

²

Φ

₁

Rearranging, we get the first two-group equation for Group 1:

D

₁

∇

²

Φ

₁

– (Σ

_a1

+ Σ

_R1

)Φ

₁

+ ν(Σ

_f1

Φ

₁

+ Σ

_f2

Φ

₂

) = 0

Often, Σ

_a1

+ Σ

_R1

is written as Σ

₁

, the total removal cross section. In many reactors, fast fission (Σ

_f1

) is negligible, simplifying the equation further.

Equation for Group 2 (Thermal Neutrons):

• Production (Source Term): No neutrons are born thermal. The source for this group is the neutrons slowing down from Group 1. This source rate is exactly the removal rate from Group 1, which is Σ _R1 Φ ₁ .
• Losses:
  • Leakage: The leakage rate is -D ₂ ∇ ² Φ ₂ .
  • Absorption: The absorption rate in the thermal group is Σ _a2 Φ ₂ . This term includes both fission and non-fission capture.

The balance equation for the thermal group is:

Σ

_R1

Φ

₁

= Σ

_a2

Φ

₂

– D

₂

∇

²

Φ

₂

Rearranging, we get the second two-group equation for Group 2:

D

₂

∇

²

Φ

₂

– Σ

_a2

Φ

₂

+ Σ

_R1

Φ

₁

= 0

These two coupled differential equations form the two-group diffusion model for a steady-state, critical, homogeneous reactor. They describe the spatial distribution of both fast and thermal neutron fluxes in the reactor.