## Atomic Number Density

**macroscopic cross-section**is derived from

**microscopic cross-section**and the

**atomic number density (N)**:

**Σ=σ.N**

In this equation, the **atomic number density** plays the crucial role as the microscopic cross-section, because in the reactor core the atomic number density of certain materials (e.g. water as the moderator) can be simply changed leading into certain **reactivity changes**. In order to understand the nature of these **reactivity changes**, we must understand the term the atomic number density.

**The atomic number density** (N; atoms/cm^{3}) is the number of atoms of a given type per unit volume (V; cm^{3}) of the material. The atomic number density (N; atoms/cm^{3}) of a pure material having **atomic or molecular weight** (M; grams/mol) and the **material density** (⍴; gram/cm^{3}) is easily computed from the following equation using Avogadro’s number (**N _{A} = 6.022×10^{23}** atoms or molecules per mole):

## Atomic Number Density of Mixtures and Compounds

## Example – Atomic number densities of boron carbide

**control rod**usually contains solid

**boron carbide**with natural boron. Natural boron consists primarily of two stable isotopes,

**(80.1%) and**

^{11}B**(19.9%). Boron carbide has a density of**

^{10}B**2.52 g/cm**.

^{3}Determine the **atomic number densities** of indivitual constituents.

Density:

M_{B} = 10.8

M_{C} = 12

M_{Mixture} = 4 x 10.8 + 1×12 g/mol

N_{B4C} = ρ . N_{a} / M_{Mixture}

= (2.52 g/cm^{3})x(6.02×10^{23} nuclei/mol)/ (4 x 10.8 + 1×12 g/mol)

= **2.75×10 ^{22} molecules of B4C/cm^{3}**

N_{B} = 4 x 2.75×10^{22} atoms of boron/cm^{3}

N_{C} = 1 x 2.75×10^{22} atoms of carbon/cm^{3}

N_{B10} = 0.199 x 4 x 2.75×10^{22} = 2.18×10^{22} atoms of 10B/cm^{3}

N_{B11} = 0.801 x 4 x 2.75×10^{22} = 8.80×10^{22} atoms of 11B/cm^{3}

N_{C} = 2.75×10^{22} atoms of 12C/cm^{3}

## Example – Fuel enrichment

**weight percent or weight fraction**, ω

_{i}, of isotope i:

The atomic number density of isotope i is:

Most of PWRs use the **uranium fuel**, which is in the form of **uranium dioxide **(UO_{2}). Typically, the fuel have enrichment of ω_{235} = 4% [grams of ^{235}U per gram of uranium] of isotope ^{235}U.

**Calculate the atomic number density of ^{235}U**

**(N235U), when:**

- the molecular weight of the enriched uranium M
_{UO2}= 237.9 + 32 =**269.9 g/mol** - the uranium density ⍴
_{UO2}=**10.5 g/cm**^{3}

**N _{UO2} = ⍴_{UO2} . N_{A} / M_{UO2}**

**N _{UO2 }**= (10.5 g/cm

^{3}) x (6.02×10

^{23}nuclei/mol)/ 269.9

**N**= 2.34 x 10

_{UO2 }^{22}molecules of UO2/cm

^{3}

N_{U} = 1 x 2.34×10^{22} atoms of uranium/cm^{3}

N_{O} = 2 x 2.34×10^{22} atoms of oxide/cm^{3}

**N _{235U}** = ω

_{235}.N

_{A}.⍴

_{UO2}/M

_{235U}x (M

_{U}/M

_{UO2})

**N _{235U}** = 0.04 x 6.02×10

^{23}x 10.5 / 235 x 237.9 / 269.9 =

**9.48 x 10**

^{20}atoms of 235U/cm^{3}