Uranium oxide (UO2) has a theoretical density of 10.96 g/cm3 . Calculate the number density (nuclei/cm3 ) of uranium and oxygen if a sample of UO2 has a density equal to theoretical density. Calculate the number densities for the fuel in a reactor that has usually a density equal to 0.95 the theoretical density.
The Correct Answer and Explanation is :
To calculate the number densities of uranium and oxygen in UO2, we need to start by understanding the theoretical density, atomic mass, and the relationship between mass, volume, and number of atoms. Let’s break this down step by step:
Step 1: Theoretical Density of UO2
Given:
- Theoretical density of UO2 = 10.96 g/cm³.
- The molecular weight of UO2:
- Uranium (U) has an atomic mass of approximately 238.03 g/mol.
- Oxygen (O) has an atomic mass of approximately 16.00 g/mol.
- Thus, the molar mass of UO2 = 238.03 + (2 × 16.00) = 270.03 g/mol.
Step 2: Number Density of UO2 (Theoretical Case)
To find the number density of uranium and oxygen nuclei, we need the number of atoms per unit volume. Using the density and molar mass, we can calculate the number of moles of UO2 in 1 cm³:
[
\text{Moles of UO2 in 1 cm³} = \frac{\text{Density of UO2}}{\text{Molar Mass of UO2}} = \frac{10.96 \, \text{g/cm³}}{270.03 \, \text{g/mol}} = 0.0406 \, \text{mol/cm³}
]
Next, we convert this to the number of formula units (UO2) per cm³ by multiplying by Avogadro’s number (N_A = 6.022 \times 10^{23} \, \text{atoms/mol}):
[
\text{Number of UO2 formula units per cm³} = 0.0406 \, \text{mol/cm³} \times 6.022 \times 10^{23} \, \text{atoms/mol} = 2.44 \times 10^{22} \, \text{UO2 formula units/cm³}
]
Since each UO2 unit contains 1 uranium atom and 2 oxygen atoms, the number densities are:
- Uranium: ( 2.44 \times 10^{22} \, \text{atoms of U/cm³} )
- Oxygen: ( 2 \times 2.44 \times 10^{22} = 4.88 \times 10^{22} \, \text{atoms of O/cm³} )
Step 3: Number Density at 95% of Theoretical Density
In a reactor, the fuel density is often 95% of the theoretical density. Thus, the new density becomes:
[
\text{Density of UO2 in reactor} = 0.95 \times 10.96 \, \text{g/cm³} = 10.41 \, \text{g/cm³}
]
Now, using the same process as above:
[
\text{Moles of UO2 in 1 cm³} = \frac{10.41 \, \text{g/cm³}}{270.03 \, \text{g/mol}} = 0.0385 \, \text{mol/cm³}
]
[
\text{Number of UO2 formula units per cm³} = 0.0385 \, \text{mol/cm³} \times 6.022 \times 10^{23} \, \text{atoms/mol} = 2.32 \times 10^{22} \, \text{UO2 formula units/cm³}
]
Thus, the number densities are:
- Uranium: ( 2.32 \times 10^{22} \, \text{atoms of U/cm³} )
- Oxygen: ( 2 \times 2.32 \times 10^{22} = 4.64 \times 10^{22} \, \text{atoms of O/cm³} )
Final Answer:
- Theoretical Number Densities (10.96 g/cm³):
- Uranium: ( 2.44 \times 10^{22} \, \text{atoms of U/cm³} )
- Oxygen: ( 4.88 \times 10^{22} \, \text{atoms of O/cm³} )
- Reactor Number Densities (95% of theoretical density):
- Uranium: ( 2.32 \times 10^{22} \, \text{atoms of U/cm³} )
- Oxygen: ( 4.64 \times 10^{22} \, \text{atoms of O/cm³} )
Explanation:
The number densities of uranium and oxygen in UO2 are determined by the mass density, the molar mass of UO2, and the atomic composition. When the fuel’s density in a reactor is reduced to 95% of its theoretical value, the number densities of the uranium and oxygen nuclei decrease proportionally, reflecting the reduction in the total mass of UO2 available in the same volume.