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Q2.27:
Estimate the average mass density of a sodium atom assuming its size to be about 2.5 (Å). (Use the known values of Avogadro’s number and the atomic mass of sodium). Compare it with the mass density of sodium in its crystalline phase : 970 (kg m^{-3}). Are the two densities of the same order of magnitude ? If so, why ?

Solution :

Step 1: Identifying Known Parameters

To calculate the average mass density of a single sodium atom, we utilize the following standard physical constants:

  • Atomic mass of sodium ($M$) = $23.0 \text{ g/mol} = 0.023 \text{ kg/mol}$
  • Avogadro's number ($N_A$) = $6.022 \times 10^{23} \text{ atoms/mol}$
  • Radius of sodium atom ($r$) = $2.5 \text{ \AA} = 2.5 \times 10^{-10} \text{ m}$
  • Density of crystalline sodium ($\rho_{c}$) = $970 \text{ kg/m}^3$

Step 2: Calculation of Mass of a Single Sodium Atom

The mass of one sodium atom ($m$) is derived by dividing the molar mass by Avogadro's number [Per the definition of the mole]:

$m = \frac{M}{N_A} = \frac{0.023 \text{ kg/mol}}{6.022 \times 10^{23} \text{ atoms/mol}}$

$m \approx 3.82 \times 10^{-26} \text{ kg}$

Step 3: Calculation of Volume of a Sodium Atom

Assuming the sodium atom is a sphere, the volume ($V$) is calculated using the geometric formula $V = \frac{4}{3}\pi r^3$:

$V = \frac{4}{3} \times 3.14159 \times (2.5 \times 10^{-10} \text{ m})^3$

$V \approx 1.333 \times 3.14159 \times 15.625 \times 10^{-30} \text{ m}^3$

$V \approx 6.545 \times 10^{-29} \text{ m}^3$

Step 4: Calculation of Average Atomic Mass Density

The average mass density of the atom ($\rho_{a}$) is the ratio of mass to volume [By the definition of density, $\rho = \frac{m}{V}$]:

$\rho_{a} = \frac{3.82 \times 10^{-26} \text{ kg}}{6.545 \times 10^{-29} \text{ m}^3}$

$\rho_{a} \approx 583.65 \text{ kg/m}^3 \approx 5.8 \times 10^2 \text{ kg/m}^3$

Step 5: Visual Representation of Atomic Volume

r = 2.5 Ã… Spherical Approximation of Sodium Atom

Step 6: Comparative Analysis and Conclusion

Parameter Calculated Value
Atomic Density ($\rho_{a}$) $\approx 5.8 \times 10^2 \text{ kg/m}^3$
Crystalline Density ($\rho_{c}$) $9.7 \times 10^2 \text{ kg/m}^3$

Evaluation of Order of Magnitude:

Both densities are of the order of $10^2 \text{ kg/m}^3$. This indicates that in the crystalline phase, the atoms are packed relatively close to each other without significant empty space beyond the intrinsic atomic volume. The discrepancy arises because the crystalline density accounts for the inter-atomic spacing within the crystal lattice structure, whereas the atomic density calculation assumes a hard-sphere model for an isolated particle.

Final Solution: The estimated atomic density is $5.8 \times 10^2 \text{ kg/m}^3$. Yes, both values are of the same order of magnitude ($10^2 \text{ kg/m}^3$), implying that the mass is distributed with similar density in both the individual atom and the crystalline solid state.


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