All Exams Test series for 1 year @ ₹349 only
Question

The theoretical density of a metallic solid crystal structure is given by:

\( \rho = \frac{nA}{V_C N_A} \)

where \( n \) is the number of atoms associated with each unit cell, \( A \) is atomic weight, \( V_C \) is the volume of the unit cell, and \( N_A \) represents 

The correct answer is
Avogadro's number

Understanding Theoretical Density in Crystal Structures

The theoretical density of a metallic solid crystal structure is a fundamental property that relates the mass of atoms within a unit cell to the volume of the unit cell. The formula provided is a standard way to calculate this density:

\( \rho = \frac{nA}{V_C N_A} \)

Let's break down each term in this formula to understand its meaning:

  • \( \rho \) represents the theoretical density of the material. Density is typically measured in units like grams per cubic centimeter (\(\text{g/cm}^3\)).
  • \( n \) is the number of atoms associated with each unit cell. This value depends on the specific crystal structure (e.g., simple cubic, body-centered cubic, face-centered cubic).
  • \( A \) is the atomic weight (or molar mass) of the element. This is the mass of one mole of atoms, usually expressed in grams per mole (\(\text{g/mol}\)).
  • \( V_C \) is the volume of the unit cell. The unit cell is the smallest repeating unit in a crystal lattice. Its volume is typically calculated based on the lattice parameters.
  • \( N_A \) is the term we need to identify. Looking at the units of the other terms, we have:
    • \( n \): atoms/unit cell (dimensionless number)
    • \( A \): g/mol
    • \( V_C \): cm\(^3\)/unit cell
    The formula becomes \( \rho = \frac{(\text{atoms/unit cell}) \times (\text{g/mol})}{(\text{cm}^3\text{/unit cell}) \times N_A} \). To get density in g/cm\(^3\), \( N_A \) must have units that cancel out 'mol' from the numerator and 'atoms' from the denominator, leaving units of 'atoms/mol'. The standard unit for \( N_A \) is atoms per mole.

What does \( N_A \) Represent?

The term \( N_A \) in the formula converts the atomic weight (mass per mole) into the mass per atom when considered within the unit cell context. Specifically, atomic weight (\( A \)) is the mass of one mole of atoms. A mole is defined as containing a specific number of elementary entities (like atoms or molecules).

This specific number is known as Avogadro's number. Avogadro's number is defined as the number of constituent particles, usually atoms or molecules, that are contained in one mole of a given substance. Its approximate value is \(6.022 \times 10^{23}\) particles per mole.

In the context of the theoretical density formula, \( N_A \) is used to convert the atomic weight \( A \) (in g/mol) into the mass of 'n' atoms within the unit cell. The mass of 'n' atoms is \( n \times \frac{A}{N_A} \). Dividing this mass by the volume of the unit cell \( V_C \) gives the density:

\( \rho = \frac{n \times (A / N_A)}{V_C} = \frac{nA}{V_C N_A} \)

Therefore, \( N_A \) represents Avogadro's number.

Analyzing the Options

Let's look at the given options for what \( N_A \) represents:

  1. force in newton

    Force is a physical interaction that can change the motion of an object. It is measured in Newtons (N). This is unrelated to the number of atoms in a mole.

  2. number of particles

    While Avogadro's number is a number of particles, the term "number of particles" is too general. \( N_A \) specifically refers to the number of particles in a mole.

  3. Avogadro's number

    This matches our understanding. Avogadro's number is the specific constant used to relate the number of atoms or molecules to the macroscopic quantity of a mole.

  4. number of turns

    Number of turns is a term used in contexts like coils or screws. It is unrelated to crystal structures or density calculations.

Based on the analysis of the formula and the options, \( N_A \) represents Avogadro's number.

Term Represents Typical Unit
\( \rho \) Theoretical Density g/cm\(^3\)
\( n \) Number of atoms per unit cell Dimensionless
\( A \) Atomic Weight (Molar Mass) g/mol
\( V_C \) Volume of the unit cell cm\(^3\)/unit cell
\( N_A \) Avogadro's number atoms/mol (or particles/mol)

Revision Table: Key Concepts in Crystal Structure Density

Concept Description
Unit Cell The smallest repeating structural unit in a crystal lattice.
Theoretical Density (\(\rho\)) Mass of atoms in a unit cell divided by the volume of the unit cell. Represents the expected density assuming a perfect crystal lattice.
Atomic Weight (\(A\)) The mass of one mole of an element's atoms.
Avogadro's Number (\(N_A\)) The number of constituent particles (atoms, molecules, etc.) in one mole of a substance, approximately \(6.022 \times 10^{23}\) particles/mol.
Number of atoms per unit cell (\(n\)) Depends on the crystal structure type (e.g., BCC \(n=2\), FCC \(n=4\), Simple Cubic \(n=1\)).

Additional Information: Calculating Theoretical Density

To calculate the theoretical density of a metallic element, you need the following information:

  1. The crystal structure type (e.g., FCC, BCC, SC) to determine \( n \).
  2. The atomic weight \( A \) of the element.
  3. The lattice parameter (edge length of the unit cell, 'a') to calculate \( V_C \).
  4. Avogadro's number, \( N_A \).

The volume of the unit cell (\( V_C \)) depends on the crystal structure and lattice parameter ('a'):

  • Simple Cubic (SC): \( V_C = a^3 \)
  • Body-Centered Cubic (BCC): \( V_C = a^3 \)
  • Face-Centered Cubic (FCC): \( V_C = a^3 \)

Note that for BCC and FCC, the lattice parameter 'a' is related to the atomic radius 'r' differently: For BCC, \( a = \frac{4r}{\sqrt{3}} \); for FCC, \( a = \frac{4r}{\sqrt{2}} \). For SC, \( a = 2r \). These relationships are used if the atomic radius is given instead of the lattice parameter.

Once \( n \), \( A \), \( V_C \), and \( N_A \) are known, the theoretical density \( \rho \) can be calculated using the formula \( \rho = \frac{nA}{V_C N_A} \).

Was this answer helpful?

Important Questions from Miscellaneous-Engineering

  1. A well foundation of 6 m external diameter and of 5 m internal diameter is sunk to a depth of 15 m in a deep deposit of sand. If the average N value of sand is 20, the load that the well can carry by bearing alone will be nearly
  2. What is the maximum depth to which a trench of vertical sides can be excavated in a clay stratum with c = 50 kN/m², y = 16 kN/m³, β = 90°, ¢ = 0°, Fc = 1 and N = 0-261?

  3. Which one of the following is an example of reinforced earth wall?
  4. Which one of the following data is not required for design of a weir or a barrage?
  5. In case of design of a weir or a barrage by providing a higher afflux, the waterway and, therefore, the length of the weir can be reduced, but it will result in
Need Expert Advice?

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App