Consider a three-dimensional crystal of $N$ inert gas atoms. The total energy is given by $U(R) = 2N\epsilon \left[p \left(\frac{\sigma}{R}\right)^{12} - q \left(\frac{\sigma}{R}\right)^6\right]$, where $p = 12.13$, $q = 14.45$, and $R$ is the nearest neighbour distance between two atoms. The two constants, $\epsilon$ and $R$, have the dimensions of energy and length, respectively. The equilibrium separation between two nearest neighbour atoms in units of $\sigma$ (rounded off to two decimal places) is ________
Equilibrium Separation Condition
The total energy $U(R)$ for $N$ inert gas atoms is given by:
$ U(R) = 2N\epsilon \left[p \left(\frac{\sigma}{R}\right)^{12} - q \left(\frac{\sigma}{R}\right)^6\right] $
Equilibrium separation occurs when the net force between atoms is zero. This happens when the derivative of the energy $U(R)$ with respect to the distance $R$ is zero ($\frac{dU}{dR} = 0$).
Energy Derivative for Equilibrium
We find the derivative of $U(R)$ with respect to $R$:
$ \frac{dU}{dR} = 2N\epsilon \left[ p \frac{d}{dR}\left(\frac{\sigma}{R}\right)^{12} - q \frac{d}{dR}\left(\frac{\sigma}{R}\right)^6 \right] $
Applying the chain rule for differentiation:
- $ \frac{d}{dR}\left(\frac{\sigma}{R}\right)^{12} = 12 \left(\frac{\sigma}{R}\right)^{11} \cdot \left(-\frac{\sigma}{R^2}\right) = -12 \frac{\sigma^{12}}{R^{13}} $
- $ \frac{d}{dR}\left(\frac{\sigma}{R}\right)^6 = 6 \left(\frac{\sigma}{R}\right)^{5} \cdot \left(-\frac{\sigma}{R^2}\right) = -6 \frac{\sigma^6}{R^{7}} $
Substituting these results back into the derivative expression:
$ \frac{dU}{dR} = 2N\epsilon \left[ p \left(-12 \frac{\sigma^{12}}{R^{13}}\right) - q \left(-6 \frac{\sigma^6}{R^{7}}\right) \right] $
$ \frac{dU}{dR} = -2N\epsilon \left[ 12p \frac{\sigma^{12}}{R^{13}} - 6q \frac{\sigma^6}{R^{7}} \right] $
Solving for R/sigma Equilibrium
To find the equilibrium separation, we set $\frac{dU}{dR} = 0$:
$ -2N\epsilon \left[ 12p \frac{\sigma^{12}}{R^{13}} - 6q \frac{\sigma^6}{R^{7}} \right] = 0 $
This simplifies to:
$ 12p \frac{\sigma^{12}}{R^{13}} = 6q \frac{\sigma^6}{R^{7}} $
We rearrange the equation to solve for the ratio $\frac{R}{\sigma}$:
$ \frac{12p}{6q} = \frac{R^{13}\sigma^6}{R^{7}\sigma^{12}} $
$ 2\frac{p}{q} = \frac{R^6}{\sigma^6} $
Taking the sixth root of both sides gives the equilibrium separation in units of $\sigma$:
$ \frac{R}{\sigma} = \left(\frac{2p}{q}\right)^{1/6} $
Numerical Calculation of Separation
The given constants are $p = 12.13$ and $q = 14.45$. Substitute these values:
$ \frac{R}{\sigma} = \left(\frac{2 \times 12.13}{14.45}\right)^{1/6} $
First, calculate the term inside the parenthesis:
$ \frac{24.26}{14.45} \approx 1.67889 $
Now, calculate the sixth root:
$ \frac{R}{\sigma} \approx (1.67889)^{1/6} \approx 1.0905 $
Rounding the result to two decimal places, the equilibrium separation is $1.09 \sigma$.
Important Questions from Crystal Structure Bravais Lattices Unit Cell
- For a two-dimensional hexagonal lattice with lattice constant $ a $, the atomic density is
- Consider a crystal that has a basis of one atom. Its primitive vectors are $ \vec{a_1} = a\hat{i} $, $ \vec{a_2} = a\hat{j} $, $ \vec{a_3} = \frac{a}{2}(\hat{i} + \hat{j} + \hat{k}) $, where $ \hat{i}, \hat{j}, \hat{k} $ are the unit vectors in the $ x, y $ and $ z $ directions of the Cartesian coordinate system and $ a $ is a positive constant. Which one of the following is the correct option regarding the type of the Bravais lattice?
- A compound consists of three ions X, Y and Z. The Z ions are arranged in an FCC arrangement. The X ions occupy $\frac{1}{6}$ of the tetrahedral voids and the Y ions occupy $\frac{1}{3}$ of the octahedral voids. Which one of the following is the CORRECT chemical formula of the compound?
For the given unit cells of a two dimensional square lattice, which option lists all the primitive cells?
The number of distinct ways the primitive unit cell can be constructed for the two dimensional lattice as shown in the figure is ______.
