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Chapter 7: Problem 50

In ionic solids such as \(\mathrm{NaCl}\) (salt), the potential energy of a pair of ions takes the form \(U=b / r^{n}-a / r,\) where \(r\) is the separation of the ions. For \(\mathrm{NaCl}, a\) and \(b\) have the SI values \(4.04 \times 10^{-28}\) and \(5.52 \times 10^{-98},\) respectively, and \(n=8.22 .\) Find the equilibrium separation in NaCl.

Short Answer

Expert verified
The equilibrium separation of ions in NaCl is found by substituting the given values into the equation \(r = \left( a / (nb) \right)^{\frac{1}{n - 1}}\).

Step by step solution

01

Write down the given potential energy function

The potential energy of a pair of ions in NaCl is given by the equation \(U=b / r^{n}-a / r,\) where the values of \(a\), \(b\), and \(n\), in this case, are \(4.04 \times 10^{-28}\), \(5.52 \times 10^{-98},\) and \(8.22\) respectively.
02

Find the derivative of the potential energy function

The equilibrium position is where the potential energy is at its minimum, and at the minimum, the derivative of the function is equal to 0. We find the derivative using the power rule of differentiation, resulting in \(-nb / r^{n+1}+a / r^{2} = 0\).
03

Solve for the separation \(r\)

Equating the result from step 2 to zero yields \(nb / r^{n+1} = a / r^{2}\). One can simplify this equation to \(nb \cdot r^{n-1} = a\). Solving for \(r\) gives us \(r = \left( a / (nb) \right)^{\frac{1}{n - 1}}\). Substitute the given values of \(a\), \(b\), and \(n\) into this equation to find the equilibrium separation of ions in NaCl.

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Most popular questions from this chapter

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