Question
Show that the ionic cohesive energy of an ionically bonded solid is given by Equation 43.18 . Suggestion: Start with Equation 43.17 and note that $d U / d r=0$ at $r=r_{0}$ .
Step 1
Step 1: We start with the given total potential energy equation: \[U_{\text{total}} = -\frac{\alpha k_{e} e^{2}}{r} + \frac{b}{r^{m}}\] Show more…
Show all steps
Your feedback will help us improve your experience
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Starting with Equation 43.17, show that the ionic cohesive energy of an ionically bonded solid is given by Equation 43.18.
Show that the minimum potential energy of an ion pair in an ionic solid is $$U_{0}=-\alpha k \frac{e^{2}}{r_{0}}\left(1-\frac{1}{m}\right)$$ where $r_{0}$ is the equilibrium separation between ions, $\alpha$ is the Madelung constant, and the repulsive potential energy between ions is $B / r^{m} .\left|U_{0}\right|$ is called the ionic cohesive energy and is the energy required to separate to infinity an ion pair originally at a separation of $r_{0}$ in the crystal. Hint: $$\left.\frac{d U}{d r}\right|_{r=r_{0}}=0$$
Use Equation 43.18 to calculate the ionic cohesive energy for NaCl. Take $\alpha=1.7476, r_{0}=0.281 \mathrm{nm},$ and $m=8 .$
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD