Solution Found!

Physics / Sears and Zemansky's University Physics with Modern Physics 13 / Chapter 42 / Problem 54P

Sears and Zemansky's University Physics with Modern Physics | 13th Edition | ISBN: 9780321696861 | Authors: Hugh D. Young; Roger A. Freedman; A. Lewis Ford

The one-dimensional calculation of Example 42.4 (Section

Chapter 42, Problem 54P

(choose chapter or problem)

QUESTION:

The one-dimensional calculation of Example 42.4 (Section 42.3) can be extended to three dimensions. For the three-dimensional fcc NaCl lattice, the result for the potential energy of a pair of $\mathrm{Na}^{+}$ and $\mathrm{Cl}^{-}$ ions due to the electrostatic interaction with all of the ions in the crystal is $U=-\alpha e^{2} / 4 \pi \epsilon_{0} r$, where $\alpha=1.75$ is the Madelung constant. Another contribution to the potential energy is a repulsive interaction at small ionic separation $r$ due to overlap of the electron clouds. This contribution can be represented by $A / r^{8}$, where $A$ is a positive constant, so the expression for the total potential energy is

$U_{t o t}=-\frac{\alpha e^{2}}{4 \pi \epsilon_{0} r}+\frac{A}{r^{8}}$

(a) Let $r_{0}$ be the value of the ionic separation $r$ for which $U_{\text {tot }}$ is a minimum. Use this definition to find an equation that relates $r_{0}$ and $A$ ,and use this to write $U_{\text {tot }}$ in terms of $r_{0}$. For NaCl, $r_{0}=0.281$ nm. Obtain a numerical value (in electron volts) of $U_{\text {tot }}$ for NaCl. (b) The quantity $-U_{\text {tot }}$ is the energy required to remove a $\mathrm{Na}^{+}$ ion and a $\mathrm{Cl}^{-}$ ion from the crystal. Forming a pair of neutral atoms from this pair of ions involves the release of 5.14 eV (the ionization energy of Na) and the expenditure of 3.61 eV (the electron affinity of Cl). Use the result of part (a) to calculate the energy required to remove a pair of neutral Na and Cl atoms from the crystal. The experimental value for this quantity is 6.39 eV; how well does your calculation agree?

Equation Transcription:

$NaCl$

$N{a}^{+}$

$C{l}^{-}$

$U=-\alpha{}{e}^{2}/4\pi{}{\epsilon{}}_{0}r$

$\alpha{}=1.75$

$A/{r}^{8}$

${U}_{tot}=-\frac{\alpha{}{e}^{2}}{4\pi{}{\epsilon{}}_{0}r}+\frac{A}{{r}^{8}}$

${r}_{0}$

${U}_{tot}$

${r}_{0}$

${U}_{tot}$

${r}_{0}$

$NaCl$

${r}_{0}=0.281$

${U}_{tot}$

$NaCl$

$-{U}_{tot}$

$N{a}^{+}$

$C{l}^{-}$

$Na$

$Cl$

$Na$

$Cl$

Text Transcription:

NaCl

Na^+

Cl^-

U=-alpha e^2/4pi epsilin_0r

alpha=1.75

A/r8

U_tot=-alpha e^2 over 4pi epsilon_0r+A over r^8

r_0

U_tot

NaCl

r_0

U_tot

r_0

r_0=0.281

U_tot

NaCl

-U_tot

Na^+

Cl^-

Questions & Answers

QUESTION:

$U_{t o t}=-\frac{\alpha e^{2}}{4 \pi \epsilon_{0} r}+\frac{A}{r^{8}}$

Equation Transcription:

$NaCl$

$N{a}^{+}$

$C{l}^{-}$

$U=-\alpha{}{e}^{2}/4\pi{}{\epsilon{}}_{0}r$

$\alpha{}=1.75$

$A/{r}^{8}$

${U}_{tot}=-\frac{\alpha{}{e}^{2}}{4\pi{}{\epsilon{}}_{0}r}+\frac{A}{{r}^{8}}$

${r}_{0}$

${U}_{tot}$

${r}_{0}$

${U}_{tot}$

${r}_{0}$

$NaCl$

${r}_{0}=0.281$

${U}_{tot}$

$NaCl$

$-{U}_{tot}$

$N{a}^{+}$

$C{l}^{-}$

$Na$

$Cl$

$Na$

$Cl$

Text Transcription:

NaCl

Na^+

Cl^-

U=-alpha e^2/4pi epsilin_0r

alpha=1.75

A/r8

U_tot=-alpha e^2 over 4pi epsilon_0r+A over r^8

r_0

U_tot

NaCl

r_0

U_tot

r_0

r_0=0.281

U_tot

NaCl

-U_tot

Na^+

Cl^-

ANSWER:

Introduction

First we have to find the expression for the for the NaCl crystal. Then we have to obtain the numerical value of using . Then we have to calculate the total energy required to remove one pair of neutral Na and Cl atom from the crystal.