第二章 固体结合2.1、证明两种一价离子组成的一维晶格的马德隆常数,设离子的总数为。<解> 设想一个由正负两种离子相间排列的无限长的离子键,取任一负离子作参考离子(这样马德隆常数中的正负号可以这样取,即遇正离子取正号,遇负离子取负号),用r表示相邻离子间的距离,于是有前边的因子2是因为存在着两个相等距离α=∑(±1)/fU=2 [1/r-1/2r+1/3r-1/4r+···]-|||-,-|||-alpha =sum dfrac ((pm 1))(r)=2[ dfrac (1)(r)-dfrac (1)(2r)+dfrac (1)(3r)-dfrac (1)(4r)+... ] 的离子,一个在参考离子左面,一个在其右面,故对一边求和后要乘2,马德隆常数为α=∑(±1)/fU=2 [1/r-1/2r+1/3r-1/4r+···]-|||-,-|||-alpha =sum dfrac ((pm 1))(r)=2[ dfrac (1)(r)-dfrac (1)(2r)+dfrac (1)(3r)-dfrac (1)(4r)+... ] α=∑(±1)/fU=2 [1/r-1/2r+1/3r-1/4r+···]-|||-,-|||-alpha =sum dfrac ((pm 1))(r)=2[ dfrac (1)(r)-dfrac (1)(2r)+dfrac (1)(3r)-dfrac (1)(4r)+... ] 当X=1时,有α=∑(±1)/fU=2 [1/r-1/2r+1/3r-1/4r+···]-|||-,-|||-alpha =sum dfrac ((pm 1))(r)=2[ dfrac (1)(r)-dfrac (1)(2r)+dfrac (1)(3r)-dfrac (1)(4r)+... ] 2.3、若一晶体的相互作用能可以表示为试求:(1)平衡间距;(2)结合能α=∑(±1)/fU=2 [1/r-1/2r+1/3r-1/4r+···]-|||-,-|||-alpha =sum dfrac ((pm 1))(r)=2[ dfrac (1)(r)-dfrac (1)(2r)+dfrac (1)(3r)-dfrac (1)(4r)+... ] (单个原子的);(3)体弹性模量;(4)若取α=∑(±1)/fU=2 [1/r-1/2r+1/3r-1/4r+···]-|||-,-|||-alpha =sum dfrac ((pm 1))(r)=2[ dfrac (1)(r)-dfrac (1)(2r)+dfrac (1)(3r)-dfrac (1)(4r)+... ] ,计算及的值。