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ÔÎÄÏ× If d(=1.421°£)is the interatomic distance,the surface density of atoms in a graphite layer is n =4/[3*(sqrt3)*d*d] and the charge density xi=6*e*n=0.6405a.u.¡¾here we use hartree units(a.u.) such that h=m=e=1a.u.¡¿ ÎÒÔõôҲËã²»³öÀ´ÊÇ0.6405 ÎÒÊÇÕâôËãµÃ£ºn=4/(3*1.732*1.421*1.421)=0.3812 ÄÇôe=1£¬xi=2.287a.u. µÃ²»³öÀ´0.6405a.u. ÄÄλÄܸæËßÎÒÊÇÄĶùµÄÎÊÌâ°¡£¿£¿£¿£¿£¿ [ Last edited by ddx-k on 2008-12-20 at 13:07 ] |
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nucleus01
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