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ÕâÊÇÎÄÏ×£¨J. Phys. Chem. B 2004, 108, 11927-11934£©Öиù¾Ý¹«Ê½¼ÆËãIn2S3 ¼¤×Ó²¨¶û°ë¾¶µÄÀý×Ó£¬ÎÒ°´ÕÕËü¸øµÄÊýÖµ´øÈ빫ʽ¼ÆË㣬²¢Î´µÃµ½ÎÄÏ×ÖÐËã³öÀ´µÄ33.8nm¡£¶øÊÇ381m£¬µ¥Î»ÊÇm£¬ºÍnm²îµÄҲ̫¶àÁË£¡£¡£¡ÎÒÊÇÓùú¼Ê±ê×¼µ¥Î»×ª»»³ÉÒ»¼¶µ¥Î»´øÈ빫ʽÖе¼³öµÄµ¥Î»Ã×£¬¼ÆËãҲû·¢ÏÖ³ö´í£¬¶øÇÒÓÃÁíÍâһƪÎÄÏ×Ò²¼ÆËãûµÃµ½ÎÄÏ×±¨µÀµÄÊýÖµ£¬µ¥Î»ÒÀÈ»ÊÇm ÎÒÏÖÔÚÖØ¸´ÎÄÏ×µÄËã·¨¶¼Ëã²»¶Ô£¬¸üû°ì·¨¼ÆËãÎҵĶ«Î÷ÁË¡£ ÎÒ²»ÖªµÀ×Ô¼ºÄĶù³ö´íÁË£¬ÓÈÆäÊǵ¥Î»Ôõô²îÄÇô¶à£¿£¿ÄѵÀÊÇÎÄÏ×ÓÐÎÊÌ⣿£¿»¹ÊÇ»¹ÓÐʲôÒþº¬µÄ¶«Î÷~~~~~~~ Çë´ó¼Ò°ï°ï棬¶®µÄÈ˰ïæ¿´¿´£¡£¡£¡Ëãһϣ¬²»»áµ¢Îó´ó¼Ò¶àÉÙʱ¼äµÄ£¡ÕæµÄºÜÆæ¹Ö£¡£¡£¡ The Bohr radiusof the exciton in In2S3 may be calculated by ¦Á= h^2(ƽ·½) ¦Å/e^2(ƽ·½) ¦Ì where ¦Å is the dielectric constant (=11), h is the Planck constant and me and mh are the electron and hole effective mass, respectively. Assuming that me = mh = ¦Ì= 0.25 * 10^-28 g,£¨1/¦Ì=1/me+1/mh£©¦ÌÊǵç×ӺͿÕѨµÄÕÛËãÖÊÁ¿ the Bohr radius of the exciton in In2S3 is calculated to be 33.8 nm. [ Last edited by magy22 on 2009-7-6 at 12:28 ] |
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