| ²é¿´: 494 | »Ø¸´: 4 | |||
| µ±Ç°Ö÷ÌâÒѾ´æµµ¡£ | |||
| ¡¾Óн±½»Á÷¡¿»ý¼«»Ø¸´±¾Ìû×Ó£¬²ÎÓë½»Á÷£¬¾ÍÓлú»á·ÖµÃ×÷Õß linuxeno µÄ 9 ¸ö½ð±Ò £¬»ØÌû¾ÍÁ¢¼´»ñµÃ 1 ¸ö½ð±Ò£¬Ã¿ÈËÓÐ 1 ´Î»ú»á | |||
| µ±Ç°Ö»ÏÔʾÂú×ãÖ¸¶¨Ìõ¼þµÄ»ØÌû£¬µã»÷ÕâÀï²é¿´±¾»°ÌâµÄËùÓлØÌû | |||
linuxenoÌú¸Ëľ³æ (ÕýʽдÊÖ)
|
[½»Á÷]
¡¾ÇóÖú¡¿ÎÊnernst-einstern equationµÄ¼ÆËã´¦Àí
|
||
|
˵ʵ»°£¬»¯Ñ§³öÉíµÄ£¬ÕâЩµ¥Î»¼ÆËãÖ®ÀàµÄÕæµÄÕâô¶àÄêÒѾÍüµÄ²î²»¶àÁË¡£¡£¡£ ÏÖÔÚÐèÒªÓÃÕâ¸ö·½³Ì¼ÆËãÒ»µã¶«Î÷£¬¿ÉÊÇÔõô¶¼²»ÖªµÀµ¥Î»ÉϸÃÔõô´¦Àí£¬Ö»ºÃÇë½ÌÎïÀí·½ÃæµÄÈËÀ´Ö¸µãһϡ£¡£¡£¡£ ÏÖÔÚÒ»¸öÄ£ÐÍÐèÒª´¦ÀíÔØÌåµçºÉºÍÆäÀ©É¢Ô˶¯²úÉúµÄµçµ¼ÂÊÖ®¼äµÄ¹ØÏµ£¬ Ï໥֮¼äµÄ¹ØÏµÒѾ±»Å£ÈËÃǽ¨Á¢ÆðÀ´ÁË¡£¡£¡£ ËÄܸæËßÎÒÏÖÔÚ¼ÆËãʱ£¬¸÷±äÁ¿µÄµ¥Î»¶¼ÊÇÔõô´¦ÀíµÄ°¡£¿ÒÔ¼°ÓÃʲô³£ÊýÖµ°¡£¿ ʽÖеĵçºÉµ¥Î»¸ÃÊÇÊ²Ã´ÄØ£¿e»¹ÊÇ1.6*10^-19C£¿ ²»ÖªµÀ¸ÃÔõô´¦Àí£¬ÊÔËãÁ˼¸´Î½á¹û¶¼Ïà²î10¸öÊýÁ¿¼¶ÒÔÉÏ¡£¡£¡£ ʵÔÚÊDzÑÀ¢¡£¡£¡£¸÷λָµãһϰɡ£¡£¡£×îºÃÏêϸһµã¡£¡£ÒÔǰѧµÄ¶«Î÷ÕæµÄÍüµÄûʲô¶«Î÷ÁË (ָǰÒò×ÓAÊÇûÓе¥Î»µÄ£»µçµ¼Âʵ¥Î»ÊÇS(Î÷ÃÅ×Ó)/m;À©É¢ÏµÊý±ê×¼µ¥Î»ÊÇm^2/S(Ãë))£»Å¨¶ÈcµÄµ¥Î»±ÈÈçppmÓ¦¸ÃÊǦÌg/g£¬ËùÒÔÓ°Ïì²»´ó°É£»Ö÷ÒªÊÇÕâ¸öcharge carrierºÍboltzmann constant¡£¡£¡£×Ô¼ºÔõôҲ´Õ²»³öÀ´ÁË(ÍüÁ˵絼ÂʺÍÕâЩ֮¼äµÄµ¥Î»×ª»»ÁË)¡£¡£¡£ [ Last edited by linuxeno on 2009-9-19 at 23:23 ] |
» ²ÂÄãϲ»¶
288Çóµ÷¼Á Ò»Ö¾Ô¸¹þ¹¤´ó ²ÄÁÏÓ뻯¹¤
ÒѾÓÐ10È˻ظ´
µ÷¼Á0855-288
ÒѾÓÐ5È˻ظ´
µç×ÓÐÅÏ¢µ÷¼Á½»²æÑ§¿ÆÓÐÍÆ¼öÂð
ÒѾÓÐ6È˻ظ´
266Çóµ÷¼Á
ÒѾÓÐ6È˻ظ´
²ÄÁÏ¿ÆÑ§Ó빤³Ì¿¼ÑÐ
ÒѾÓÐ10È˻ظ´
283·Ö²ÄÁÏÓ뻯¹¤Çóµ÷¼Á
ÒѾÓÐ29È˻ظ´
Ò»Ö¾Ô¸¹þ¶û±õ¹¤Òµ´óѧ085600Ó¢Ò»Êý¶þ337·ÖÇóµ÷¼Á
ÒѾÓÐ4È˻ظ´
²ÄÁϵ÷¼Á
ÒѾÓÐ11È˻ظ´
268Çóµ÷¼Á
ÒѾÓÐ8È˻ظ´
085600²ÄÁÏÓ뻯¹¤µ÷¼Á
ÒѾÓÐ23È˻ظ´
linuxeno
Ìú¸Ëľ³æ (ÕýʽдÊÖ)
- Ó¦Öú: 1 (Ó×¶ùÔ°)
- ½ð±Ò: 5273.6
- ºì»¨: 87
- Ìû×Ó: 354
- ÔÚÏß: 40.8Сʱ
- ³æºÅ: 427975
- ×¢²á: 2007-07-31
- רҵ: ÓîÖæÑ§
|
thanks for your explanation. In the original form of the Nernst-Einstein Equation, the unit of c seems that of electron. However, in this revised equation, which relates the concentration and mobility (diffusion) of the charged species. In some papers, the electrical conductivity can be enhanced by the dissolution of H (proton) into solid materials, and in some papers which I am reading, the unit of c is ppm H and diffusion coefficient is also for H. I am now trying to calculate the relation between H content, temperature and also electrical conductivity.... However, I always fail to get the right results... In the paper, for example 100 ppm H, T=1573K, D=2.4E-10 m2/s, the charger carrier caused by H (q=+e) results in a conductivity of 0.01 S/m (A=unit 1). Can you help me to use the above values for the evaluation? [ Last edited by linuxeno on 2009-9-20 at 17:18 ] |
3Â¥2009-09-20 17:05:01
the unit of c is C/cm3, but not "ppm". it is
¡ï ¡ï
linuxeno(½ð±Ò+1):лл²ÎÓë
ddx-k(½ð±Ò+1,VIP+0):лл 9-22 19:42
linuxeno(½ð±Ò+1):лл²ÎÓë
ddx-k(½ð±Ò+1,VIP+0):лл 9-22 19:42
|
the electron concentration but not the impurity content concentration in a material. Boltzmann constant: J/K (SI) charge: C (coulomb) (SI) Just sent you a note but not showed up here. Don't know why. inform me of your receiving or not. ths. Find the absolute concentration of "H" in your sample. That is the quantity entering your expression of conductivity. absolute concentration=[density of your sample/(mole mass*1.67*10^(-27))]*(relative concentration of H in ppm) 1.67*10^(-27) is the mass of a proton or neutron in the expression. [ Last edited by bote on 2009-9-21 at 06:34 ] |
2Â¥2009-09-20 13:21:32
4Â¥2009-09-21 06:47:03














»Ø¸´´ËÂ¥