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°æÖ÷ (ÎÄ̳¾«Ó¢)
- ²©Ñ§EPI: 2910
- Ó¦Öú: 120 (¸ßÖÐÉú)
- ¹ó±ö: 0.414
- ½ð±Ò: 85739.6
- Ìû×Ó: 39490
- ÔÚÏß: 4852.7Сʱ
- ³æºÅ: 797958
zhangweilc(½ð±Ò+5, ²©Ñ§EPI+1): 2011-01-13 08:12:49
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²Î¿¼ÎÄÏ×µÚ¶þÒ³ËùÊö£º https://docs.google.com/viewer?a ... mpvQSZaBLSYPARp0l-A |
2Â¥2011-01-10 16:00:04
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½ð³æ (ÕýʽдÊÖ)
- ²©Ñ§EPI: 5
- Ó¦Öú: 2 (Ó×¶ùÔ°)
- ½ð±Ò: 1323.7
- Ìû×Ó: 568
- ÔÚÏß: 128.3Сʱ
- ³æºÅ: 1166788
3Â¥2011-01-10 17:14:12
|
Epc=E0+RTln(RT ks /¦ÁnF) /¦ÁnF - RTlnv/¦ÁnF Epa=E0+RTln[RT ks /(1-¦Á)nF] /(1-¦Á)nF - RTlnv/(1-¦Á)nF A graph of Ep = f (log v) yields two straight lines with a slope equal to -2.3 RT/¦ÁnF for the cathodic peak, and 2.3 RT/(1-¦Á)nF for the anodic peak. |
4Â¥2011-01-10 21:45:53









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