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kingleaves

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ÎÊÌâÊÇÕâÑùµÄ£¬Ò»ÆªÎÄÕÂдµ½excited at 340nm£¬ÄÇÎÒ×ö¼ÆËãµÄ»°£¬¼¤·¢Ì¬ÊDz»ÊǸÃÑ¡ÓëÕâ¸öÖµ±È½Ï½Ó½üµÄrootÀ´×ö¼¤·¢Ì¬µÄÓÅ»¯£¬½ø¶ø¼ÆËãÓ«¹â£¿¶ø²»È¥¿¼ÂǼÆËãÖеÄгÕñÇ¿¶ÈfµÄ´óС¡£Ò²¾ÍÊÇ˵ÒѾ­ÓÐÁËʵÑ鼤·¢²¨³¤µÄ»°¸ÃÈçºÎ¼ÆËãÓ«¹âÄØ£¿ÔÚ´ËÏÈл¹ý¸÷λÁË¡£
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tomhy

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5Â¥: Originally posted by kingleaves at 2011-12-28 19:27:45:
ллÄãµÄ»Ø¸´£¬¿ÉÊÇ»¹ÓиöÒÉÎÊÊÇ£¬¾ÍÏñÄãËù˵µÄ£¬Æäʵӫ¹âÊÇ´ÓµÚÒ»¼¤·¢Ì¬ÏÂÀ´µÄ£¬ÄÇÏñ¸ß˹ÀïµÄ¶ÔÓÚÆäËû̬µÄÓÅ»¯£¬ÒâÒåÓÖÔÚÓÚÄÄÀïÄØ£¿Ð»Ð»

The absorption and emission of light direct correlate to the energy differences between the state. Think about how can you get the full uv-vis absorption spectrum from calculation for a molecule. For emission, we match the first excitation to the experimental data. (I might be wrong.) If you read something about dual photon excitation, that will give you a better understanding to your question.
6Â¥2011-12-28 21:56:10
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nwwolfchj

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2Â¥2011-12-26 01:28:13
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kingleaves

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ÒýÓûØÌû:
Â¥: Originally posted by nwwolfchj at 2011-12-26 01:28:13:
²»Êǰɣ¬Òª¸ù¾ÝÕâ¸ö̬µÄ¶Ô³ÆÐÔ£¬µÚ¼¸¸ö¼¤·¢Ì¬×ۺϿ¼Âǰɣ¬ÊµÑéºÍÀíÂÛ²»Ò»¶¨ÍêÈ«ÎǺϣ¬ÀíÂÛÉÏÓÐʱºòÓжà¸ö½Ó½üµÄ̬£¬ÕâʱºòÓÖÒªÀíÂÛºÍʵÑéµÄ¹éÊôÒ»Ö£¬²ÅÄܱȽϡ£Ò»¶¨Òª±¾ÖÊÒ»Ö£¬¸ºÔðÄøöÖµÏà½üÆäʵ¸ù±¾²»ÊÇÒ»»ØÊ ...

ÄÇÇëÎÊʵÑéµÄ·½·¨ÔõôÀ´È·¶¨µç×Ó̬µÄ¶Ô³ÆÐÔÄØ£¿Ð»Ð»£¬¶ÔÓÚʵÑé²»ÊǺÜÁ˽⡣¿´µ½ÎÄÕÂÖж¼Ö»ÊÇËãÁ˵ÚÒ»¼¤·¢Ì¬£¬ÇÒfÖµ¶¼Í¦´óµÄ£¬µ«ÊÇÔÚÎÒ×Ô¸öµÄ¼ÆËãÖеÚÒ»¼¤·¢Ì¬ÓÐЩfÖµºÜС£¬²»ÖªµÀ¸ÃÈçºÎÏÂÊÖÁË¡£²»ÖªµÀÎÒ˵Ã÷°×ÁËû¡£Ð»Ð»
3Â¥2011-12-26 16:43:25
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tomhy

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in the experiment, we usaually use single excitation to irradiate the sample to get the emission spectrum, such as flourescence and phosphorescence. In this case, I will calculate the UV-Vis absorption of the ground state molecule to see if there is overlap between the calculated absorption band and exictation light at 340nm. I believe there is.
The phosphorescence is the emission spectrum from triplet state T1 to ground state. so, no matter what the excitation light you used, if there is phosphorescence emission, the spectrum should be the same for the same molecule.

this is my 2 cents.
4Â¥2011-12-28 00:24:38
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