Ask your own question, for FREE!
Chemistry 15 Online
OpenStudy (anonymous):

Hi guys., i wanna discuss about Raman spectroscopy, i have ever read that when electric field applied along the axis is approximately twice as large as that induced by the same field applied across the axis?? can you tell me why ??

OpenStudy (chmvijay):

to create polrizability in the molecule i think!!!

OpenStudy (anonymous):

i mean the axis of molecule

OpenStudy (anonymous):

yes of course.., but why the greatest polarizability in these molecules is achieved when the electric field is applied parallel to the molecule rather than perpendicular to the molecule. ??

OpenStudy (anonymous):

I can't imagine ..,

OpenStudy (anonymous):

have you idea @chmvijay ??

OpenStudy (chmvijay):

humnnn i have to think over it ! let u know after some time

OpenStudy (anonymous):

i'm sorry., but would you like to explain this figure http://photonicswiki.org/images/thumb/9/91/Anisotropic_polar.JPG/800px-Anisotropic_polar.JPG it's anisotropic polarizability..,

OpenStudy (chmvijay):

it is related to the something molecular bond axis and the electric field axis

OpenStudy (chmvijay):

if the molecular axis and the electric field parallel then only polarizability in molecule can be created i think!!!

OpenStudy (anonymous):

do u know what is the reason??

OpenStudy (chmvijay):

ok lket me read and come back to you right now i dont no clearly ok!!

OpenStudy (anonymous):

ok..., thank you so much :)

OpenStudy (anonymous):

@chmvijay look at this one , Xue et al using "Average Polarization theory" to make a model for effective thermal conductivity of nanofluids..,

OpenStudy (anonymous):

I'm making timeline ideas for this paper.., can you help me?

OpenStudy (chmvijay):

yaa let me read and come back

OpenStudy (anonymous):

ok

OpenStudy (chmvijay):

how this si related to raman spectroscopy

OpenStudy (anonymous):

no.., it's different assignment..,

OpenStudy (chmvijay):

what i have to do in this assigmenemrnt its more phys and maths and less chemistry

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!