- #1

JoJoQuinoa

- 17

- 0

I was wondering if someone could help clarifying this question.

The question asks to estimate the energy state difference between the vibrational ground state of S

_{0,v=0}and the first excited vibrational ground state S

_{0,v=1}of the spectra below.

The given solution: S

_{1,v=1}-> S

_{0,v=1}at [itex]\lambda [/itex] = 400 nm and S

_{1,v=2}-> S

_{0,v=1}at [itex]\lambda [/itex] = 380 nm.

There are two things I'm confused about the solution:

1) From Figure B, I would assume that the first Fluorescence peak at 380 nm in Figure A corresponds to S

_{1,v=0}-> S

_{0,v=1}and the second peak at 400 nm corresponds to S

_{1,v=0}-> S

_{0,v=2}. Larger transition results in higher emission energy or smaller wavelength.

2) Why is S

_{0,v=1}being used as the final state for both peaks?

The S

_{1,v=1}-> S

_{0,v=1}would occur as S

_{1,v=1}->S

_{1,v=0}->S

_{0,v=1}and

S

_{1,v=2}->S

_{1,v=0}->S

_{0,v=1}.

Since S

_{1,v=2 or 1}->S

_{1,v=0}is internal conversion, wouldn't S

_{1,v=0}->S

_{0,v=1}for the two transitions give off the same energy?

Thanks in advance!