1. The problem statement, all variables and given/known data Consider fluorescence (light emission from an electronically excited state of a molecule) in the presence of a quenching molecule M. Quenching refers to the removal of energy from an excited electronic state by collision, and so without light emission. A mechanism for such a system can be given as follows: A + hν → A* rate = ke [A] (2.1) A* + M → A + M rate = kq [A*] [M] (2.2) A* → A + hν' rate = kf [A*] (2.3) where the rates of each individual step have been given. In the above expressions ke is the rate constant for formation of electronically excited molecules, is the intensity of the light source used to excite molecules, kq is the rate constant for quenching of molecules, and kf is the rate constant for fluorescence. a) Find an expression for the steady state concentration of electronically excited molecules, [A*]ss. b) We may assume that If, the intensity of fluorescence, is proportional to [A*]ss, the steady state concentration of electronically excited molecules. Call the (unknown) constant of proportionality c (so If = c [A*]). Find an expression for If in terms of c and other constants. c) Based on your answer in b, suggest an experimental method for finding kf/kq, the ratio of the rate constants for fluorescence and quenching. d) Use your answers above to analyze the data given below. Note that you can use the half-life measured for fluorescence in the absence of quenching molecules to find kf, which makes it possible to find a value for kq. M (mol/L) If (au) au = arbitrary units 0.0010 0.41 0.0050 0.25 0.0100 0.16 In the absence of any quenching molecules M the half-life for fluorescence is t1/2 = 29. μs. 2. Relevant equations -N/A 3. The attempt at a solution (a)d/dt=ke[A]-kq[A*][M]-kf[A*]=0 (b)I'm completely lost on this part, but I feel that I have to do some sort of substitution and then algebraic manipulation. (c) I have no idea. (d)Completely lost. I only want to know if I'm right on part (a). And if I'm headed in the right direction for (b).