Lecture 27

Added: 2026-05-09

[00:00:20]Hello everyone, welcome back to the MOO's course on advanced analytical technique.

[00:00:24]Today I'm going to finish the second unit of this course molecular luminous spectroscopy. It is the fifth lecture of the sixth week. the and the lectures that you have attended so far has completed two important topics. First is the absorption and emission spectroscopy in which we have discussed about atomic absorption and atomic emission spectroscopy and how the techniques are uh and how the analysis is being done, what are the components, what are the detectors and how important those techniques are.

[00:01:02]And then we have started the second unit the molecular luminance spectroscopy in which we have studied two different phenomena the photooluminescences and the chemoluminescence. In photooluminance we have learned about fluoresence as well as phosphorusence.

[00:01:16]After completing the instrumentation the principle and related topic to the fluesence and phosphorusence in the previous uh after completing some related topic after completing few related topics of fluoresence and phosphorusence including their instrumentation working principle etc. I have discussed with you the chem luminous and how measurements are being done using a spectrum. Now in today's lecture I'm going to discuss some examples with you how we analyze a spectra, how we do some numerical problems based on the data we obtain from fluoresence spectra.

[00:01:56]Let us start with today's discussion.

[00:02:00]As in all spectroscopic techniques, we first of all prepare a calibration curve which is needed for the quantitative analysis.

[00:02:12]If I hope you remember that in case of atomic emission and atomic absorption spectroscopy as well I told you the importance of calibration. I told you about the calibration curve. What is it and what what is the importance of a calibration curve and I told you there that how a calibration curve is prepared.

[00:02:31]Again for your uh just again just so just a recall calibration curve is basically based on some known concentrations.

[00:02:44]For these known concentrations we perform the spectrum. For these known concentrations we perform the study generate the spectrum and for these known concentrations we performed the experiment obtained the spectra and from the spectra we obtained some useful parameters. For example, we determine the intensity of the fluoresence emission and which is which is in the form of a electrical signal because we use photo detector a photo multiplier tube or SSD solid state device in this case as well. So what we get at the end is the light intensity is converted into a electrical signal that electrical signal gives the intensity of the emitted light. So that signal that the value of that signal is taken and then a graph is plotted between those concentrations the known concentrations generally we sometimes we go for concentrations in ppm for example sometimes we go for concentrations in ppm like we have 10 20 30 and so far we prepare five to six solutions in ppm however with techniques like a AES specifically or with fluorescent spectroscopy since they are very sensitive so 10 ppm or 20 50 ppm it's quite high concentration for these samples and as you all know that with concentrated sample we have some sort of interferences so the results may not be reliable and in order to minimize those interferences we always go for a dilute solution so instead of taking 10 ppm 20 30 ppm for these sensitive instruments we take 10 20 30 and so on but in ppb level we take concentrations for calibration curve in parts per billion level and 1 ppb is equal to 1 micro g per liter 1 ppb while ppm is 1,000 time more concentrated 1 ppm is equal to 1 mg per liter so for making these standard solution we take a stock solution for example in this particular case I'm focusing on riboflavven which is the vitamin B2 2. So for analyzing this vitamin riboplavin in a given sample for example milk contains a significant amount of this riboflavin.

[00:05:05]It is used for energy production. It is used for skin health and it is very useful for the body. So if we are going to check the amount of riboflavven B2 in the milk that is given to us as a sample first of all we prepare. First of all we we take up your standard solution of ribof 11 which is called as a stock solution because its concentration is high is its concentration is 100 ppm.

[00:05:31]Now we first of all take this solution from the refrigerator. Keep it keep it at room temperature and allow it to just comes to the room temperature or allow it to equilibriate to the room temperature. So that volumetric measurements can be done consistently because at low temperature the volume may have some difference.

[00:05:54]While if uh if you compare it the with the volume at for example in case of water as well at 4° the volume is a little bit higher. So we allow the solution of this ra of leaven to equilibriate to the room temperature and then we take 1 ml of this solution and dilute it to 10 ml. For example we have a standard flask and in this standard flask we have this 100 ppm solution which is called as stock solution. Now this is stock of 100 ppm of B2. From this we take 1 ml in a small standard flask and then we add 9 ml deionized water or double distilled water.

[00:06:43]The volume of this flask is 10 ml. So this way we dilute it to we this. So this way we dilute it 10 times. So earlier it was 100 ppm. Now it becomes 10 ppm.

[00:07:03]So in this way a diluted solution is prepared. The solution of 10 ppm that we have just prepared. We again dilute it to 100 times. We again dilute it 100 times by taking 1 ml of this solution. By taking 1 ml of this solution and adding 99 ml distilled water or deionized water then it becomes 100 times dilution its concentration will go 10 upon 100 ppm will be equal to 0.1 ppm. Now this 0.1 ppm can also be written as 100 ppb.

[00:07:37]Now this is our main solution. From this solution, we will prepare five different solutions. And those solutions are numbered as 1 2 3 4 5. And those solutions are numbered as 10, 20, 30, 40, 50 ppm, 50 ppb.

[00:08:03]and one more solution 0 PPP which does not contain the sample which contains de deionized water only.

[00:08:12]Now these solutions are prepared in 10 ml standard flask.

[00:08:20]Now these solutions are prepared in 10 ml standard flask in this way.

[00:08:44]And in these 10 10 ml standard flasks, in the first flask we put 10 ml distilled water only. While in the second we put 1 ml of this solution, 9 ml distilled water. Here we put 2 ml of this solution. 8 ml distilled water.

[00:09:03]Then we put 3 + 7 ml distilled water.

[00:09:07]Then 4 ml of this solution and 10 ml distilled and 6 ml distilled water.

[00:09:12]While in the last one we put 5 ml of the above solution.

[00:09:17]While in the last one we put 5 ml of the above solution plus 5 ml distilled water. We just make two times diluted.

[00:09:24]So the concentration comes from 100 pp.

[00:09:26]It is when it is diluted twice only it comes to 50 pbp. So we have five different solutions. And now we'll do the fluent measurements by using a spectrum.

[00:09:37]It will come out to be like this. Here we have intensity of the light radiation which is in the form of mill volts because it's an electrical signal. Here we put these solutions. We have zero here we so then we'll plot this uh data for calibration curve preparation. This is intensity in mill volts. And then we have 0 10 20 30 40 and 50B ppb.

[00:10:19]The corresponding fluorescent in the corresponding signals will be measured from the fluorescent measurements using spectrometers. Those field signals are plotted here and finally a straight line is observed. Some of the points lie exactly on this straight line while some of the point falls little bit outside the line. But still we get a straight line showing a linear relationship between the concentration of ribbof 11 or B2 with intensity it is linear.

[00:10:58]Now our calibration curve is ready.

[00:11:01]The next step is to take the sample of the milk, remove some interfering agents like fats, do some filtration and then the sample is ready for analysis. We put the sample again in the cubit for measurement of fluoresence spectra. We put the sample again in the cubit for measurement of flu sensor for measurement of fluescent for measurement of fluesence spectrum. And from there we obtain the electrical signal again. Now the electrical signal that is obtained falls on this particular line. Say for example the signal falls somewhere here.

[00:11:41]Now we draw a perpendicular to the x yaxis because the signal falls here. And now then we draw a perpendicular to the x-axis and the point sorry. So now we draw a perpendicular to the X. Now we draw a perpendicular to the X-axis just to know the point at which it cuts the X-axis. And the point at which this perpendicular cuts the X-axis is the concentration of analyte in the sample. This gives us the concentration of riboflavin or vitamin B2 in milk. For example, it is cutting at somewhere around 36 ppm. So 36 ppb.

[00:12:22]This means that our milk sample contains 36 ppb of riboflavin. This is just for an example. So this is again proportion of unknown solution means this is our sample. So in this particular case we have as I told you we have to identify the concentration or the quantity of B2 vitamin B2 or riboflavin in milk. So it is a dry milk sample. We take this dry milk reconstitute again in water and then we remove interfering fats and proteins so that they should not interfere with the data.

[00:12:57]They should not in interfere with the light that is being used as a source of excitation and to have a reproducible and reliable results.

[00:13:11]Once these interfering fats and proteins are removed, a small procedure is carried out on the milk sample. We take 10 ml of the milk sample, add exactly 75 ml of the solution consisting of equal parts of this is estic acid and nine. So we prepare a solution with three molar acetic acid CH3 CO. They are added as agents to support the process of excitation. And then we have three molar NCL. they just added in the solution. This is part of procedure. Sometimes just to remove interfering agent, sometime do some um there's a support from the excitation as well. So this solution is prepared and then 75% of ml of this solution is added and then 75 ml of this solution is added while 25 ml of the milk sample. So milk sample is diluted four times. From 25 ml we are making it 25 + 75 the total volume becomes 100. So we are taking from 25 to 100 ml. So we are making four times dilution of this.

[00:14:27]Again we we do this in a 250 ml or 400 ml beaker to dilute. We do it in a 250.

[00:14:35]We do it in 250 ml or 400 ml beaker for dilution of the sample. Then we strain the solution again. We filter it just to avoid any suspended impurities or any particle of the fat or the protein. We use buckner funnel for this purpose. I hope you have all have seen a buckner funnel which is like this and which has a small holes inside it.

[00:15:01]And this funnel we attach to a flask. We pour the sample here.

[00:15:12]The advantage of using a buckner funnel is that it has a provision for creating vacuum by running tab water. So we create vacuum by running tab water. We pass tap water which removes all the air and a vacuum is created and that vacuum pulls the solvent and that vacuum pulls the liquid through it and that vacuum pulls the liquid through these pores resulting into a fast filtration process.

[00:15:48]The difference between a normal funnel and a buckner funnel is that a normal funnel operates on the principle of gravity means it uses gravity through which the liquid particle moves through it downwards while the solid particles are retained. While in case of bucknar funnel we it uses vacuum the power of vacuum to pull the liquid through these uh pores a little bit quicker resulting into the filtration done in a short period resulting into the filtration done in a short period of time. So this and this vacuum is created by running tap water which removes all the air. So automatically there's a vacuum a vacuum is created. We connect this rubber pipe. We connect a rubber pipe to this tube. We connect a rubber pipe to this buckner funnel for passing tab water.

[00:16:48]So once the filtration is done, this filtrate is taken. Now it is ready for analysis means it is ready for measuring the fluoresence spectrum. Again we go for dilution. 5 ml of this elote is diluted to 9 is diluted to 100 ml and for this we add 95 ml distilled water.

[00:17:15]Means for diluting it to 100 ml from 5 ml we add 95 ml of distilled water. So that the total volume becomes 100 ml.

[00:17:24]And then we can say that the dilution is from five since we are going from 5 ml to 100 ml the dilution is 20 times. So we have done 20 * dilution in this case. So we have done 20 times dilution in this case. Now this diluted sample is taken in the cubette placed in the and the spectrum is obtained. From there we obtained the intensity reading in the form of an electrical signal in mill volt. We just put that mill volt reading in and we just we just put the mill volt reading in the calibration curve that we have prepared and we can identify from this calibration curve the actual concentration of the riboplavin in our sample. So in this way the analysis has so in this way the analysis will be done. So remember in all cases for quantitative analysis whether we are using molecular lemon sens spectroscopy means the fluid sense spectroscopy or even atomic absorption spectroscopy or atomic emission spectroscopy in all cases for quantitative analysis we need a calibration curve and for calibration curve we need pure sample we prepare different diluted samples from the stock solution of the pure sample and then we perform the analysis obtain data put it on y-axis against concentration on x-axis so that a linear curve is obtained which is the calibration curve and then we take the real sample dilute it up to the extent and then we perform the analysis obtain the reading put it on the y-axis and then correspondingly measure the reading on the x-axis to know its concentration or its actual so to know its concentration or its actual concentration in the sample this in this way the analysis the quantitative analysis is being Then there are two rules that I would like to tell you before moving on moving on to some problems. First is called as the Kasha rule. Kasha rule states that in case of photochemistry when we have when an molecule is excited by absorption of light energy.

[00:19:38]It is excited from ground level to maybe to the first excited state or sometimes it happens that it may reach to the second excited state and as you all know that we have different levels vibrational levels attached to these electronic states. So according to this rule there will be some vibrational relaxation first which brings the molecule from higher vibrational level to the lower vibrational levels of the excited states only and when the electron reaches to the lower vibrational level of this excited state then only it comes back and emit light radiations.

[00:20:22]Clear? It will not happen that electron in the higher vibrational level of this S1 excited state from here directly comes to ground state.

[00:20:36]According to Kasha's rule, this is not possible.

[00:20:41]What is possible according to this rule is that first the electron will undergo vibrational relaxation which is a form of non-radiative deactivation process and comes to the lowest vibrational level of that excited state and then it comes back to the ground state emit light in the visible region. This rule also states that even from the second excited state, the electron does not come back directly and by emitting energy. Means if the electron is in the second excited state by absorbing large amount of energy, it first comes to the first excited state by internal conversion. Then only it comes back to ground state causing H new emissions.

[00:21:29]So in a nutshell I can say that Kasha rules says that both internal convergence as well as vibrational relaxations happens in the excited state. So that the electron from higher excited states comes to the first excited state by internal conversion and electron also changes the higher vibrational levels and comes back to the lower vibrational levels of the excited state through vibrational relaxations. Both these processes, both these deactivations means the internal conversion for for coming from S2 to S1 as well as vibrational relaxation for coming from higher vibrational level of S1 to lower vibration level of S1. Both are non-radiative energy loss phenomena.

[00:22:11]Both are non-radiative deactivation phenomena means they do not emit light radiation. they emit heat radiation or some sort of kinetic energy is being lost thereby molecule thereby electrons comes to those lower level and then finally so both these deactivation phenomena means the internal conversion which brings the molecule from S2 to S1 as well as vibrational relaxation which brings molecule which brings the electron from higher vibrational level to the lower vibrational level of the excited state only. They are non-radiative deactivation phenomena means they do not release light radiations. They only causes loss of energy in the form of heat or kinetic energy.

[00:23:04]And when this happen the remaining amount of energy is then released when electrons comes back finally to the ground state. And this is the reason because some of the energy is already lost via vibrational relaxation and internal conversion. Therefore, the light energy that is being emitted at the end when electron comes from S1 to S not is always lesser than the amount of light that is being absorbed or I should say the wavelength of the light emitted is always greater than the wavelength of the light absorbed because this long this long wavelength has lower energy and the lower and the reason for this lower energy is that some of the energy has already been lost while changing the relaxation because some of the energy has already been lost during vibrational relaxations as well as internal conversions in the excited state only and the remaining amount of energy is being released when the electron comes back to the ground state only and that energy is released the rem and that remaining amount of energy is released in the form of light radiations so this is the kasha's rule clear so it says that the excess energy vibrational energy because the electron reaches to the higher vibrational level of the excited state this energy Energ is lost via vibrational relaxation. And if the electron is in S_2 state, it first of all comes to S_sub_1. Then only it comes to S not by emitting light radiation. Here it is only either heat or kinetic energy.

[00:24:32]This is called as inver internal. This is called as internal convergence. So both these phenomena vibrational relaxation and internal convergence occur in the excited state as per the kashia's rule. Then only the electron comes from excited first. Then only electron comes from first excited state to the ground state with simultaneous release of energy in the visible region showing the phenomena of fluesence clear this now there is another rule web's rule which states that the wavelength of light at which the excitation is taking place means the excitation wavelength does not influence the wavelength of light at which emission is taking place because some of the energy has already been lost in the excited state. Clear?

[00:25:27]It does not we can never say that since light of very high energy has been absorbed now the light energy that is being emitted will also be high. No, this is not pos. No, this is not the case because even if the higher energy radiation is absorbed and molecule gains large energy causing electrons to go even to S3 level but from S3 they cannot directly come to S not no this is not possible according to Kasha's rule from S3 it will first go to S_sub_2 then to S1 and finally it comes back to S not so during these transition from S3 to S2 S2 to S1 the molecules loses the molecules lose some energy and therefore the energy that is being released in the form of light wavelength in the form in the form of light wavelength when it comes from S1 to S not is totally independent upon the energy that is being absorbed. So this is where we love rule. So we have learned two rules. Kasha's rule that relaxation that vibrational relaxation as well as internal conversion takes place and which causes the electron in the excited state to come to the first excited state and the lowest vibrational level of the first excited state and then electron comes back to ground state emitting light radiations.

[00:26:50]Second villo's rule says that the wavelength of light emitted is independent of the wavelength of light that is being used for excitation.

[00:26:58]Because when excitation takes place, electrons go to the higher excited state. The excess energy is already being lost in the excited state only and the very small amount of energy that is being left that that is being that is being left is emit and the small amount of energy that is being left is emitted when electron comes back finally to the ground state. So this wavelength is independent of the excitation wavelength.

[00:27:28]Now very famous is turn volmer equation.

[00:27:31]This equation is basically used when we are using quenching methods.

[00:27:37]As you all know that quenching methods are very useful methods for studying fluoresence. And the beauty of these methods is that even compounds that do not flourish naturally they can be studied by these methods.

[00:27:52]And we all know that for example we have a molecule which is not a quench. For example, we have a molecule which is a fluorescent molecule. It under goes excitation that excited state is shown by a steric.

[00:28:09]Now we add a quencher over here. This quencher forms a complex and molecule comes back to the ground state or sometimes what happen this quencher gets excited while molecules again comes to the ground state. So this happens in some cases because because so this happens in case of quenching studies.

[00:28:32]Clear? Even if a compound which is not a fluorescent compound, we can study by by this method. How? For example, we have a compound A which is not a fluorescent compound. Now we consider this compound as quencher.

[00:28:49]Suppose we have a compound X which is not showing fluesence. Now we consider this as a quencher.

[00:28:58]Now we consider this as a quencher and we add it to a fluorescent molecule.

[00:29:07]Now this fluorescent molecule is showing fluoresence emitting light at some fluorescent intensity.

[00:29:13]When it when it combines it forms a complex with it thereby causes reduction in intensity in intensity or it is called as or it causes reduction in intensity also known as quenching.

[00:29:34]So in this way we can identify the presence of this compound as a quencher. If the quenching intensity going if the fluent intensity is going down that means a quencher is present. If there is no effect on the fluent intensity that means there is no compound then that means that there is no non-fllorescent compound present. Now quenching is of two types. One is called as dynamic quenching other is called as static quenching. In case of static quenching the quencher forms a complex with the fluorescent molecule in the ground state only so that it will not undergo excitation resulting into emission of light. While in case of dynamic quenching the fluoroscent molecule undergo excitation and in the excited state the quencher goes and forms a complex. So this is the difference. When the quenching stops in the excited state it is called as dynamic quenching. When quenching occur at excited state, it is dynamic quenching. When it occurs at ground state, this is static quenching.

[00:30:33]Now, this quenching is related to some parameters. For example, I not so this quenching is related to some parameters and this equation is called as and this relation. So, this quenching is related to some parameters and this relation. So this quenching is related to some parameters and this relation or this equation is called as the very famous stern wmer equation. According to according to this equation for dynamic quenching I upon I is equal to to not upon toao which is equal to 1 + kq to kn square bracket q means then concentration of quencher and it is also equal to 1 + ksv. KSV is the stern wulmer constant. KSV is a stone volmer constant into Q again in square bracket means the concentration of quencher for static quenching we have a simplified relation I not upon I is equal to 1 + KSV Q KSV is the WMER constant I not is the intensity of light in absence of quencher I is the intensity of light in presence of quencher t to not is the lifetime of the excited state in absence of quencher means for how long the light is being emitted the molecule for how long the molecule stays in that excited state in absence of quencher. While toao is the time or the lifetime of the excited state in presence of quencher. Then Q in square bracket is the concentration of the quencher. KQ is the quenching constant. KQ is the quenching constant and KSV as I already told you is the strong WMER constant. Then sometimes we have static as well as dynamic quenching. So in that case the intensity ratio I not upon I means intensity in absence of quencher to the intensity in presence of quencher is equal to 1 + ksvq mean 1 + stern constant cq it comes from dynamic quenching again we have 1 plus ksvq that comes from static quenching since both are taking place simultaneously this causes deviation from the linear relation.

[00:32:46]So for static quenching it is KS into Q.

[00:32:49]KS is called as the static quenching equilibrium constant. KS is the static equilibrium. KS is the static quenching equilibrium constant. And for combined static and dynamic quenching we have I upon I means intensity ratio. I not is the intensity in absence of quencher while I is the intensity in presence of quencher is equal to 1 + ksvq.

[00:33:10]That comes from dynamic quenching. Why while 1 + ksq that comes from static quenching because both the quenching are taking place we have to take the combined uh and that causes a deviation from linear relation of quenching and fluence intensity. So these are the explanations of these terms. I not is the florence intensity in absence of quencher and I is the florence intensity in presence of quencher. TOAO KN is the fluesence lifetime without a quencher means for how long the fluesence is being the light is being emitted. TOAO is the fluence lifetime in presence of quencher. Then we have Q is the concentration of quencher. KQ is the biomolecular quenching rate constant molar inverse second inverse. Then KSV is the stern wmer constant. For dynamic quenching we have KSC is equal to KQ to means the equilibrium quenching equilibrium constant and to not the lifetime of fluesence or the time duration up to which fluesence lighting light is emitted to not into KQ the equilibrium quenching constant and finally KS is the static quenching equilibrium constant. These these are the explanation of these terms. The key concept as you all know fluesence quenching means the reduction in the intensity of the light being emitted during the phenomena of fluoresence and it happens because of the presence of a quencher. Now dynamic quenching as I told you these quenchings are divided into two category the static quenching and dynamic quenching. In case of dynamic wenching the fluoro or the fluorescent molecule undergo excitation by absorption of energy and it reaches to the excited state and in the excited state the quencher comes and form a complex so that in the excited state only the molecule undergo complexation with the quencher and then the energy is released in the form of a non-radiative deactivation. Clear? No visible light is emitted. This is called as dynamic quenching. So it results in a non-radiative process non-radiative deactivation and to explain this stern wmer equation is mainly used constant which with KSV which is defined for for dynamic quenching the stern wmer constant is equal to the product of quenching constant and the tow not which is defined as the lifetime of the fluesence means the time duration for which the fluoresence light is emitted in absence of quencher at time duration to not when multiplied by KQ the equilibrium constant it gives us the value of KSV the stern volmer constant in case of static quenching as I told you it involves a ground state complexation in which the quencher reacts with the fluorescent molecule or the floror in the ground state and does not allow the molecule to go into excited state clear and in this case a modified strenmer equation is used that modified Walmer equation is I upon I equ= = to 1 + ks q where ks is the static quenching equilibrium constant where ks where ks is the static where ks is the static quenching equilibrium constant.

[00:36:28]Again if we draw a plot in which we plot to not upon toao on y-axis while we put quencher concentration on x-axis. This is the ratio of lifetime of fluesence in presence of in absence of quencher and presence of quencher means to not refers to the fluence lifetime in absence of quencher. Tao represents the fluence lifetime in presence of a quencher comes as a straight line. But due to these dynamic and static quenching we may have some deviations from this linear relation.

[00:37:07]Although we we have a linear relation between quencher concentration and the intensity ratio.

[00:37:15]Although we have a linear relation between quencher concentration and ratio of lifetimes of fluence in absence and presence of quencher. But sometimes as I told you we have some deviations from this linear relation. This is upward deviation then we have a downward deviation.

[00:37:32]Clear? And one more thing the slope of this graph the slope of this graph the slope which is obtained from tan theta.

[00:37:43]Theta is the angle or it is obtained from taking the ratio of for example this if this is P this is Q then tan theta or the slope is equal to P upon Q and this is called as stern wmer constant. So if we plot a graph between toao versus Q quenture concentration the slope of that graph gives us the stern wmer constant value.

[00:38:11]In some cases we have deviations from this linear relation. So the causes of nonlinear deviations.

[00:38:20]So the causes of these nonlinearity.

[00:38:23]So the causes of these nonlinearity are in the stern warmer plot. In the stmer plot these deviation occur due to quenching.

[00:38:38]Even for a purely static quenching we observed a linear relationship between strong wmer constant.

[00:38:50]This but because static quenching does not affect fluoresence lifetime because it is a static quenching. Here the quenching is taking place in the ground state only. The molecule is not going to the excited state. But in case of dynamic quenching it reduces the lifetime of the because if dynamic quenching is not taking place it exists for a longer period of time in excited state but since dynamic quenching is taking place the lifetime changes. So we can say that dynamic quenching changes the lifetime of excited state while static while static quenching does not affect the lifetime of flloresence and therefore we observe two types of deviation. The upward deviation is because of the simultaneous presence of both static and dynamic quenching.

[00:39:32]Although static quench quenching does not affect much but dynamic quenching affect the linearity because it changes the lifetime. And then a downward deviation exist when we have two types of fluoor means two types of fluence emitters and we those emitters have different stern warmer constant and because of those because it creates a complicated phenomena as we have two flu fluoro force like fluence emitters and we have two different KSVS that created a downward deviation. So we can say that although this graph from which we calculate the estron wmer constant the this is linear the estron wmer relationship is linear means the graph of to not by toao versus q or since to by toa is also equal to i by i. So we can say that t to not by tow or i by i versus q gives a straight line because and that from the slope we obtain ksv.

[00:40:36]But in some cases when both static dynamic quenching is taking place. We have some deviations. Although if only static devi although if stat only static quenching is taking place no deviation from this model because static quenching does not affect the lifetime means it does not affect the value of toao because since since the quenching takes place in the ground state the molecule does not the fluorophor does not reaches the excited state. While in case of dynamic quenching since the fluoro4 reaches to the excited state if dynamic quenching is not taking place it may have a higher life longer lifetime but if dynamic quenching is taking place quickly it may have a shorter lifetime. So we can say that dynamic quenching affect the lifetime of fluoro4 and since it is affecting the lifetime it is affecting the value of toao we may have deviation and two types of deviation upward deviation and downward deviation. Now upward deviation is now upward deviation occurs when we have both these quenching static and dynamic quenching taking place together. We have this upward deviation while in case of downward deviation we actually have in a sample two fluoro force. We have two fluorescent molecule. We have two fluorescent molecules which causes the emission of light. And because of these two fluoro force we have two stern wulmer constant here in this case KS V1 and KSV V2. Because of these two constant we have a complicated phenomena and we are and therefore the relation between I by I versus Q or to not by Q does not remains linear. it goes nonlinear and we observed a downward deviation. So this is all about the stronmer constant. So this is all about equation as well as the kasha rule and wavov's rule. Now we'll uh now we'll discuss few examples or few now we'll discuss about few problems relating to fluoresence spectrum and one more thing in many systems at low concentration of quencher means if the value of Q is low we have a linear relationship but at high Q concent we have nonlinear nonlinear Here it may be upward or low downward. But for low cube low quenture concentration we have a linear relationship. Clear? So this is also an important point that if the quencher concentration is low we have a linear relationship while we have a linear strand relationship. On the other hand, if the quenture concentration is high, the linear relationship gets converted into nonlinear relationship and we may have upward or downward deviation from stern wmer relationship. Clear? So let us discuss few problems relating to So let us discuss few problems relating to fluosense spectroscopy. The first problem the so the first scenario a chemist is using fluence spectroscopy to measure the concentration of a fluorescent die. A standard solution of the D with a concentration of 10^ - 6 molar low concentration gives an emission intensity of 4.8 m volts. An unknown sample of the same D means this is for calibration purposes. An unknown sample of the same D produces a reading of 2.2 molt. Under the same instrument, what is the concentration of the unknown sample? Because he the concentration at this concentration the signal is 4.5.

[00:44:15]What is the concentration at 22 m volts?

[00:44:19]The question is the concentration.

[00:44:22]So for this we use the concept of calibration curve. From there we can have this relation means because the fluence intensity with concentration there's a direct relationship. They are linearly related. So we can say that fluesence for sample upon f fluesence for standard is equal to concentration of the sample upon concentration of the standard. This formula has been applied 2.2 m volts is the intensity and earlier we have 4.8 while we need to know the concentration of sample while it is 1 into 10^ - 6. We do the calculation and we finally obtain that the concentration of the sample is 4.53 into 10 ^ -7 mill 4.53 into 10 ^ - 7 moles or it can also be written as 0.453 10 ^ -6 molar while in case of the standard it is 1 into 10^ - 6. So this is having less concentration as compared to the standard.

[00:45:30]Then another question look at this scenario of fluorescent molecule in solution has an initial fluoresence intensity of 8 m volts. After adding a quenching agent of since we are adding a quenching agent of course there should be a decrease in intensity the intensity drops to 5.5 m.

[00:45:49]So we put this I not is 8 m volts when then I is 5.5 m volt 1 + ksv the concentration of Q is 6 molers we divide it we get 1.45 this is 1.45 1 plus KSV6 molar then we subtract it we get 0.45 45 this we solve it and the value of KSV is 0.75 per mole the value of his volmer constant comes out to be 0.75 mole inverse again there's another scenario you obtained the fluence excitation and emission spectra for a new compound the excitation peak comes at 350 nanome while emission peaks comes at 410 nanome ter. The question is what is the stroke shifts in nanometers for this compound?

[00:46:42]As we all know that when we have spectrum like this, this is excitation.

[00:46:48]This is emission. The difference between these two maxima is the stroke shift.

[00:46:59]We subtract these four 10 minus 350. The answer is 60. Then the second question, why is the emission peak at a longer wavelength than excitation peak? We all know that due to relaxation due to we all know that due to vibrational relaxations in the excited state as well as internal conversion, some energy is lost as a non radiative energy and the remaining amount of energy is emitted when electron comes back. So this energy is always lesser than the energy which is being absorbed and therefore the emission occurs at longer wavelength.

[00:47:31]If you want to perform a quantitative measurement, what excitation would you choose and why? This we'll discuss. So again the stroke shifts just by subtracting the excitation and emission wavelengths we get 60 nanometer. And for the second question, why it is emitting light at longer wavelength? Because some of the light because some of the energy is lost as non-radiative process because some of the energy is lost as non-radiative process like vibrational relaxation and even due to internal conversion. This causes energy of the molecule to go to a lesser value and that lesser energy is released when it comes back to ground state in the form of visible light. Therefore less energy means longer wavelength and therefore we observed at longer wavelength.

[00:48:22]So this so this as I just mentioned that due to internal conversion and vibrational relaxation some energy is lost the remaining energy is released. Of course since the energy is low it release with longer wavelength. Now the third question is what wavelength would you choose and why? The optimal excitation wavelength to choose is 350 nanometer which is already given in this question.

[00:48:48]This is because the wavelength as I said according to WLO's rule the excitation wavelength does not affect the wavelength of the emitted light. It only affects the intensity of the signal. So this is because the wavelength correspond to excitation spectrum exam.

[00:49:04]Exciting the molecule at this wavelength ensure the most efficient absorption which in turn produces most intense fluent signal and the best sensitivity for the quantitative analysis.

[00:49:15]Therefore, we are not going to change the wavelength as it is already given that this is the optimal excitation wavelength for this sample. Means it is the lambda max at which maximum absorption is occurring for this particular molecule. If we change the wavelength lesser absorption will occur and therefore the in fluoresence in and therefore the intensity of the emitted signal would be low. Although this wavelength does not affect the wavelength of the emitted light means it is not going to change the x-axis value but of course it changes the signal height means the intensity of the fluorescent light. Therefore, if we choose some other wavelength, the absorption will not be very high because this particular wavelength is the optimal wavelength already given in the equation which means it is the lambda max at which maximum absorption occur.

[00:50:05]When maximum absorption occur, the intensity of the emitted signal would also be high and for high signal V means better sensitivity of the spectrum for this particular quantitative analysis. So we'll keep this excitation at this same value only.

[00:50:26]Now we have another question.

[00:50:29]If we have we record an excitation and emission is maxima for a fluoro4 means for the flu and sample at 330 nanome and 465 nanome. So there's a big difference means at 330 it is absorbing light causing excitation and it is emitting light at 465 nanometers. What is the stroke shift again and then then convert stock shift into wave number. First of all we calculate the stock shifts just by subtracting 465 and 330 we'll get 15 465 and 330 we'll get 135 nanome. So this 195 135 nanometer is the stock shift. So 135 nanometer is the stock shift means it is the difference in the wavelength maxima. Now in the second part we have been asked to convert this into wave number and as we all know that l wavelength and wave numbers are reciprocals of each other means lambda is equal to new bar. New bar is the wave number or if you want to calculate new bar it is equal to 1 upon lambda. So in this case it is 1 upon 135 nanome and we all know that 1 nanome is equal to 10 ^ - 9 m or it is 10 ^ -7 cm. So finally it will become 1 upon 135 into 10 to the power - 7 which ultimately goes in the numerator 10^ 7 upon 135 which turns out to be something around 8,000 which which turns out to be something around 8,000 8 7 98 of course cm inverse Its unit is centimeter inverse because we have put this conversion here.

[00:52:30]So we have two ways to do this calculation. First of all either we can convert both these wavelengths into wave number and then take the difference or we first of all take the difference. The answer is 135 and then we can convert this 135 into wave number. So I have explained you by noting down the difference in terms of wave length and then convert it into wave number. In this solution we have given the other method 335 nanome again 330 into 10 ^ - 7 then it then it is being converted 3.3 10^ - 5 cm again 4.75 it is converted in terms of cm. This wavelength has been converted into centimeter first of all.

[00:53:17]Then because we are going to calculate wave number which is reciprocal of a wavelength. So wave number is calculated by for excitation this wave number is 1 upon 3.3 10^ - 5 comes out to be 3,00 it comes out to be 30,3.

[00:53:35]Then for emission again reciprocal of wavelength 1 upon 4.65 65 10^ - 5 cm. It comes out to be 21,000. When we take the difference, the difference is 8798. A little bit long process. Therefore, it is better first of all to just to subtract the two wavelengths to obtain a stroke shift which is 105 nanometer. Then directly convert it by this relation. 10^ 7 upon 105 will give you the value of game number which is 8 7 98. This is the easiest way of solving this question.

[00:54:16]Another question. A fluence peak comes at 500 nanometers. Floence peak means it is the emission peak. Emission peak.

[00:54:28]Now what is the wave number in centimeter inverse? As I just told you, if you want to convert wavelength into wave number, just do the reciprocal and since it is in nanometer, answer is required in centimeter. Just put 10 ^ 7 in numerator while the value of wavelength in nanometer. Here do the calculations. It turns out to be 10 to the^ 3 into 10 to the power 4. Just for solving our question, this 10^ 3 becomes 1,000.

[00:55:00]upon 500. We have this is 10 days to ^ 4.

[00:55:06]This turns out to be 2. Then we have 2 into 10 days to ^ 4 cm inverse is the answer. 2 into 10 ^ 4 cm inverse means 20,000 cm inverse is the wave number corresponding to this wavelength of 500 nanometers.

[00:55:28]The last question light with wave number 25,000 speed of light we all know around three lakhs we consider in kilometers/s but it is exactly equal to 2.998 10 to the power 8 m/s or it is 2.998 into 10 ^ 5 kilome/s Although it is given in centm. So from meter we convert it into cm. Again we multiply by 10^2. So it becomes 10^ 10 cm/s.

[00:56:11]Now wave number is given, speed is given. We have been asked to calculate the frequency and we all know that v is equal to n lambda.

[00:56:21]Okay. And we all know that V is equal to And we all know that C which is the speed of light it is equal to frequency into lambda the wavelength.

[00:56:39]And since lambda is equal to 1 upon new bar wave number because they are reciprocal we convert this equation into the form of wave number. So this is speed of light. Then we have frequency and then lambda is replaced with new bar.

[00:56:58]We do cross multiply.

[00:57:02]Wave number comes here and equation becomes new upon C. Now frequency is being asked. Frequency will be equal to C into wave number. So it is 2. So it is 2.998 10 raised to the power 10 * 25,000 means 25 into 10 days to the power 3 it will become 7.495 into 10 days to the power 14.

[00:57:38]It will become 74.95 into 10 days to the power 13 or frequency will be 7.495 into 10 days to the power 14. This is the answer mean per second or you we can also do it like that that we first convert this 25,000 into wavelength. This 25,000 can be calculated into wavelength by 1 upon 25,000.

[00:58:11]So it will become so it will become 4 into 10 to the power -4 become 4 into 10 to the power minus 5 the wavelength means it is 400 nanometers.

[00:58:30]So we put this value because it is in cm. Here velocity is equal to frequency into lambda velocity is equal to frequency into lambda and frequency is equal to velocity upon lambda. So in that way this frequency will be equal to 2.998 10 raised to the power 10 divided by 4 into 10 to the power - 5 again it turns out to be 7.495 into 10 to the power 14 second inverse.

[00:59:13]So we can calculate it the either way either by first of all changing new bar wave number into wavelength and then applying the formula C is equal to frequency into wavelength or we directly apply the formula C is equal to frequency into wavelength and then convert wavelength into 1 upon wave number and then by the relation of frequency is equal to speed C into wave number we multiply and get the answer. So in this way we can calculate uh so in this way we can calculate the value of frequency and this is how we perform uh and this is how we for and this is how we solve problems relating to flow sensor spectrum with this I hope I have covered all the aspect of and so with this I have uh so with this I hope I have covered all all the aspects of molecular luminance spectroscopy including the photooluminescence fluesence phosphorusence including the photooluminescence as well as chemoluminence. In photooluminence I've covered fluesence phosphorusence the instrumentation of fluesence the principle methods of taking fluesence measurement like and methods of take and methods of measuring the fluesence intensity of a sample by direct method for naturally fluorescent compound and indirect method of derivatization or quenching methods for non-flloroscent compounds. Then we have studied about few rules like the Kasha rule, Stern Wulmer equation as well as the Villo's rule and finally we have solved few questions. So with this I hope that you have learned about these molecular luminance spectroscopy and now you will be able to solve questions related to this particular topic.

[01:01:05]With this I am again very thankful to all of you for patience viewing this.

[01:01:12]With this I am again thankful to all of you for patiently watching my lectures and I hope these lectures would help you to improve your understanding about the topics of advanced analytical techniques and you will get success in your career.

[01:01:28]Take care. Thank you very much.