Chapter 21, Problem 21.14QAP

Interpretation Introduction

(a)

Interpretation:

The factors related to the analyte influencing the emission intensity needs to be identified.

Concept introduction:

X-Ray photoelectron spectroscopy or XPS is a quantitative spectroscopic technique that is surface-sensitive. It measures the composition of materials with respect to elements in the range of parts per thousand. It also helps in determining the empirical formula, chemical state and electronic state of elements present in the materials.

It is now widely used because with the help of this technique not only the element within the films, but the other elements bonded to it can be determined.

For example, with the help of XPS technique, the oxidation state of metal in the metal oxide can be determined.

Explanation of Solution

The equation relating the intensity of emission and atomic concentration (density) is represented as follows:

$I=n\varphi \sigma \epsilon \eta ATl$

Here, $n$ is number density of atoms $\left(\text{atoms}/{\text{cm}}^{-3}\right)$ of the sample, $\varphi$ is the flux of incident X-ray beam $\left({\text{photons cm}}^{-3}{\text{ s}}^{-1}\right)$, $\sigma$ is the photoelectric cross-section for the transition $\left({\text{cm}}^2/\text{atom}\right)$, $\epsilon$ is the angular efficiency factor for the instrument, $\eta$ is the efficiency of producing photoelectrons, $A$ area of sample from which photoelectrons are detected $\left({\text{cm}}^2\right)$, $T$ is the efficiency of detection of photoelectrons and $l$ is the mean free path of the photoelectrons in the sample $\left({\text{cm}}^2\right)$.

In spectroscopy, the analyte is the molecule for which the concentration needs to be measured using spectroscopic techniques. In the above formula, the factors related to analyte are $n$ -number of density of atoms of the sample measured in atoms per cubic centimeter, $\eta$ -efficiency of producing photoelectrons, A -area of the sample from which photoelectrons are detected in centimeter square and l is the mean free path of the photoelectrons in the sample in nm.

Interpretation Introduction

(b)

Interpretation:

The factors related to the spectrometer influencing the emission intensity needs to be identified.

Concept introduction:

X-Ray photoelectron spectroscopy or XPS is a quantitative spectroscopic technique that is surface-sensitive. It measures the composition of materials with respect to elements in the range of parts per thousand. It also helps in determining the empirical formula, chemical state and electronic state of elements present in the materials.

It is now widely used because with the help of this technique not only the element within the films, but the other elements bonded to it can be determined.

For example, with the help of XPS technique, the oxidation state of metal in the metal oxide can be determined.

Explanation of Solution

The equation relating the intensity of emission and atomic concentration (density) is represented as follows:

$I=n\varphi \sigma \epsilon \eta ATl$

Here, $n$ is number density of atoms $\left(\text{atoms}/{\text{cm}}^{-3}\right)$ of the sample, $\varphi$ is the flux of incident X-ray beam $\left({\text{photons cm}}^{-3}{\text{ s}}^{-1}\right)$, $\sigma$ is the photoelectric cross-section for the transition $\left({\text{cm}}^2/\text{atom}\right)$, $\epsilon$ is the angular efficiency factor for the instrument, $\eta$ is the efficiency of producing photoelectrons, $A$ area of sample from which photoelectrons are detected $\left({\text{cm}}^2\right)$, $T$ is the efficiency of detection of photoelectrons and $l$ is the mean free path of the photoelectrons in the sample $\left({\text{cm}}^2\right)$.

Now, the factors influencing the emission intensity related to the spectrometer or instrument are $\varphi$ -flux of incident X-ray beam measured in photons per cubic centimeter per second, $\sigma$ -photoelectric cross-section for the transition, $\epsilon$ -angular efficiency factor for the instrument and $T$ -efficiency of detection of photoelectrons.

Interpretation Introduction

(c)

Interpretation:

The relation between the ratio of measured quantity in XPS for analyte to the internal standard and their atomic concentrations needs to be determined.

Concept introduction:

X-Ray photoelectron spectroscopy or XPS is a quantitative spectroscopic technique that is surface-sensitive. It measures the composition of materials with respect to elements in the range of parts per thousand. It also helps in determining the empirical formula, chemical state and electronic state of elements present in the materials.

It is now widely used because with the help of this technique not only the element within the films, but the other elements bonded to it can be determined.

For example, with the help of XPS technique, the oxidation state of metal in the metal oxide can be determined.

Explanation of Solution

In the quantitative XPS, the measured quantity is usually I/S. Here, I is the peak area and S is the sensitivity factor. For the analyte, the quantity measured is ${\left(I/S\right)}_a$ and that of internal standard is ${\left(I/S\right)}_s$.

The internal standard is the concentration of a substance present in every sample needed to be analyzed.

The ratio I/S is directly proportional to the concentration on the surface.

Or,

${\left(\frac{I}{S}\right)}_a=C_a$

Similarly,

${\left(\frac{I}{S}\right)}_s=C_s$

Thus, the relation between the ratio $\frac{{\left(I/S\right)}_a}{{\left(I/S\right)}_s}$ with the atomic concentration ratio of analyte to the internal standard is as follows:

$\frac{{\left(I/S\right)}_a}{{\left(I/S\right)}_s}=\frac{C_a}{C_s}$

Interpretation Introduction

(d)

Interpretation:

The given equation for the fractional atomic concentration of element A needs to be proved.

Concept introduction:

X-Ray photoelectron spectroscopy or XPS is a quantitative spectroscopic technique that is surface-sensitive. It measures the composition of materials with respect to elements in the range of parts per thousand. It also helps in determining the empirical formula, chemical state and electronic state of elements present in the materials.

It is now widely used because with the help of this technique not only the element within the films, but the other elements bonded to it can be determined.

For example, with the help of XPS technique, the oxidation state of metal in the metal oxide can be determined.

Explanation of Solution

The atomic concentration for an element (say 1) measured by XPS technique can be represented as ${\left(\frac{I}{S}\right)}_1$. Similarly, for all the second element the atomic concentration measured will be ${\left(\frac{I}{S}\right)}_2$. The fractional atomic concentration can be determined by taking the ratio of the atomic concentration of that element to the total number of atomic concentrations of all the elements present in the sample.

If the total number of elements is n, the fractional atomic concentration can be represented as follows:

$f_1=\frac{{\left(\frac{I}{S}\right)}_1}{{\left(\frac{I}{S}\right)}_1+{\left(\frac{I}{S}\right)}_2+\mathrm{.....}+{\left(\frac{I}{S}\right)}_n}$

Or,

$f_1=\frac{{\left(\frac{I}{S}\right)}_1}{{\displaystyle \sum {\left(\frac{I}{S}\right)}_n}}$

Thus, for all the elements measured by XPS technique, the fractional atomic concentration for element A can be represented as follows:

$f_A=\frac{I_A/S_A}{{\displaystyle \sum \left(I_n/S_n\right)}}$

Here, $I_n$ is the measured peak for n element and $S_n$ is the sensitivity factor for the peak.

Interpretation Introduction

(e)

Interpretation:

The atomic concentration of three elements that is C, N and O in the polyurethane sample needs to be determined.

Concept introduction:

In the quantitative XPS, the measured quantity is usually I/S.

Here, I is the peak area and S is the sensitivity factor. For the analyte, the quantity measured is ${\left(I/S\right)}_a$ and that of internal standard is ${\left(I/S\right)}_s$.

The internal standard is the concentration of a substance present in every sample needed to be analyzed.

The ratio I/S is directly proportional to the concentration on the surface.

Or,

${\left(\frac{I}{S}\right)}_a=C_a$

Similarly,

${\left(\frac{I}{S}\right)}_s=C_s$

Explanation of Solution

The given sample is polyurethane containing C, N and O elements that need to be detected by XPS.

The sensitivity factor for C, N and O is 0.25, 0.42 and 0.66 respectively.

Also, the peak area for C, N and O is 26550, 4475 and 13222 respectively.

The value of the atomic concentration of C, N and O can be calculated as follows:

$A_C=\frac{I_C}{S_C}$

Putting the values,

$A_C=\frac{26550}{0.25}=106200$

Similarly,

$\begin{array}{c}{A}_N=\frac{I_N}{S_N}\\ =\frac{4475}{0.42}\\ =10654.76\end{array}$

And,

$\begin{array}{c}{A}_O=\frac{I_O}{S_O}\\ =\frac{13222}{0.66}\\ =20033.33\end{array}$

Now, the value of the fractional atomic concentration of C can be calculated as follows:

$f_C=\frac{I_C/S_C}{I_C/S_C+I_N/S_N+I_O/S_O}$

Putting all the values,

$f_C=\frac{\left(26550/0.25\right)}{\left(26550/0.25\right)+\left(4475/0.42\right)+\left(13222/0.66\right)}=0.77$

Similarly, for N and O the atomic concentration can be calculated as follows:

$\begin{array}{c}{f}_N=\frac{I_N/S_N}{I_C/S_C+I_N/S_N+I_O/S_O}\\ =\frac{\left(4475/0.42\right)}{\left(26550/0.25\right)+\left(4475/0.42\right)+\left(13222/0.66\right)}\\ =0.0778\end{array}$

And,

$f_O=\frac{I_O/S_O}{I_C/S_C+I_N/S_N+I_O/S_O}$

Putting the values,

$\begin{array}{c}{f}_O=\frac{\left(13222/0.66\right)}{\left(26550/0.25\right)+\left(4475/0.42\right)+\left(13222/0.66\right)}\\ =0.146\end{array}$

Thus, the fractional percentage concentration of C, N and O will be:

$\%f_C=0.77\times 100\%=77\%$

Or,

$\%f_N=0.0778\times 100\%=7.78\%$

Or,

$\%f_O=0.146\times 100\%=14.6\%$

Interpretation Introduction

(f)

Interpretation:

The limitations of quantitative analysis with XPS needs to be explained. The reason for atomic concentration measured to not correspond to the bulk composition needs to be explained.

Concept introduction:

X-Ray photoelectron spectroscopy or XPS is a quantitative spectroscopic technique that is surface-sensitive. It measures the composition of materials with respect to elements in the range of parts per thousand. It also helps in determining the empirical formula, chemical state and electronic state of elements present in the materials.

It is now widely used because with the help of this technique not only the element within the films, but the other elements bonded to it can be determined.

For example, with the help of XPS technique, the oxidation state of metal in the metal oxide can be determined.

Explanation of Solution

The limitation of quantitative analysis with XPS is as follows:

XPS is a surface technique, thus, it provides a limited amount of the organic information i.e. it is limited to elements with atomic Number 3 or more than 3 thus it is not able to detect the hydrogen and helium atom. The method also takes a long time to display. In this Spectroscopic technique the size of sample matters. It is associated with 10% of the relative error.

The reason for the atomic concentrations measured not corresponding to the bulk composition is as follows: -

XPS can determine the quantity of elements present within the top 1-12 nm of the surface of the sample. XPS can only detect those electrons which get escaped from the sample and reach the detector for that photoelectron must travel through the sample. As depth increases, the signal detected from analytes at the surface is much stronger than the signals detected deeper to the surface of the sample.

Interpretation Introduction

(g)

Interpretation:

The percentage error in the atomic concentration of C, N and O needs to be determined.

Concept introduction:

For all the elements measured by XPS technique, the fractional atomic concentration for element A can be represented as follows:

$f_A=\frac{I_A/S_A}{{\displaystyle \sum \left(I_n/S_n\right)}}$

Here, $I_n$ is the measured peak for n element and $S_n$ is the sensitivity factor for the peak.

The atomic concentration of C, N and O for the polyurethane sample is given 76%, 8.0% and 16% respectively.

The percentage error from the values calculated in part (e) can be calculated as follows:

$\begin{array}{l}\%C=\left(\left|77-76\right|\right)\%=1\%\hfill \\ \%N=\left(\left|7.78-8\right|\right)\%=0.22\%\hfill \\ \%O=\left(\left|14.6-16\right|\right)\%=1.4\%\hfill \end{array}$

Thus, the percentage error in the atomic concentration of C, N and O is 1%, 0.22% and 1.4% respectively.

Explanation of Solution

The atomic concentration of C, N and O for the polyurethane sample is given 76%, 8.0% and 16% respectively.

The percentage error from the values calculated in part (e) can be calculated as follows:

$\begin{array}{l}\%C=\left(\left|77-76\right|\right)\%=1\%\hfill \\ \%N=\left(\left|7.78-8\right|\right)\%=0.22\%\hfill \\ \%O=\left(\left|14.6-16\right|\right)\%=1.4\%\hfill \end{array}$

Thus, the percentage error in the atomic concentration of C, N and O is 1%, 0.22% and 1.4% respectively.