Chapter 2.9, Problem 70SEP

Most modern scanning electron microscopes (SEMs) are equipped with energy-dispersive X-ray detectors for the purpose of chemical analysis of the specimens. This X-ray analysis is a natural extension of the capability of the SEM because the electrons that are used to form the image are also capable of creating characteristic X-rays in the sample. When the electron beam hits the specimen, X-rays specific to the elements in the specimen are created. These can be detected and used to deduce the composition of the specimen from the well-known wavelengths of the characteristic X-rays of the elements. For example:

Suppose a metallic alloy is examined in an SEM and three different X-ray energies are detected. If the three energies are 7492, 5426, and 6417 eV, what elements are present in the sample? What would you call such an alloy? (Look ahead to Chap. 9 in the textbook.)

To determine

The element present in the sample when the X-ray energy is 7462 eV.

Answer to Problem 70SEP

The element present in the sample when the X-ray energy is 7462 eV is $\text{nickel}$.

Explanation of Solution

Write the equation to calculate the wavelength of the characteristic X-rays of the element $\left(\lambda \right)$.

 $\lambda =\frac{hc}{E}$                                                                                                                      (I)

Here, energy of the radiation is $E$, Planck's constant is $h$ and speed of light is c.

Conclusion:

The values of Planck constant and speed of light are $6.63\times {10}^{-34}\text{J}\cdot \text{s}$ and $3\times {10}^8\text{m}/\text{s}$ respectively.

Substitute $7462\text{eV}$ for $E$, $6.63\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$ and $3\times {10}^8\text{m}/\text{s}$ for c in Equation (I).

 $\begin{array}{c}\lambda =\frac{\left(6.63\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{7462\text{eV}}\\ =\frac{\left(6.63\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(7462\text{eV}\times \frac{1.602\times {10}^{-19}\text{J}}{1\text{eV}}\right)}\\ =1.659\times {10}^{-10}\text{m}\\ =0.1659\text{nm}\end{array}$

From the given table, the wavelength value of $0.1659\text{nm}$ corresponds to nickel $\left(\text{Ni}\right)$.

Thus, the element present in the sample when the X-ray energy is 7462 eV is $\text{nickel}$.

To determine

The element present in the sample when the X-ray energy is 5426 eV.

Answer to Problem 70SEP

The element present in the sample when the X-ray energy is 5426 eV is $\text{chromium}$.

Explanation of Solution

Conclusion:

Substitute $5426\text{eV}$ for $E$, $6.63\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$ and $3\times {10}^8\text{m}/\text{s}$ for c in Equation (I).

 $\begin{array}{c}\lambda =\frac{\left(6.63\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{5426\text{eV}}\\ =\frac{\left(6.63\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(5426\text{eV}\times \frac{1.602\times {10}^{-19}\text{J}}{1\text{eV}}\right)}\\ =2.291\times {10}^{-10}\text{m}\\ =0.2291\text{nm}\end{array}$

From the given table, the wavelength value of $0.2291\text{nm}$ corresponds to chromium $\left(\text{Cr}\right)$.

Thus, the element present in the sample when the X-ray energy is 5426 eV is $\text{chromium}$.

To determine

The element present in the sample when the X-ray energy is 6417 eV.

Answer to Problem 70SEP

The element present in the sample when the X-ray energy is 6417 eV is $\text{iron}$.

Explanation of Solution

Conclusion:

Substitute $6417\text{eV}$ for $E$, $6.63\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$ and $3\times {10}^8\text{m}/\text{s}$ for c in Equation (I).

 $\begin{array}{c}\lambda =\frac{\left(6.63\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{6417\text{eV}}\\ =\frac{\left(6.63\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(6417\text{eV}\times \frac{1.602\times {10}^{-19}\text{J}}{1\text{eV}}\right)}\\ =1.937\times {10}^{-10}\text{m}\\ =0.1937\text{nm}\end{array}$

From the given table, the wavelength value of $0.1937\text{nm}$ corresponds to iron $\left(\text{Fe}\right)$.

Thus, the element present in the sample when the X-ray energy is 6417 eV is $\text{iron}$.

The elements, chromium, iron and nickel present in the sample represent the principal constituents of the austenitic stainless steels.