Material Testing Methods Questions and Answers

Answer: C

The first peak is at about 44.43°.

For bcc, reflections appear only when h + k + l is even, so the first is (110). With a = 2.880 Å:

\[ \begin{aligned} d_{110} &= \frac{2.880}{\sqrt{2}} = 2.036\ \text{Å} \\ \sin\theta &= \frac{1.54}{2(2.036)} = 0.378 \\ \theta &= 22.2^{\circ},\ 2\theta = 44.4^{\circ} \end{aligned} \]

21.76° is close to the Bragg angle \(\theta\) itself, a reminder that diffractometers report 2θ. 33.05° and 66.10° match no allowed bcc reflection for this lattice parameter; the next peak, (200), falls near 64.7°.

Answer: B

The signal produced by elastic scattering is the Backscattered electron.

Backscattered electrons are beam electrons deflected back out of the specimen by the nuclei of its atoms, through large-angle elastic (Rutherford) scattering, with little loss of energy. Because heavier atoms scatter more strongly, they give atomic-number contrast in the SEM.

The other signals come from inelastic processes. Secondary electrons are low-energy electrons knocked out of the specimen's atoms by the beam. Auger electrons are emitted when an inner-shell vacancy is filled and the energy is passed to another electron. Photoelectrons are ejected by incident photons, as in XPS, not by electron scattering.

Answer: D

Microstrain is measured from the peak Position and broadening.

Non-uniform microstrain spreads the lattice spacings around their mean, which broadens each diffraction peak. Strain broadening grows with angle as \(\beta_{\varepsilon} = 4\varepsilon\tan\theta\), while size broadening varies as \(1/\cos\theta\). Separating the two, as in a Williamson-Hall plot of \(\beta\cos\theta\) against \(\sin\theta\), needs both the breadth of each peak and its position \(\theta\).

Peak area and intensity relate to the amount of a phase and its structure factor, not to strain, so the options built on them are wrong.

Answer: A

The match is P-4, Q-2, R-5, S-1.

• Scanning tunnelling microscopy relies on an atomically sharp conducting tip, through whose apex the tunnelling current flows.
• Scanning electron microscopy images with secondary and backscattered electrons.
• Transmission electron microscopy, especially aberration-corrected high-resolution TEM, reaches sub-angstrom resolution.
• Atomic force microscopy senses forces on a cantilever tip and works in air or liquid, with no vacuum required.

Photoelectrons belong to X-ray photoelectron spectroscopy, which is not in the list, so any option using item 3 is wrong.

Answer: C

The spike appears when heating pure ice from 263 K to 284 K.

A DSC curve shows a peak only where a phase transition absorbs or releases latent heat. Ice melts at 273 K, inside the range 263-284 K, giving an endothermic peak.

The other ranges contain no transition: copper melts at about 1358 K, far above 673 K; water cooled from 323 K to 278 K stays liquid, since it freezes at 273 K; and the Pb-Sn eutectic solidifies at about 456 K (183 °C), above the whole range 323-273 K, so the alloy is already solid.

Answer: B

The missing peak is (220).

For a cubic crystal, \(\sin^{2}\theta = \frac{\lambda^{2}}{4a^{2}}(h^{2} + k^{2} + l^{2})\). In FCC, \(h, k, l\) must be all odd or all even, so the first reflections are (111), (200), (220), (311), (222), with \(h^{2} + k^{2} + l^{2}\) = 3, 4, 8, 11, 12.

Divide each measured value by these numbers to find the constant \(\lambda^{2}/4a^{2}\):

\[ \begin{aligned} 0.136/3 &= 0.0453 \\ 0.185/4 &= 0.0463 \\ 0.504/11 &= 0.0458 \\ 0.544/12 &= 0.0453 \end{aligned} \]

All four give the same constant, about 0.0455, when matched to (111), (200), (311) and (222). The reflection with sum 8 is the one not recorded: (220), expected at \(\sin^{2}\theta \approx 8 \times 0.0455 = 0.364\). (200), (311) and (222) are already in the list.

Answer: C

Differential scanning calorimetry measures heat.

In DSC the sample and an inert reference are heated or cooled on the same temperature programme, and the instrument records the difference in heat flow needed to keep them at the same temperature. Peaks and steps in that heat flow reveal melting, crystallization, glass transitions and reaction enthalpies.

Weight change is measured by thermogravimetric analysis (TGA). Entropy is not measured directly, though it can be derived from DSC enthalpies. Vapour pressure needs a different apparatus, such as an isoteniscope or effusion cell.

Answer: D

The match is P-3; Q-4; R-1; S-2.

• Secondary electrons are low-energy electrons escaping from a thin surface layer, so their yield depends on surface tilt: the SE detector gives a topographic image.
• The backscattered yield rises with atomic number, so the BSE detector gives a compositional contrast image.
• EDS sorts the characteristic X-rays by energy: elemental composition analysis.
• EBSD records the diffraction of backscattered electrons as Kikuchi lines, used to find crystal orientation.

The other options misassign at least one detector; for example P-4 gives secondary electrons the atomic-number contrast that belongs to BSE.

Answer: A

The correct matching is P → 2; Q → 1; R → 4; S → 3.

• Electrical conductivity is measured by the four-probe method, which passes current through the outer probes and measures voltage across the inner ones, removing contact resistance (2).
• Band gap energy comes from UV-Vis absorption spectroscopy: the absorption edge, often analysed with a Tauc plot, gives E_g (1).
• Young's modulus can be found acoustically, from the speed of sound in the material (for a thin rod, \(v = \sqrt{E/\rho}\)) or from resonance frequencies (4).
• Heat of fusion is the area of the melting peak in differential scanning calorimetry (3).

The other options pair conductivity with spectroscopy or band gap with calorimetry, which do not measure those properties.