Materials Science Questions and Answers

Answer: B

The material is diamagnetic.

A diamagnetic material has a small negative susceptibility: the applied field induces a magnetisation opposing it, so the material is pushed towards weaker field, that is, repelled.

Paramagnetic and ferromagnetic materials have positive susceptibility and are attracted, ferromagnets strongly. Antiferromagnetic materials have a small positive susceptibility as well, so they are weakly attracted, not repelled.

Answer: C

The true statement is a phonon is annihilated in the process.

In an indirect band gap semiconductor the bottom of the conduction band lies at a different crystal momentum from the top of the valence band. A photon carries almost no momentum, so the transition can only conserve momentum if a phonon takes part, either absorbed (annihilated) or emitted. Of the options, phonon absorption is the one that can occur.

The electron's energy increases, since it moves up into the conduction band, so 'energy decreases' is wrong. A photon is absorbed, not emitted or created, in a valence-to-conduction transition; emission belongs to the reverse process, recombination.

Answer: D

A non-linear triatomic molecule has 3 vibrational degrees of freedom.

Three atoms have \(3N = 9\) degrees of freedom in all. Three are translations of the centre of mass and, for a non-linear molecule, three are rotations, leaving

\[ 3N - 6 = 9 - 6 = 3 \]

vibrations; water, for example, has a symmetric stretch, an asymmetric stretch and a bend. 9 is the total, 6 counts translations and rotations, and 4 is the vibrational count for a linear triatomic (\(3N - 5\)), such as CO₂.

Answer: A

The mass of Mg dissolved is 6 g.

By Faraday's law the charge gives the moles of electrons, and each Mg atom releases two electrons:

\[ \begin{aligned} n_{e} &= \frac{48250}{96500} = 0.5\ \text{mol} \\ n_{\text{Mg}} &= \frac{0.5}{2} = 0.25\ \text{mol} \\ m &= 0.25 \times 24 = 6\ \text{g} \end{aligned} \]

12 g forgets the factor of two electrons per Mg atom, and 24 and 48 g would need two and four times the charge given.

Answer: B

The current is 1280 mA.

In an intrinsic semiconductor the electron and hole densities are equal, so both carriers contribute to the conductivity:

\[ \begin{aligned} \sigma &= ne(\mu_{n} + \mu_{p}) \\ &= (10^{12})(1.6\times10^{-19}) \\ &\quad \times (8\times10^{4}) \\ &= 1.28\times10^{-2} \text{ S/cm} \end{aligned} \]

With 100 V across a 1 cm cube the field is \(E = 100\) V/cm, so the current density is \(J = \sigma E = 1.28\) A/cm². Across the 1 cm² face, \(I = 1.28\) A = 1280 mA.

640 mA counts only the electrons (the hole current is equal to it), and 6400 and 12800 are five and ten times too large through a power-of-ten slip in the carrier density.

Answer: C

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

• Cu_0.5Al_0.5 is an alloy of two metals: metallic bonding.
• ZnS (zinc blende) has tetrahedral, largely covalent bonds.
• Na₂O, an alkali-metal oxide, is ionic.
• Li₄SiO₄ has covalent Si-O bonds within the silicate tetrahedra and ionic bonding between Li⁺ and those groups: mixed.

Each of the other options gives the alloy or the oxide the wrong bond type, for example calling Na₂O metallic or ZnS ionic.

Answer: A

The statement taken as true is that electrical conductivity increases with increasing temperature and pressure.

In an intrinsic semiconductor the carriers are thermally excited across the band gap, and their concentration varies as \(\exp(-E_{g}/2kT)\). Raising the temperature therefore increases the conductivity sharply, which rules out both options with decreasing temperature.

For pressure, the paper follows the textbook picture in which compressing the lattice increases the overlap of atomic orbitals, broadens the bands and narrows the gap, so more carriers are excited (silicon behaves this way; some semiconductors, such as germanium, show the opposite trend). On that picture conductivity rises with pressure, which rules out 'decreasing pressure'.

Answer: A

The right step is to choose a metallic adhesion layer with very low interfacial energy with the deposited thin film.

Whether a film grows layer by layer or as islands depends on wetting: layer (Frank-van der Merwe) growth is favoured when \(\gamma_{substrate} \ge \gamma_{film} + \gamma_{interface}\). A low interfacial energy between the adhesion layer and the metal film makes the film wet the layer, so it spreads into a continuous planar film instead of balling up into islands.

Choosing the layer only for its match with SiO₂ ignores the film, whose wetting is what matters. Raising the substrate temperature and lowering the rate gives atoms time to reach their equilibrium arrangement, which is islands when wetting is poor; and annealing between stages encourages dewetting and agglomeration.

Answer: C

The material best suited to a permanent magnet is (C).

A permanent magnet must keep a large magnetization when the field is removed (high remanence, where the loop crosses the M axis) and must resist being demagnetized (high coercivity, where it crosses the H axis). Such hard magnetic materials have broad, square loops enclosing a large area.

Loop (C) is the widest, crossing the H axis at about 2.5 units, and also the tallest, so it has both the largest coercivity and the largest remanence. Loops (A), (B) and (D) are progressively narrower: a few tenths to just over one unit of field would demagnetize them. Narrow loops suit soft magnetic materials, such as transformer cores, where low hysteresis loss matters.

Answer: D

The correct curve is (D).

A photon can be absorbed by exciting an electron across the band gap only if its energy is at least \(E_{g}\). For \(E_{g} \approx 2\) eV, light below about 2 eV passes through with little absorption, and above about 2 eV absorption rises steeply to a high value. That is the step up near 2 eV in curve (D).

Curve (A) is reversed, absorbing low-energy photons and transmitting high-energy ones. The flat curves (B) and (C) show no absorption edge at all, which is how a metal (strong absorption at all energies) or a wide-gap insulator in this range might behave, not a 2 eV semiconductor.