Atomic Structure and Bonding Questions and Answers

Answer: C

The most suitable is a direct bandgap material with a large bandgap.

The photon energy at 380 nm is

\[ \begin{aligned} E &= \frac{hc}{\lambda} = \frac{1240\ \text{eV nm}}{380\ \text{nm}} \\ &\approx 3.26\ \text{eV} \end{aligned} \]

so the bandgap must be about 3.3 eV, which is large; materials such as GaN and InGaN are used for near-ultraviolet and violet LEDs.

The gap must also be direct, so that an electron and hole can recombine and emit a photon without needing a phonon to supply momentum; indirect materials such as silicon emit light very inefficiently. A small bandgap would emit in the infrared, not at 380 nm.

Answer: B

The conductivity is \(1.60 \times 10^{-1}\) S/m.

The electron mobility is the drift velocity per unit field, and conductivity follows from \(\sigma = ne\mu\):

\[ \begin{aligned} \mu &= \frac{v_{d}}{E} = \frac{100}{500} = 0.2\ \text{m}^{2}\text{/V s} \\ \sigma &= (5 \times 10^{18})(1.602 \times 10^{-19})(0.2) \\ &= 0.16\ \text{S/m} \end{aligned} \]

That is \(1.60 \times 10^{-1}\) S/m. \(0.16 \times 10^{-1}\) is ten times too small, and 2.50 and 30.05 come from using the drift velocity or field in place of the mobility.

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.