Atomic Structure and Bonding Questions and Answers

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.

Answer: C

The correct figure is (C).

An intrinsic semiconductor has a filled valence band and an empty conduction band separated by a gap with no allowed states, so the plot must show two bands with a gap. Near each band edge the density of states is proportional to \(\sqrt{E - E_{edge}}\), so \(E - E_{edge} \propto g^{2}\): plotted with E up and g(E) across, each curve leaves the edge with a horizontal tangent and then bends towards the vertical. Each band is finite, so g(E) rises to a maximum and returns to zero at the other edge, which gives the two separate lobes of (C).

(A) has one continuous band and no gap (a metal-like picture). (D) shows the two bands overlapping, which has no gap. (B) does have a gap, but its curves leave the axis steeply (the wrong curvature for a square-root edge) and have no upper limit to the band, so it is not the usual picture.

Answer: B

The correct curve is (B).

Electron occupancy follows the Fermi-Dirac function \(f(E) = 1/(1 + e^{(E - E_F)/kT})\). It is close to 1 well below \(E_F\), close to 0 well above, and exactly 0.5 at \(E_F\). At the hot end of the rod, T is higher, so the change from 1 to 0 is spread over a wider energy range of a few kT. Plot (B) shows f falling smoothly from 1 through 0.5 at \(E_F\) towards 0 as E rises.

Plot (A) has occupancy increasing with energy, which is backwards. Plot (C) is the sharp step of absolute zero, appropriate to a cold end rather than a hot one, and (D), with f = 0.5 at every energy, is not a Fermi distribution.

Answer: A

The unit is A m².

A magnetic dipole moment is the current in a loop times the loop's area, \(m = IA\), so its SI unit is ampere times square metre. Equivalently, it is torque per unit flux density, N m / T, which reduces to A m².

A m⁻¹ is the unit of magnetic field strength H and of magnetisation (dipole moment per unit volume). Wb m⁻² (tesla) is the unit of flux density B. Wb m² is not a standard magnetic quantity.

Answer: A

The Meissner effect is \(B = 0\) at \(T \le T_{c}\).

When a superconductor is cooled below its critical temperature in a magnetic field, it expels the magnetic flux from its interior: the flux density inside becomes zero. The material behaves as a perfect diamagnet (\(\chi = -1\)), which distinguishes a superconductor from a merely perfect conductor.

Above \(T_{c}\) the material is a normal conductor and the field penetrates, so \(B = 0\) at \(T \gt T_{c}\) is wrong, and \(B \ne 0\) below \(T_{c}\) denies the effect. \(\nabla B = 0\) is not a statement of the Meissner effect.

Answer: A

Optical absorption occurs when \(E_{g} \lt E_{p}\).

A photon is absorbed across the band gap by lifting an electron from the valence band into the conduction band, leaving a hole behind. That is possible only if the photon carries at least the band gap energy, so the photon energy must exceed (or at least equal) E_g. This is why a semiconductor is transparent to light with photon energy below its gap.

If \(E_{g} \gt E_{p}\) the photon cannot bridge the gap and passes through. Electron-hole recombination is the reverse process, which emits light (as in an LED) rather than absorbing it, and 'electron-electron recombination' is not a physical process.