# Phase Transformation Questions and Answers – Solid and Vapour Interfaces

This set of Phase Transformation Multiple Choice Questions & Answers (MCQs) focuses on “Solid and Vapour Interfaces”.

1. Which among the following plane has the least density of atoms?
a) {110}
b) {200}
c) {220}
d) {111}

Explanation: {220} has the least density among the following, the density of atoms in these planes decreases as (h2+k2+l2) increases (here the value of h = 2, k = 2, l = 0). Similarly substituting other values we get {220} as the highest.

2. If 1 mol of solid is vaporized 12Na broken bonds are formed. Therefore L (Latent heat of sublimation) = 12 Na*E/2(E is the bond strength of metal). Consequently the energy of a {111} surface is given as ___
a) 0.50*L/Na (J/surface atoms)
b) 0.75*L/Na (J/surface atoms)
c) 0.25*L/Na (J/surface atoms)
d) 1.25*L/Na (J/surface atoms)

Explanation: The energy of a {111} surface is given as 0.25*L/Na (J/surface atoms). This result will only be approximate since second nearest neighbours have been ignored and it has also been assumed that the strengths of the remaining bonds in the surface are unchanged from the bulk values.

3. Surface free energy is given by_____
a) γ= E +PV+TS
b) γ= E +PV/TS
c) γ= E +PV*TS
d) γ= E +PV-TS

Explanation: From the definition of Gibbs free energy the surface free energy will be given by γ= E +PV+TS
Here even if the ‘PV’ term is ignored surface entropy effects must be taken into account. It might be expected that the surface atoms will have more freedom of movement.

4. Considering the nearest neighbours it can be seen that the atoms on a {111} surface, for example, are deprived of three of their twelve neighbours.
a) True
b) False

Explanation: It can be seen that the atoms on a {111} surface, for example, are deprived of three of their twelve neighbours. If the bond strength of the metal is E each bond can be considered as lowering the internal energy of each atom by E/2. Therefore every surface atom with three ‘broken bonds’ has an excess internal energy of 3E/2 over that of the atoms in the bulk.

5. Extra configurational entropy can also be introduced into the surface by the formation of surface vacancies. Entropy effects γ is slightly dependent on temperature and is given as, γ = -2T+2. Calculate the entropy S from the given condition?
a) T
b) -2
c) 2
d) T2

Explanation: The entropy S can be determined using the equation, ∂γ/∂T = -S. Measured values of S are positive and vary between 0 and 3mJ m-2K-1.Hence once we differentiate it with respect to temperature we get the required value.

6. Melting point of Sn, Ag, Al, and Au are 232, 961, 660, and 1063(In Degree Celsius) given in their respective order. Which among the following values mentioned below corresponds to the surface energy of Sn? (Options corresponds to the surface energy of the above mentioned 4 elements in mJ m-2)
a) 680
b) 1080
c) 1390
d) 1120

Explanation: Metal with high melting point temperatures have high value of latent heat of sublimation and high surface energies hence from the given value of melting point one can predict that Sn has the least value of surface energy among the four.

7. A crystal plane is at an angle 45 to the close-packed plane. Calculate the energy of the surface, attributing E/2 energy to each broken bond? (Take the a value as 1(lattice parameter))
a) E/2
b) E
c) 2E
d) (1/√2) E

Explanation: For a crystal plane is at an angle 45 to the close-packed plane, the Surface energy is given by the equation (sinφ +cosφ)*(E/2a2), where φ here is 45 and therefore substituting the values we get the surface energy as (1/√2) E.

8. For an isolated crystal bounded by the planes A1, A2, with energies γ1 and γ2 the total surface energy will be given by _______
a) (A1 γ1) – (A2 γ2)
b) (A1 γ1) / (A2 γ2)
c) (A1- γ1)(A2- γ2)
d) (A1 γ1) + (A2 γ2)

Explanation: When several planes are bounds a single system with their respective energies, the total surface energy is given by the sum of their product with the respective system. That is (A1 γ1) + (A2 γ2) + ……. it goes on depending on the number of planes. Actually the sum of Ai*γi is minimum in the equilibrium shape .

9. If we plot φ (angle between the crystal plane and closed packed plane) vs E (total surface energy), all the low index planes should be located at________
a) Low energy cusps
b) High energy cusps
c) In between the low energy cusps and high energy cusps
d) Cannot be determined

Explanation: It should be noted that the close-packed orientation (φ = 0) lies at a cusped minimum in the energy plot. Similar arguments can be applied to any crystal structure for rotations about any axis from any reasonably close-packed plane. All low-index planes should therefore be located at low-energy cusps.

10. The equilibrium shape has the property that the summation of Ai*γi (Ai represents the planes and γi their respective surface energies) is a minimum and the shape that satisfies this condition is given by the following, so-called as _________
a) Wulff construction
b) Marks construction
c) Henrys diagram
d) Twin diagram

Explanation: Wulff construction is a technique or a method to determine the equilibrium shape of a crystal or a droplet of a constant volume inside a separate phase usually its saturated solution or a vapor.

11. A convenient method for plotting the variation of γ with surface orientation in three dimensions is to construct a surface about an origin such that the free energy of any plane is equal to the distance between the surface and the origin when measured along the normal to the plane. This type of polar representation is known as_____
a) γ-plot
b) Surface plot
c) Energy plot
d) α-γ plot

Explanation: This type of polar representation of γ is known as a γ-plot and has the useful property of being able to predict the equilibrium shape of an isolated single crystal and free energy of any plane is equal to the distance between the surface and the origin when measured along the normal to the plane.

12. Equilibrium shapes can be determined experimentally by________
a) Annealing large void inside a crystals
b) Annealing large crystals at high temperature
c) Annealing small crystals at low temperature
d) Annealing small void inside a crystals

Explanation: Equilibrium shapes can be determined experimentally by annealing small single crystals at high temperatures in an inert atmosphere, or it can be done by annealing small voids present inside a crystal.

13. When the macroscopic surface plane has a high or irrational {h k l} index the surface will appear as a stepped layer structure where each layer is a close packed plane.
a) True
b) False

Explanation: When the macroscopic surface plane has a high or irrational {h k l} index the surface will appear as a stepped layer structure where each layer is a close packed plane. This can be illustrated for a simple cubic crystal. A crystal plane at an angle φ to the close-packed plane will contain broken bonds in excess of the close-packed plane due to the atoms at the steps.

14. For the given notation {222}, the spacing of equivalent atom plane is given by________ (Take the lattice parameter a as 2)
a) 1/√3
b) 1/√2
c) √3
d) √2

Explanation: For a given notation {h k l} the spacing of the equivalent atom plane is given by the formula a/√ (h2+k2+l2), where ‘a’ is the lattice parameter. Here substituting the value of h, k, l as 2 we get 1/√3 as the answer.

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