Phase Transformation Questions and Answers – Solids Diffusional Transformations – Heterogeneous Nucleation

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This set of Phase Transformation Question Bank focuses on “Solids Diffusional Transformations – Heterogeneous Nucleation”.

1. Nucleation in solids, as in liquids, in most cases is always __________
a) Homogeneous
b) Heterogeneous
c) Mixed
d) Semi heterogeneous
View Answer

Answer: b
Explanation: This is the scenario found in most cases because they prefer the non-equilibrium defects as there nucleation sites, in many cases it occurs in the grain boundaries, excess vacancies and in the stacking faults.
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2. If the creation of a nucleus results in the destruction of a defect this will result in the_______
a) Destruction of new nuclei
b) Mass production of new nuclei
c) Change of phase
d) Reduction of activation energy barrier
View Answer

Answer: d
Explanation: The creation of a nucleus results in the destruction of a defect this will result in the production of some energy, let it be ΔGd and this will be released hence making an impact in the activation energy barrier and that too in the reduction of activation energy barrier.

3. Which among the following equation satisfies heterogeneous nucleation?
a) ΔGhet = – V*(ΔGv – ΔGs) + Aγ – ΔGd
b) ΔGhet = – V*(ΔGv – ΔGs) – ΔGd
c) ΔGhet = – V*(ΔGv – ΔGs) + Aγ + ΔGd
d) ΔGhet = – V*(ΔGv – ΔGs) + Aγ
View Answer

Answer: a
Explanation: Equation, ΔGhet = – V*(ΔGv – ΔGs) + Aγ – ΔGd corresponds to heterogeneous nucleation. Here the only difference between the homogeneous and the heterogeneous nucleation comes in the last term (-) ΔGd and this the energy released during the destruction of defects.
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4. If we ignore the effect of misfit strain energy, which among the following could minimize the interfacial free energy?
a) Optimum embryo texture
b) Optimum embryo shape
c) Optimum embryo volume
d) Optimum embryo length
View Answer

Answer: b
Explanation: Ignoring any misfit strain energy, the optimum embryo shape should be that which minimizes the total interfacial free energy. The optimum shape for an incoherent grain-boundary nucleus will consequently be two abutted spherical caps.

5. The ability of the grain boundary to reduce the value of critical free energy ΔG* depends on
a) CosΦ
b) SinΦ
c) TanΦ
d) CotΦ
View Answer

Answer: a
Explanation: The ability of a grain boundary to reduce ΔG (het), i.e. its potency as a nucleation site, depends on cosΦ. The cosΦ is actually the ratio of γαα/γαβ (assuming γαβ is isotropic and equal for both grains).
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6. Calculate the shape factor if the angle tends to 90 degree?
a) 0
b) 1
c) 2
d) 4
View Answer

Answer: b
Explanation: The shape factor will tend to unity when the angle tends to 90 degree, this can be proved easily using the following equation 0.5(2+cosΦ)*(1-cosΦ)2. Substitute the value 90 in the place of Φ then we get the required answer as 1.

7. If the ΔG* heterogeneous is given as 5kJ/mol and ΔG* homogeneous is given as 10kJ/mol. Calculate the shape factor?
a) 0.5
b) 1
c) 2
d) 2.5
View Answer

Answer: a
Explanation: Here in this case the shape factor can be determined by the ratio of ΔG*heterogeneous/ΔG*homogeneous. So here for this particular case the shape factor value is 5/10 =0.5. Actually the grain boundary has the ability to reduce ΔG* heterogeneous, it’s the power it has as a nucleation site.
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8. Excess vacancies are retained during the quench if the age hardening alloys are retained from______
a) Dark texture
b) Room temperature
c) Low temperature
d) High temperature
View Answer

Answer: d
Explanation: Excess vacancies are retained during the quench if the age hardening alloys are retained from high temperature. This can actually increase the diffusion rate or it can relieve the misfit strain energies hence it assist the nucleation.

9. What is special about the dark-field electron microscope micrograph?
a) Precipitates can be imaged bright and matrix dark
b) 360 degree rotation is possible
c) High speed imaging is possible
d) Anti roller scheme available
View Answer

Answer: a
Explanation: The so-called dark-field electron microscope micrograph in which the precipitates are imaged bright and the matrix dark. The precipitates lie in rows along dislocations. And this can be used in case of niobium carbonitride precipitates on dislocations in a ferritic iron matrix
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10. The relative magnitudes of the heterogeneous and homogeneous volume nucleation rate is given as 0.5. Calculate the factor C/C1 (the number of atoms on heterogeneous sites relative to the number within the matrix)? (Consider exp (ΔG*(hom) – ΔG*(het)/kT) to be 1)
a) 0.5
b) 1.2
c) 0.25
d) 1.25
View Answer

Answer: a
Explanation: The relative magnitudes of the heterogeneous and homogeneous volume nucleation rate is given by the equation (C/C1) *(exp (ΔG*(hom) – ΔG*(het)/kT). So in this case the nucleation rate is 0.5 which itself is the factor C/C1.

11. At very small driving forces, when activation energy barriers for nucleation are high, the highest nucleation rates will be produced by grain corner nucleation.
a) False
b) True
View Answer

Answer: b
Explanation: The driving force is the factor which determines whether the type of site gives the highest volume nucleation rate or not. At very small driving forces, when activation energy barriers for nucleation are high, the highest nucleation rates will be produced by grain corner nucleation.

12. A coherent nucleus with a negative misfit can reduce the critical volume free energy by forming a region of _________
a) Compressive strain
b) Compressive stress
c) Tensile strain
d) Tensile stress
View Answer

Answer: a
Explanation: This directly implies that the smaller volume than the matrix can reduce its ΔG* by forming a region of compressive strain above an edge dislocation, whereas if the misfit is positive it is energetically favourable for it to form below the dislocation.

13. In FCC crystals the a/2*[110] unit dislocations can dissociate to produce a ribbon of stacking fault. Which among the following is as an example of the same?
a) a/2*[110]->a/6*[121] + a/6[211]
b) a/2*[110]->a/6*[101] + a/6[201]
c) a/2*[110]->a/6*[821] + a/6[911]
d) a/2*[110]->a/6*[161] + a/6[221]
View Answer

Answer: a
Explanation: a/2*[110]->a/6*[121] + a/6*[211] is an example and this gives a stacking fault on (111) separated by two Shockley partials. Since the stacking fault is in effect four close-packed layers of hcp crystal, it can act as a very potent nucleation site for an hcp precipitate.

14. Which among the following curve represent the grain boundary?
phase-transformation-questions-answers-heterogeneous-nucleation-q14
a) 1
b) 2
c) 3
d) Cannot be predicted
View Answer

Answer: a
Explanation: Curve one represents the grain boundary and this can be determined from the effect of angle on the activation energy for grain boundary nucleation relative to homogeneous nucleation. Curve 2 and 3 represents the grain edges and grain corners respectively.

15. Nucleation on dislocations may also be assisted by solute segregation which can raise the composition of the matrix to nearer that of the precipitate.
a) True
b) False
View Answer

Answer: a
Explanation: Nucleation on dislocations may also be assisted by solute segregation which can raise the composition of the matrix to nearer that of the precipitate. The dislocation can also assist in growth of an embryo beyond the critical size by providing a diffusion pipe with a lower ΔG(migration).

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter