1. What is Nanotechnology?
a) Technology related with design and fabrication of objects of 10 to 100 nm size
b) Technology related with design and fabrication of objects of 100 to 1000 nm size
c) Technology related with materials having novel physical or chemical properties
d) Technology related with materials having novel physical and chemical properties by the virtue of their size
View Answer
Explanation: Nanomaterials have some special properties apart from bulk materials due to their specific size, which is different for different materials. There are also some materials called meta materials which shows some different properties, but those properties are not due to their size.
2. What is the biggest problem with Top Down approach?
a) Increase surface area
b) Increased internal stress
c) Imperfection in surface structure
d) Reduced Gibbs energy
View Answer
Explanation: Top Down approach uses equipment which is not so small to make a smooth surface, hence imperfections in surface structures are observed. Increased internal stress and reduced Gibbs energy are the driving force behind nanomaterials synthesis and surface area increment is the result of synthesis.
3. What is not the consequence of imperfection in surface structure in Top Down approach?
a) It introduces inelastic surface scattering which result in reduced conductivity
b) It introduces Nanowires with rough surfaces which are not suitable as sensors
c) Excessive heat in nano devices due to increased collision
d) Agglomeration of nanoparticles
View Answer
Explanation: Imperfection in surface structure introduces impurities and roughness on surface which results in more collision, inelastic surface scattering and hence more heat generates and less conductivity of electrons in nano devices.
4. What is the driving force in Bottom Up approach?
a) Increment in internal stress of material
b) Reduction in Gibbs free energy
c) Decrease in surface energy
d) Increase in surface energy
View Answer
Explanation: Reduction in Gibbs free energy produces a product which is closer to thermodynamic equilibrium state. Increment in internal stress is driving force in Top Down approach. In synthesis of nanomaterials there is always an increase in surface area and thus increase in surface energy.
5. What is not the benefit of Bottom Up approach?
a) Less crystal defects
b) Homogeneous chemical composition
c) Better short and long-range ordering
d) Less surface energy
View Answer
Explanation: In Bottom Down approach; atoms or molecules are added one by one to one so in this way composition stays homogeneous, order is maintained and less defects are produced. Surface energy always increases in nanomaterial synthesis.
6. Why nanomaterials are thermodynamically unstable in initial state of synthesis?
a) Large surface energy
b) Reduced Gibbs energy
c) Increased internal stress
d) Less surface energy
View Answer
Explanation: Large surface area increases large surface energy which in turn increases Gibbs energy and the system becomes thermodynamically unstable but later system relax itself by various methods, reducing Gibbs energy and increasing internal stress which stabilizes the system.
7. Which is not a side effect of increase in surface energy?
a) Difficulty in fabrication of nanomaterials
b) Unwanted growth in size of nanomaterials
c) Agglomeration of nano clusters
d) More active sites for catalysis
View Answer
Explanation: Increase in surface area gives more broken bonds which serves as active sites for catalytic purposes, but it also forms new bonds to impurities hence producing unwanted growth or to its neighbor molecules to form agglomeration.
8. By which method surface energy of nanomaterials cannot be minimized?
a) Surface relaxation
b) Agglomeration
c) Absorption
d) Nucleation
View Answer
Explanation: Nucleation is the starting step of formation of nanoparticles which gives rise to surface energy. This surface energy can be minimized by shrinkage of surface which is relaxation and formation of new bonds via absorption and agglomeration.
9. Which crystal plane will have more surface energy?
a) FCC {110} plane
b) Simple cubic {110} plane
c) BCC {110} plane
d) Surface energy does not depend on crystal plane
View Answer
Explanation: FCC {110} plane will have more number of atoms on plane than same plane of other crystal system. FCC have extra atoms on each surface of cube which is absent in other crystal systems.
10. Which FCC plane will have higher surface energy?
a) {100}
b) {110}
c) {111}
d) {112}
View Answer
Explanation: Surface energy is proportional to planer density. Planer density is determined by number of atoms/planer area. {100} plane in FCC has highest no of atoms among them with least surface area, hence it has highest surface energy.
11. What is the difference Between Ostwald Ripening & Sintering?
a) New interfaces are formed between particles
b) Reduction of solid-gas surface does not occur in both
c) Reduction of surface energy does not occur in both.
d) Reduction of Gibbs free energy does not occur in both
View Answer
Explanation: In Ostwald ripening, small molecules merge into bigger molecules which gives us a big molecule with no newly formed interfaces while in Sintering, molecules combine into a bigger particle by make solid-solid interfaces.
12. If R1gt; R2, which surface will have more chemical potential energy?
a) Convex surface with radius R1
b) Concave surface with radius R1
c) Concave surface with radius R2
d) Convex surface with radius R2
View Answer
Explanation: Chemical potential is inversely proportional to the radius of curvature. Radius of curvature is negative for concave surfaces and positive for convex surfaces.
13. Which method is irreversible for reducing the surface energy of a nanomaterial?
a) Ostwald Ripening
b) Sintering
c) Structure relaxation
d) Agglomeration
View Answer
Explanation: In other 3 methods molecules either get dissolve into bigger particles or forms weaker bonds, but in agglomeration stronger chemical bonds are formed which makes agglomerates non-destructible.
14. If R1>R2, which surface will have more solubility & vapor pressure?
a) Convex surface with radius R1
b) Concave surface with radius R1
c) Concave surface with radius R2
d) Convex surface with radius R2
View Answer
Explanation: Solubility & vapor pressure dependence on chemical potential, more the chemical potential more will be the solubility and vapor pressure.
15. Which is not a side effect of Ostwald Ripening in sintering of polycrystalline materials?
a) Abnormal grain growth
b) Inhomogeneous mixture
c) Inferior mechanical property
d) Narrow size distribution
View Answer
Explanation: Oswald ripening gives dissolution of small nanoparticles in large nanoparticles. If case of poly crystalline materials small molecules dissolve into different crystals thus giving inhomogeneous mixture, abnormal grain growth & inefficient mechanical strength. By dissolution of small particles to bigger particles, it eliminates the smaller particles. Hence narrow size distribution.
16. When a solid emerges in a solution of charged species, it does not stabilize itself by which mechanism?
a) Absorption of charged species
b) Dissociation of surface charged species
c) Accumulation or depletion of electrons or protons
d) Substitution of surface species by same constituent species
View Answer
Explanation: In solution containing charged species, solid has ions on its surface due to dynamic equilibrium. These charges attract counter ions, electrons or dissociate its surface ions, thus stabilizes themselves but there is no change in surface charge in substitution with same species.
17. Concentration of charge determining ions for zero surface charge is called point of zero charge. If point of zero charge for an oxide nanomaterial in water is pH 6 & has negative surface charge, what is the pH of the solution?
a) 3.2
b) 6.0
c) 1.2
d) 7.9
View Answer
Explanation: For pH below 6 will have hydronium ions in equilibrium with solid surface so surface will be positive. For pH above 6 will have hydroxyl ions in equilibrium with surface hence negative surface charge is observed.
18. What is the surface potential of a surface for a point of zero charge PZC 9 and pH of the solution is 4.
a) 0.3 volt
b) 0.4 volt
c) 0.5 volt
d) 0.1 volt
View Answer
Explanation: Surface charge potential is given by the equation E = 2.303RT (PZC-pH)/F; where R is gas constant, T is absolute temperature, F is faraday constant. Calculating the value with proper values will give 0.3 volt.
19. Van Der wall forces and Brownian motion are responsible for agglomeration, if solution has charged species then what should be done to avoid agglomeration?
a) Increase the concentration of charged counter species in solution
b) Decrease the concentration of charged counter species in solution
c) Increase the pressure & volume
d) Decrease the pressure & volume
View Answer
Explanation: According to DLVO theory, there is a potential barrier between two nanoparticles in solution. This barrier is directly proportional to Debye Huckel screening strength around solid particles in solution which increases with increase in concentration and valency of the counter ions not by the pressure.
20. If in a solution of charged species, concentration of counter ions is constant, but now counter ions are in reduced state so what effect it will have on potential barrier between nanoparticles?
a) Van der wall forces will increase
b) Van der wall forces will decrease
c) Electric double layer potential will increase
d) Electric double layer potential will decrease
View Answer
Explanation: Electric double layer potential depends inversely on Debye Huckel screening strength. It is determined by valency of counter ions, as valency has decreased here so it will increase which in turn will increase electric double layer potential.
21. Among them which is not a difference between steric stabilization & electrostatic stabilization?
a) Electrostatic stabilization is kinetically stable while steric stabilization is thermodynamically stable
b) Electrostatic stabilization not suitable for higher concentration while steric stabilization is suitable
c) Electrostatic stabilization not suitable for multi-phase system while steric stabilization is suitable
d) Electrostatic stabilization is not electrolyte sensitive while steric is sensitive to electrolyte
View Answer
Explanation: Steric stabilization involves uncharged species, so it is not sensitive to electrolyte. Gibbs energy is the driving force behind steric stabilization hence it is thermodynamically stable. Multi-phase systems have more stability in steric method.
22. Which one of these does not contribute to steric stabilization?
a) Anchored polymers
b) Physically absorbed polymers
c) Chemically absorbed polymers
d) Non-absorbing polymers
View Answer
Explanation: Any polymer which is attached to the surface of nanoparticles, by any means, will end in repulsion of particles closing to one another due to increased Gibbs energy but non-absorbing polymers are in solution hence they have no effect on that.
23. Which is the best stabilization method for preventing agglomeration?
a) Steric stabilization
b) Electrostatic stabilization
c) Electrosteric stabilization
d) Ostwald Ripening
View Answer
Explanation: Electrosteric stabilization is the combination of electrostatic and steric stabilization hence it is more efficient. Ostwald ripening is a method to reduce surface energy by accumulation of nanoparticles, so it increases the size in place of stabilizing it.
24. What is the driving force behind steric stabilization?
a) Decrease in Gibbs energy of nanoparticles closing to one another
b) Increase in Gibbs energy of nanoparticles closing to one another
c) Increase in surface charge of nanoparticles closing to one another.
d) Decrease in surface charge of nanoparticles closing to one another.
View Answer
Explanation: In steric stabilization, polymers are absorbed or anchored on the surface of nanoparticles. When particles approach one another, they increases Gibbs energy of polymers due to coiling of them (increase in entropy). Surface charges are developed in electrostatic stabilization not in steric.
25. Which nanoparticles will experience higher van der wall force potential?
a) Particles with radius 5 nm and separation distance 10 nm
b) Particles with radius 4 nm and separation distance 12 nm
c) Particles with radius 3 nm and separation distance 15 nm
d) Particles with separation distance 10 nm and radius 6 nm
View Answer
Explanation: Van der wall potential in its simplest form is given by V = -Ar/12S; where r is the radius, A is constant, and S is the separation between particles. More the negative value, more will be the attraction potential.
26. The Gibbs free energy per unit volume for formation of a nucleus of a nanoparticle, given solute concentration is 5 M, equilibrium concentration is 3 M, temperature 300 k and atomic volume is N, will be___
a) (1.95×10-23)/N
b) (1.32×10-23)/N
c) (4.32×10-23)/N
d) (2.32×10-23)/N
View Answer
Explanation: Gibbs free energy change per unit volume is given by the equation ∆G= – {(KT) ln (1+C/Co)}/N, where K is Boltzmann constant, T is absolute temperature, C is concentration at any time and Co and is initial concentration.
27. If surface energy per unit area is 5×10-2 joule, Gibbs free energy per unit volume is 1×107 joule. What will be the critical radius of the nanoparticle?
a) 10 nm
b) 50 nm
c) 100 nm
d) 5 nm
View Answer
Explanation: Critical radius is given by the equation: r* = -2Y/∆G, where Y is surface energy per unit area and ∆G is Gibbs free energy per unit volume. Gibbs energy can be calculated by the equation, ∆G= – {(KT) ln (1+C/Co)}/N. As values of Gibbs free energy change is given we can calculate directly the critical radius of particle.
28. Which will have effect no effect on critical radius of nanoparticles?
a) Increase in temperature
b) Increase in pressure
c) Increase in volume
d) Increase in pressure and decrease in volume by same fold
View Answer
Explanation: By increasing pressure, temperature and volume separately; concentration will get effected. But increasing pressure by 2-fold and decreasing volume by 2-fold, will have no effect on concentration according to ideal gas law PV=NRT. Gibbs energy change depends on concentration.
29. Nucleation in nanoparticle synthesis gives_______
a) Kinetically stable product
b) Thermodynamically stable product
c) Electrically stable product
d) Electro-sterically stable product
View Answer
Explanation: Kinetic approach for synthesis includes controlled use of precursors. While nucleation occurs in supersaturated solution, it is driven by Gibbs free energy reduction, so it is thermodynamic approach. Nucleation involving charges does not give electrically or sterically product.
30. Above which concentration nucleation does not start but subsequent growth of nucleus starts?
a) Concentration just above equilibrium solubility
b) Concentration at critical radius formation
c) Concentration at full grown nucleus
d) Concentration just below equilibrium solubility
View Answer
Explanation: Super saturation is not reached when concentration is below equilibrium solubility, so as a result neither nucleation nor growth takes place. At concentration between critical radius formation and full grown nucleus both nucleation and growth takes place. At concentration just above equilibrium solubility growth can take place but nucleation cannot as nucleation starts only at critical radius.
31. What steps are followed in subsequent growth of nucleus?
a) Generation-diffusion-absorption-growth
b) Diffusion-generation-absorption-growth
c) Generation-absorption-diffusion-growth
d) Diffusion-absorption-generation-growth
View Answer
Explanation: In subsequent growth of nucleus, first of all, generation of the growth species starts, these growth species reaches surface of nucleus by diffusion. Now absorption of these species starts and as a final step growth at surface of nucleus occurs.
32. In which mechanism particles grow linearly with time?
a) Growth controlled by generation of species
b) Growth controlled by diffusion
c) Growth controlled by surface process: monolayer growth
d) Growth controlled by surface process: poly-nuclear growth
View Answer
Explanation: In growth controlled by diffusion and monolayer growth, growth rate depends upon radius of the particle but for linear growth it should not depend on radius of particle which is observed in poly-nuclear growth. For poly-nuclear growth dR/dT = KT; where t is time, R is radius & K is a constant.
33. Which growth controlling mechanism is suitable for mono-sized nanoparticles?
a) Diffusion control
b) Surface process control: monolayer growth
c) Surface process control: poly-nuclear growth
d) Generation control
View Answer
Explanation: Generation is a necessary condition for growth of nucleus, not a control process. In surface process control, there are abundant growth species at surfaces which result in irregular sizes (kinetic approach) while in diffusion control, there are shortage of species so size is regulated.
34. If k = 1×10-7 and Initial radius is 3 nm, what will be the radius of the particle after 100 seconds in poly-nuclear growth process?
a) 4 nm
b) 3 nm
c) 5 nm
d) 6 nm
View Answer
Explanation: For poly-nuclear growth, radius of the particle at any instant during growth is given by the equation: R = (KT) + Ro; where Ro is the initial radius, K is the constant & T is the time.
35. Which method is suitable for synthesis of mono-sized nanoparticles by diffusion control process?
a) Keeping the concentration of diffusion species extremely low
b) Decreasing the viscosity of solution
c) Removal of the diffusion barrier
d) Increasing the supply of growth species
View Answer
Explanation: Keeping the concentration of diffusion species very low, we increases the distance for diffusion which results in less supply of species at surface of particle hence size is regulated. In other methods we are increasing the concentration of species at particle’s surface hence size is not regulated here.
36. Which one tells correctly how heterogeneous nucleation is different from homogeneous nucleation?
a) Different phases are produced from the precursor solution on the material
b) Same phase is produced but on a different material
c) Different phase is produced on a different material
d) Same phase is produced on same material
View Answer
Explanation: In homogeneous nucleation a new phase develops from precursor solution on the material from maternal solution but in heterogeneous nucleation a new phase develops on a new material substrate introduced in solution.
37. If precursor concentration and temperature is same, which will be same for heterogeneous nucleation and homogeneous nucleation?
a) Total surface energy
b) Gibbs free energy change ∆G
c) Gibbs potential barrier
d) Critical radius for a fix ∆G
View Answer
Explanation: As in heterogeneous nucleation a new surface is introduced hence surface energy is changed but ∆G only depends on concentration of precursors and temperature, if they are fixed, homogeneous and heterogeneous both will have same ∆G.
38. What should be the contact angle between substrate and growth species for a zero critical energy barrier heterogeneous nucleation?
a) 0 degree
b) 180 degree
c) 90 degree
d) 45 degree
View Answer
Explanation: In heterogeneous nucleation, 0 degree contact angle will give zero value for angle dependent term in the equation: ∆G heterogeneous = ∆G homogeneous × (0.5 – (0.75×Cosq) + 0.25 Cos3q), where q is the contact angle. For zero q we will have zero ∆G.
39. If ∆G of homogeneous nucleation is 10 kJ and contact angle is 30 degree, what will be the ∆G of heterogeneous nucleation?
a) 0.1 kJ
b) 0.2 kJ
c) 0.01 kJ
d) 1 kJ
View Answer
Explanation: According to the equation for ∆G of heterogeneous nucleation, ∆G heterogeneous = ∆G homogeneous × (0.5 – (0.75×Cosq) + 0.25 Cos3q) value will be 0.1 kJ.
40. To obtain monodisperse nanoparticles, what should be the sequence of nucleation and growth of the precursor?
a) First nucleation then growth and nucleation discontinue when growth starts
b) First growth then nucleation
c) Both nucleation and growth at same time
d) First nucleation then growth and nucleation continues when growth starts
View Answer
Explanation: For monodisperse particles, nucleation should be done first with a fast rate so that mono-sized nucleus is formed, then diffusion control growth should be done to maintain the size distribution of all particles and during this no new nucleus should be formed else new size particles will form.
More MCQs on Nanomaterials:
MCQs on Nanowires and Nanorods:
MCQs on Two Dimensional Structures: Thin Films:
MCQs on Special Nanomaterials:
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