L4 Synthesis & Characterisation of NP

Question 1

1. Name ONE physical or chemical property that can potentially change as you render a material from its bulk form to its nanoscale form.

Answer 1

• Melting Point, Optical, Surface Reactivity, Surface Area (or surface area to volume ratio)
  
  • related to the interactions between the atoms - both with each other any any atoms that ‘approach’ the suface; and also interactions with surface electrons (i.e. for optical)
• Not ‘shape’
• Materials will try to minimise surface energy, usually by aggregating

Question 2

2. Why do gold nanoparticles (NP) have considerably low melting temperatures compared to their bulk counterparts? (interaction between the atoms)

(A) Less energy overall is needed to break atomic bonds at the surface atoms, and there are considerably more surface atoms in nanoparticles

(B) Electrons of atoms travel faster in nanoparticle form, hence less energy is required
to melt the nanoparticles

(C) The effect of photothermal therapy is stronger in smaller nanoparticles ← this is actually
true, but not related to melting temperature

(D) The surface plasmon resonance effect allows for additional energy during heating.

Answer 2

(A) Less energy overall is needed to break atomic bonds at the surface atoms, and there are considerably more surface atoms in nanoparticles

Question 3

3. Which of the following NP synthesis techniques is NOT a ‘top-down’ approach?

(A) Ball milling
(B) Photolithography
(C) Laser ablation
(D) Liquid phase synthesis

Answer 3

• Top-down: breaking down a bulk material to its nanoparticles (either by mechanical or temperature + force)
• Bottom-up: using atoms to build NP

(A) Ball milling - i.e. crushing the bulk material
(B) Photolithography - using UV or electron beams to shape NP (through a mask)
(C) Laser ablation - using laser to blast material off
(D) Liquid phase synthesis - dissolving precursors → atoms → atoms build to form NP

Question 4

4. During ball milling, the longer the milling time employed, the resulting average particle size:

(A) Decreases
(B) Increases
(C) Does not change
(D) Increases during a certain time range, and then decreases

Answer 4

(A) Decreases

(NB - Surfactants: agents used to minimise surface energy (stabilise the surface) → thus preventing the likelihood of particles aggregating with each other)

Question 5

5. Photolithography uses UV light to crosslink specified areas of a photosensitive material, after which the non-crosslinked photosensitive material is washed away, leaving behind solid particles where the crosslink has occurred. The shape and size of these particles are controlled by:

(A) altering the power of the UV source

(B) altering the shape of the UV lamp

(C) using a mask that allows UV light to pass through specified patterns (i.e. control
where the UV is going)

(D) pre-shaping the underlying substrate material to match the desired final shape and
size of NP

Answer 5

(C) using a mask that allows UV light to pass through specified patterns (i.e. control
where the UV is going)

• Photolithography - ~printing/carving
• crosslink specified areas of a photosensitive materia - photoresist
• Expensive $$$$$$ to get high resolution and high amounts
• UV/e-beam crosslinks certain areas of a crosslinkable polymer (these areas ‘stay’); uncrosslinked areas are washed away
• Sacrificial layer - allows for easier removal

Question 6

6. Which of the following factors can affect the final NP product during laser ablation?

(A) Laser characteristics (wavelength, duration, energy, etc.)
(B) Type of solvent used to collect NP
(C) Temperature and pressure
(D) All of the above

Answer 6

(D) All of the above

i.e. Laser characteristics (wavelength, duration, energy, etc.),
Type of solvent used to collect NP, and
Temperature and pressure
(Why does pressure affect it? Pressure for what??? ← for energy)

Laser Ablation:

• Fire laser which then releases very small (NP) particles from the surface into a solution
• Initial release – a lot of very small particles – depending on the system setup parameters (i.e. options below) – the final size will differ

Question 7

7. Name TWO generic disadvantages of top-down approaches during NP synthesis

(2 marks)

Answer 7

• Not cheap or quick manufacturing
  
  • Slow and complex (time-consuming)
  • Scalability
• Large size distribution compared to bottom up
• Aggregation
• Hard to control shape (there’s an element of randomness in most of these top down approaches) (except photolithography)
• More energy required
• Large-scale synthesis is not feasible (or even possible) and difficult to achieve

Generic advantage of top-down = ‘simpler’ approaches (conceptually and experimentally), less need for potentially toxic solvents, less complex machinery/experimental setup

Question 8

8. Name ONE way to control the average size of Au NP during liquid phase synthesis process.

Answer 8

(again, controlling the energy in the system)
● Reaction temperatures
● Reductant concentration/strength
● Surfactant concentration
● Solvent
● Agitation (speed etc.)

Question 9

9. Which of the following statements regarding Au NP formation is FALSE?

(A) Au³⁺ is converted into Au⁰ monomers in the presence of a reducing agent in solution
(B) Nucleation of Au NP occurs once the concentration of the Au⁰ monomers is supersaturated
(C) The Au NP stop growing in size once all the monomers are consumed during
Nucleation
(D) The size of Au NP continue to grow by diffusion until all the smaller particles
aggregate and reach equilibrium and a narrow distribution of NP size

Answer 9

FALSE: (C) The Au NP stop growing in size once all the monomers are consumed during
Nucleation
(once the monomers are all consumed → there is a ‘population’ of different sized Au NP → leads onto (D))

● Monomers = not the polymer monomer but the ‘base’ material/atom

True:

(A) Au³⁺ is converted into Au⁰ monomers (that build up to make NP) in the presence of a
reducing agent in solution

(B) Nucleation (try not to use the term ‘precipitation’ over ‘nucleation’ - but if it helps you
understand what’s going on, then you can think of this as ‘precipitation’) of Au NP
occurs once the concentration of the Au 0 monomers is supersaturated

(D) The size of Au NP continue to grow by diffusion until all the smaller particles
aggregate and reach equilibrium and a narrow distribution of NP size (Ostwald ripening -
a thermodynamic process where atoms of a smaller particle diffuse onto a larger particle)
(- the ‘steady-state size’ depends on the parameters brought up in Q8)

Question 10

10. Thermal decomposition involves the use of temperature to ‘decompose’ or ‘break down’ a starting material to create a new material. Which of the following regarding thermal decomposition is FALSE, in the context of nanoparticle production?

(A) The temperatures employed for thermal decomposition is generally higher than
temperatures used for liquid phase synthesis

(B) Thermal decomposition needs to be done in an inert environment (i.e. chamber with inert gases such as N 2 or argon) to prevent any unwanted reactions

(C) Size & shape of the final NP product cannot be controlled using thermal
decomposition

(D) The method of thermal decomposition can be used to fabricate nanoparticles from a range of inorganic materials

Answer 10

• *(C) Size & shape of the final NP product cannot be controlled using thermal
  decomposition. **
• False - there are parameters that you can employ:*
• Reaction temperature
• Precursor to surfactant/reducing-agent ratio
• Reaction time

Thermal Decomposition: extension of wet liquid phase synthesis, except with the addition of heat. Most suitable for a wide variety of NPs made from metal, semiconductor and magnetic materials (including composite NPs)

True:

(A) The temperatures employed for thermal decomposition is generally higher than
temperatures used for liquid phase synthesis

(B) Thermal decomposition needs to be done in an inert environment (i.e. chamber with inert gases such as N 2 or argon) to prevent any unwanted reactions (metals will form
unwanted oxides in the presence of oxygen - often uncontrolled as well)

(D) The method of thermal decomposition can be used to fabricate nanoparticles from a range of inorganic materials (a lot of the other inorganic NP cannot be synthesized in room temperature)

Key Notes:
● Gold NP synthesis using liquid phase synthesis can be done at room temperature
● A lot of the other inorganic NP cannot be synthesized in room temperature
● Certain material phases can only be formed in certain temperatures

Question 11

11. What is the primary factor during the synthesis of hybrid nanoparticles that determines
    whether the hybrid nanoparticle is heteroepitaxial (Upper Figure) or homoepitaxial (Lower Figure)?

(A) Polarity of the medium used to synthesize the hybrid NP
(B) The elements used in the hybrid NP
(C) Temperature during synthesis of NP
(D) All of the above

Answer 11

(A) Polarity of the medium used to synthesize the hybrid NP

Polar - homoepitaxial (deposition of second phase surrounds the initial nanoparticle)
Non-polar - heteroepitaxial (deposition of second phase at a local point on the initial NP)

Question 12

12. Which of the following statements is FALSE regarding scanning electron microscopy
    (SEM)?

(A) SEM has a higher image resolution than light microscopy (true - beats the diffraction
limit)

(B) The electron beam in the SEM is fired towards the sample; subsequent interactions and electrons/photons emitted from the sample are detected which is then processed to generate an image

(C) Non-conductive materials can be imaged without any pre-treatment

(D) SEM is primarily used to image the surface of objects

Answer 12

False: (C) Non-conductive materials can be imaged without any pre-treatment

True:

(A) SEM has a higher image resolution than light microscopy (beats the diffraction limit)

(B) The electron beam in the SEM is fired towards the sample; subsequent interactions and electrons/photons emitted from the sample are detected which is then processed to generate an image

(D) SEM is primarily used to image the surface of objects

Question 13

13. Name ONE difference between SEM and transmission electron microscopy (TEM).

Answer 13

TEM - samples need to be very thin for the electrons to pass through
SEM - surface of samples need to be conductive in order for the electrons to ‘interact’

• TEM magnification is a lot more precise = 0.05 nm
• 3D analysis can be done using TEM and not SEM
  
  • Cross-section views of targets
• TEM can examine shape, size of material (nanosized), while SEM only surface
• TEM can show distribution of atoms in NP
• TEM requires sample to be transparent for electron beams to pass through

Question 14

14. Dynamic light scattering (DLS) relies on which of the following phenomena when
    determining the particle size distribution?

(A) Angles at which the lights reflect back from stationary particles suspended in solution, whereby the recorded angles are dependent on the particle size

(B) Intensity of reflected incident light as a function of time that fluctuates with Brownian
Motion, whereby such fluctuation profiles are dependent on the particle size

(C) The shadow effect, where particles block the transmission of light through a channel. Patterns detected on the opposite side can be converted to particle size data

(D) The light causes the particles to heat up – amount of heat generated corresponds to the particle size; subsequently generated heat map can be converted into particle size distribution profiles

Answer 14

(B) Intensity of reflected incident light as a function of time that fluctuates with Brownian
Motion(motion of particles in a medium), whereby such fluctuation profiles are dependent
on the particle size

Question 15

15. Which of the following reasons best describes why sizes of NP sometimes appear to be
    larger when measured with DLS compared to images obtained from SEM/TEM?

(A) DLS measurements are affected by diffraction limits

(B) Moving NP in solution during DLS analysis makes the particle seem bigger than it
actually is, compared to the stationary particles in SEM/TEM

(C) Agglomeration does not occur in SEM/TEM samples, and has the potential to occur in solution samples for DLS, hence the larger sizes detected in DLS

(D) DLS measures the hydrodynamic diameter (or hydrodynamic radius) which includes any organic coating applied on the surface of core nanoparticles, that cannot be detected using SEM/TEM

Answer 15

(D) DLS measures the hydrodynamic diameter (or hydrodynamic radius) which includes any organic coating applied on the surface of core nanoparticles, that cannot be detected using SEM/TEM

• ​SEM/TEM can’t see surfactant layer
• The body sees the hydrodynamic diameter of the NP, as NP is in blood/body fluids that is aqueous

Question 16

16. You have a solution of gold nanoparticles that you have recently synthesized and functionalized, and you wish to measure the size distribution of your particles. Your DLS reading of your particles shows a nice peak at size of ~5 nm. Six hours later before you head home, you decide to measure the particle size of that same sample again. You notice now that you have two peaks – one at ~5 nm, but this peak size is shorter than before, and another one at ~20 nm.

Which of the following statements best describes what has most likely happened to your nanoparticles?

(A) The initial reading was wrong, and you actually had a sample with bimodal distribution from the very beginning

(B) The Au NP had degraded in solution over the six hour period

(C) Some of the Au NP had aggregated over the six hour period

(D) There were some leftover reducing agents and unreacted gold ions in solution, which meant that nucleation and growth continued during that six hour period.

Answer 16

(C) Some of the Au NP had aggregated over the six hour period

Note for (D) “There were some leftover reducing agents and unreacted gold ions in solution, which meant that nucleation and growth continued during that six hour period” - for this you would see more of a peak shift towards higher sizes, rather than two distinct peaks

[Regarding functionalization preventing aggregation to some extent - Yes it does, depends on the functionalization, but if not properly done (as in if not all NP are functionalized adequately), aggregation can still occur]

Question 17

How does Melting Point change from bulk material to nanoscale?

Answer 17

decreases with decreased size because you need less energy to break the atomic bonds (atoms on the surface have less atomic bonds i.e. ‘coordination number’ than non-surface atoms; there are a lot more surface atoms in smaller nanoparticles, hence) less energy required to break the particle and melt.

Question 18

Describe briefly the fluorescence mechanism associated with nanoparticles.

Answer 18

Fluorescence‘colour’ corresponds to the wavelength of light emitted due to the (to simply put) drop of the electron from a higher energy state (often called ‘excited state’) to a lower energy state (often this is the ‘ground state’)

Question 19

Explain the disadvantages of Ball Milling.

Answer 19

Time-consuming to make particles nanosized.

Lack of control on size and shape destribution; relies mostly on random grinding.

Particles tend to agglomerate and are difficult to dispeerse (high surface energy without any added surfactants).

Question 20

Describe the process of Wet Chemical Synthesis

Answer 20

metal ion + reductant –> NP

• “Precursor” = “ingredient” that is the source of the element(s) needed for the production of NP
• Reductants provide electrons to metal ions to form metal monomers
• Surfactants used to control the surface energy (specifically - to lower the surface energy) - this means that more surfactants mean smaller NP, less surfactants mean larger NP
• Can also be used to control shape of NP
• Reaction speed highly dependent on reductant concentration
• Lots of reductant = high reaction speed, smaller NP (provided that these smaller NP are in a stabilised condition i.e. through surfactants - otherwise aggregation)
• “Monomers” here does not mean the polymer monomer sense. Here, the “monomers” refer to the individual atoms in solution generated by the reaction of the precursors (i.e. Au⁰)

Question 21

Describe the Sol-Gel synthesis process of NP

Answer 21

In simple terms, all the precursors are dissolve din typically an organic solvent. These dissolved precursors thenf orm colloid particles evenly disperesed in a solvent (which is now called a ‘sol’)

• This sol (i.e. liquid with all those tiny colloids) is then either used to coat surfaces, or just left as is
• This sol then undergoes gelation (i.e. turns into ‘gel’ i.e. the ‘sol-gel’ process)
• Heat treating this gel leaves behind very small particles, whose composition depends on the precursor and the type of heat treatment received

Question 22

Describe the process of Flame Spray Pyrolysis and its material-limitation

Answer 22

• Dissolve precursor in solvent –> pressure into forming aerosol –> fire the aerosol –> decomposition/nucleation/aggregation –> form NP
• Can’t do for metals because it will form oxides (need an oxygenated environment)

Question 23

Describe (simplified) the process of Gas phase NP production

Answer 23

• Heat the ‘target material’ to its melting temperature –> inert gas environment and pressure gradient –> forces target material atoms to ‘fly off’ the surface –> collide to form NP
• Size/mass control can be done through a ‘filter’