Topic 6: Nano-scale non-carbon electronic materials

Question 1

• How does a gold nanoparticle differ to bulk gold?

Answer 1

• Lower MP; as higher proportion of atoms at surface, reducing cohesion energy
• Catalytically active, where it is inert in bulk
• Work function higher
• Colour different

Question 2

• What are the two types of size effects that effect metal-nano particles?

Answer 2

• Metal nanoparticles <50 nm is subject to SA/V effects
• Those that are < 2 nm in diameter are subject to quantum size effects

Question 3

• What nanoparticle properties are affected by SA/V size effects?

Answer 3

• WF/MP/reactivity are all likely to change as a result of a larger SA/V ratio

Question 4

• What are quantum size effects and what properties are they likely to affect in nanoparticles?

Answer 4

• Electrons are confined to small potential wells, causing splitting of energy levels and discrete states of molecules to form
• Will see stepwise changes in properties such as ionisation potential or magnetic properties

Question 5

(?) Explain why gold exhibits a colour change at small length scales despite having the same bonding arrangement as bulk gold. Use a sketch to aid your answer

Answer 5

• In a conductor, free electrons in conduction band move in response to an electric field
• They can also move in response to an oscillating electric field associated with electromagnetic radiation.
• Green and blue light are absorbed, however longer red wavelength light undergoes resonance conduction as diameter of nanoparticle << λ_red.

Question 6

(?) What is resonance conduction? Include a sketch to support your answer

Answer 6

• When frequency of incident light matches natural frequency of surface electron oscillating against restoring force of positive nuclei lattice.
• Oscillating field associated with light causing movement of CB electrons to oscillation of field

Question 7

• Describe plasmon-induced hot electron injection in single molecule devices

Answer 7

• Light shone on a nanoparticle may lead to surface plasmon resonances in the particle when certain frequencies match natural electron freuqnecy of surface electrons
• These have a strong adsorption and the energy of the excitation relaxes back and donates to an electron in CB forming a hot electron
• Decay/relaxation of surface plasmon can lead to creation of e^-h⁺ pair (h⁺ at electrode and e^- above fermi level)
• Hot electron may have enough energy to surmount barrier injection into LUMO, which then allows nanoparticle to be used as a means of increasing conductance in a system

Question 8

(?) Why is plasmon-induced hot electron injection a problem and how can it be avoided

Answer 8

• Injection in to LUMO may not be desired result from nanoparticle
• To avoid, must do experiments in the dark, which makes conducting them difficult
• Toss up between new functionality of nanoparticle and difficult conditions to use it under.

Question 9

• … … separating an electron from a positive charged sphere is
• ϕ(R) = ϕ_bulk + 5.4/…
• … - metal sphere radius (…)

Answer 9

• Work done separating an electron from a positive charged sphere is
• ϕ(R) = ϕ_bulk + 5.4/R
• R - metal sphere radius (angstroms)

Question 10

• Discuss the reasoning for the work function of a particle changing with its size

Answer 10

• SA/V ratio greater in nanoparticel
• Coulombs law says the coulombic force of attraction greatest when positive charge is highly localised (not spread out)
• Overall energy needed to remove electron from particle (with positive charge distributed over surface) is therefore smaller than a point charge (macrosurface)

Question 11

• Discuss the size dependence of a nanoparticle it’s the melting point.

Answer 11

• Increasing proportion of atoms at surface where coordination # is reduced compared to bulk atoms, as atoms bound to fewer neighbours
• Mean cohesive energy reduced
• Melting point goes down

Question 12

• Describe the surface plasmon resonance sensor application

Answer 12

• I = I₀e^-σ[n]L
• σ – abs cross section
• n – conc of NPs
• ε₁ = -2ε_m
• σ (ω) = 9ωVε_m^3/2/c * ε₂(ω)/[ε₁(ω) + 2ε_m]²[ε₂(ω)]²
• ε_m = dielectric constant of medium surrounding nanoparticle (NP)
• ε₁ = real part of NP dielectric constant
• ε₂ = imaginary part of NP dielectric constant
• measure adsorption ond look at resonance
• dependence of ε_m can be used as the basis of analytical technique to detect biological molecules adsorbed at NP surface

Question 13

• Outline the nanoparticle strain gauge application of NPs

Answer 13

• Flexible substrate with Au NPs can exhibit a change in conduction as a function of movement in the wrist
• Increase when strained and vice versa
• Flexibility of substrate means separation of ligands increases with bending, meaning electrons must move further to cross circuit.