Inorganic Materials

Question 1

Synthesis Methods: Outline the high temperature solid state methods - ‘shake ‘n’ bake’ method.

Answer 1

> Reactants = solids which are ground into a powder to reduce particle size (1 μm) then pressed into pellets (to increase size of interface).

> Pellets are then heated in an inert container (500-2000°C). This method is slow because although the reactants may be well mixed at the μm level, on an atomic level they are inhomogeneous ( therefore, particles need to be mixed well in order to bring together atoms of the different elements, in the correct ratios to form the desired products).

> Reaction can only take place at the interface between reactant particles. Diffusion in solids = slow (atoms need to diffuse from one reactant to the interface) - transport of ions may be aided by liquid phase or gas phase media. GENERALLY, oxygen ions will not diffuse.

> Initial step: Nucleation of small crystals with correct product stoichiometry and crystal structure. For nuclei to be stable they must usually be several tens of Å across (i.e. considerable larger than a single unit cell).

> The critical size of nuclei represents the balance between the negative free energy of formation of the spinel product and the positive surface energy of the nuclei.

NOTE: Spinel = mineral name for MgAl2O4 (AB2O4).

Question 2

What is an inverse spinel structure and when is it favoured?

Answer 2

B3+ → 1/8 tetrahedral holes
A2+ → 1/4 octahedral holes
B3+ → 1/4 octahedral holes

> Normal spinel structure is thermodynamically favoured but formation of the inverse structure can be rationalised using a crystal field theory argument when TM are involved.

Question 3

Describe the nucleation process that occurs during spinel formation.

Answer 3

> At the MgO/MgAl2O4 interface, the oxide ion arrangement can continue unchanged since it is CCP in both structures with ABC stacking sequences.

> Formation of the spinel nucleus on top of MgO = easy due to structural similarity. This is TOPOTAXIAL NUCLEATION (TOPOTAXY- where structural similarity extends to 3D).

> The oxygen stacking sequence changes from HCP (AB) in Al2O3 to CCP in spinel. Structural similarity between substrate and nucleus = limited to 2D interface- EPITAXIAL NUCLEATION (epitaxy).

Question 4

Spinel Formation: The first few layers of product nuclei form easily but subsequent growth of the product becomes difficult as the 2 reactants are pushed apart by the newly formed impenetrable spinel layer. How is this problem overcome?

Answer 4

> Complex counter diffusion process:
→Mg 2+ ions diffuse away from MgO
→Al3+ ions diffuse towards MgO/MgAl2O4

> As reaction proceeds, the spinel layer thickens and the diffusion pathway increases and hence the reaction slows down.

> In order to preserve local electroneutrality (in reactant and product), for every 3Mg2+ ions that diffuse one way, 2Al3+ ions must diffuse the other.

> High reaction temperature is required for ion mobility- ions move via vacancies or interstitial sites often provided by defects.

> Reaction can be accelerated by frequent grinding of reactant/product mixture. For some reactions, precursors such as carbonates or nitrates can be used (decompose to give oxides with smaller particle size).

Question 5

Synthesis Methods: What is the biggest drawback of solid state reactions?

Answer 5

> Reactants are not mixed on an atomic timescale.

Question 6

Synthesis Methods: Outline chimie douce (soft chemistry) methods. What are the disadvantages?

Answer 6

> Low temperature methods which can lead to improved product purity (especially if prolonged heating at high temperatures can be avoided).

> Disadvantages of these methods:
→expensive reagents that may be difficult to handle on large scales.
→ considerable time/effort often needed to optimise reaction conditions.
(solid state reactions which are relatively quick, easy and versatile are usually tried first.)

Question 7

Synthesis Methods: Outline Sol-gel methods.

Answer 7

> Use metal-organic precursor compounds e.g. alkoxides used to generate oxides (NOT organometallics). - usually liquids and are dissolved in ater using alcohols to promote miscibility.

> Method:
(i) Prepare a homogeneous solution containing cationic components with appropriate stoichiometry.

(ii) Slowly dry out solution to give a viscous sol (contains particles of colloidal dimensions) and finally a homogeneous amorphous solid (transparent) known as a gel. (precipitation of any crystalline phase does not occur).
(ii) Heat the gel (calcinate) to burn off any organic, volatile byproducts that may be trapped in the pores of the gel/chemically bonded organic side groups leaving the final oxide product (least volatile component is very ionic).

> Water = key reagent, it hydrolyses the alkoxides, usually in the presence of an acid or base catalyst. Conversion of the alkoxide into oxide (hydrolysis) occurs via 2 steps; Hydrolysis and Condensation.

Question 8

Synthesis Methods: Outline Hydrothermal (solvothermal) methods.

Answer 8

> Reactants are heated in water/steam at modest temperatures and elevated pressures in a sealed reaction vessel.

> The water has 2 functions i) as a pressure-transmitting medium and ii) as a solvent (in which the solubility of the reactants is P,T dependent).

> Product can be crystallised due to temperature gradient in vessel and seeding off the product crystals or by controlled cooling of the reaction solution.

Question 9

Synthesis Methods: Outline 2 applications of hydrothermal methods.

Answer 9

(i) Quartz, SiO2
→NaOH is used as a mineralizer to increase solubility of SiO2 in water (ESSENTIAL).
→ SiO2 dissolves most readily at the hot end and is transported by convection to the cooler end and crystallizes on the suspended seed.

(ii) Zeolites (microporous aluminosilicates)
→ Constructed from sources of tetrahedral building units NaAlO2 and SiO2.
→ A templating agent is required to control synthesis of a particular zeolite- typically an alkyl ammonium cation.

Question 10

What is intercalation/deintercalation?

Answer 10

> Synthesising new materials from an existing crystalline solid by either selective introduction or removal of atoms/ions. The overall structure remains intact. There is a strong 3D similarity between starting phase and product. (Topotactic reaction).

Question 11

Describe the Lithium ion battery.

Answer 11

> Electrolyte needs to be polar and involatile (usually LiPF6 in ethylene carbonate).

> Li+ ions move from graphite cathode (deintercalation) and intercalate into LixMn2O4 anode.

> Anode: xLi+ + Mn2O4 + xe- → LixMn2O4
Cathode: LixC6 → xLi+ + 6C + xe-
Overall: LixMn2O4 + 6C → LixC6 + Mn2O4

Question 12

What are silicates?

Answer 12

> Silicates are natural minerals (including layered clays, micas and talcs, and fibrous materials such as asbestos). Synthetic silicates = zeolites

Question 13

What are the building blocks of SiO4 4- (silicates/orthosilicates)?

Answer 13

(i) almost all silicate structures are built of SiO4 4- tetrahedral.
(ii) The tetrahedral link by sharing corners to form larger polymeric units.
(iii) No more than 2 SiO4 tetrahedral may share a common corner (oxygen)
(iv) SiO4 tetrahedral never share edges or faces.

> Si:O ratio 1:4

> no. of O per Si - Terminal: 4 Bridging:0

> examples: Mg2SiO4 (olivine), Be2SiO4 (phenacite), ZrSiO4 (zircon).

Question 14

Describe the Si2O7 6- building blocks (sorosilicates/pyrosilicates).

Answer 14

> Si:O ratio 1:3.5

> no. of O per Si- Terminal: 3 Bridging:1

> examples: (Sc,Y)2Si2O7 (thortveitite), CaAl2Si2O7(OH)2.H2O (lawsonite).

Question 15

Describe the [SiO3]n 2n- building blocks?

Answer 15

> form a single infinite chain

> Si:O ratio 1:3

> no. of O per Si - Terminal: 2 Bridging:2

> Minerals containing isolated [SiO3]n 2n- chains = single chain inosilicates, most common type are pyroxenes e.g. MgSiO3 (enstatite), CaMg (SiO3)2 (diopside).

Question 16

What is piezoelectricity?

Answer 16

> Under the action of an applied mechanical stress, piezoelectric crystals polarise and develop electrical charges on opposite crystal face i.e. application of pressure causes a change in the electric field.

> P= dσ
where P=polarisation
d=piezoelectric coefficient
σ= stress

Question 17

What does piezoelectricity depend on?

Answer 17

> The occurrence of piezoelectricity depends on the crystal structure of the material and the direction of the applied stress e.g. quartz develops a polarisation when subjected to a compressive stress along [100] but not along [001].

Question 18

What occurs when a centrosymmetric crystal is placed under pressure?

Answer 18

> The piezoelectric effects cancel out and there is no net change in electric field.

Question 19

Describe the structure of zeolites.

Answer 19

> formed from linked tetrahedra in which MAlO2 or M’0.5AlO2 replaces some SiO2 units in a silicate structure (aluminosilicate).

> Zeolites have channel structures and are typically hydrated within the channels. Pore size ranges from 210 pm → 410 pm → 740 pm. Pore size allows for the van der Waals radius of the oxygen atoms around the ring.

> Rings join to form 𝛃 cage structures (truncated tetrahedron).

> Sodalite: Na8Cl2[Al6Si6O24] = each 𝛃 cage is surrounded by 6 𝛃 cages.

> Na-Seolite A: Na12[Al12Si12O48].27H2O

> Falijasite: Na2Ca[Al4Si10O28].20H2O

Question 20

How are zeolites prepared?

Answer 20

> Preparation: dehydration through heating under vacuum.

> Templated Synthesis: leads to control of porosity in structures (hydrothermal methods).

Question 21

Describe the 3 key features of Zeolites that make them applicable to industry.

Answer 21

1) ABSORPTION → depends upon: channel/pore size, cation content in channels (e.g. Na-Zeolite-A as a molecular sieve for water removal).

ZSM-5 has large channels (540 pm) -determined by van der Waals radii of atoms. Silicate derivatives readily absorb non-polar molecules (held in pore by vdW interactions).

2) ION EXCHANGE → Na-Zeolite-A readily exchanges Na+ for Ca2+ in aq. solution → used in water softeners.

Regeneration conducted by using a concentrated NaCl solution.

3) CATALYSIS→ makes use of the acidic environment in zeolite channels (due to highly polarised H2O molecules or due to use of acid form of zeolite i.e. H+ replaces M+).

Important in the oil industry: used in dehydration rearrangements and for shape selective analysis.

Question 22

How is conductivity calculated?

Answer 22

σ= R/L
where σ = conductivity
R=resistance
L= Length

Units = S= 1/Ω

Question 23

What values of conductivity correspond to conductors/semi conductors and insulators?

Answer 23

Metal (conductors) → ~10^2-10^6 S cm-1
Semiconductors → ~10^-5-10^2 S cm-1 (Bandgap < 3.5 eV)
Insulators → ~ 10^-21 - 10^-5 S cm-1

Question 24

Describe the properties of conductors.

Answer 24

> Conductors are typically metallic (metals, metal oxides and metal sulphides).

> Valence electrons are delocalised and therefore free to move throughout the structure. The valence band is half-filled.

> Resistance occurs and increases with temperature due to phonons (collisions between electrons and lattice vibrations).

Note: 0 resistance = superconductivity.

Question 25

Describe the conductivity of semiconductors.

Answer 25

> Conductivity increases with increasing temperature and can be modified by doping.

Question 26

Describe the conductivity of insulators.

Answer 26

> Conductivity increases with increasing temperature.

> Valence electrons are tightly bound to atoms or in bonds between them. The valence band is full and the conduction band is empty with a large band gap.

> examples: many covalent network solids and ionic solids.

Question 27

Describe band theory.

Answer 27

> The difference between conductors/semiconductors/insulators depends on:

(i) band structure of each
(ii) whether the valence bands are full or only partly full
(iii) the magnitude of any energy gap between the full and empty bands.

> The ‘chemical approach’ to band theory is to take molecular orbital theory and extend it to infinite, 3D structures. → As the number os MO’s increases =, the average band gap between adjacent molecular orbitals decreases. (MOs converge into a continuous band with increasing no. of atoms). If the atoms involved have p orbitals available aswell, then multiple bands are available.

> In each band the lowest level = fully bonding while the highest level = fully antibonding.

> Band gaps= areas between band where no orbitals are present.

> Bands have a finite band width, which depend upon the strength of the interaction. (strong interaction → wide bands, weak interaction → narrow bands).

Question 28

What is the Fermi level?

Answer 28

> Energy levels are filled from the lowest to highest. The highest occupied energy level at 0 K = the Fermi level (Ef).

Question 29

What is an intrinsic semiconductor?

Answer 29

> An intrinsic semiconductor is an undoped semiconductor. It contains an equal number of holes and electrons.

> At T > 0K, the population, P of the energy levels follows a Fermi-Dirac distribution.

Question 30

What is the effect of temperature on intrinsic semiconduction?

Answer 30

> Conductivity increases with temperature due to increased population of the conduction band and therefore the number of charge carriers.

Note: even at r.t. there is some population of the conduction band due to the ‘Boltzmann tail’ of the population distribution.

Question 31

What is an extrinsic semiconductor?

Answer 31

> An extrinsic semiconductor = modified due to the addition of dopant atoms.

Question 32

How do the conductivities of intrinsic and extrinsic semiconductors compare?

Answer 32

> Conductivity is usually greater in extrinsic semiconductors due to a decrease in the size of the band gap.

Note: extrinsic semiconductors still exhibit intrinsic semiconductor behaviour as well (minority charge carriers) - intrinsic dominates at higher temperatures.

Question 33

What are n-type semiconductors?

Answer 33

> dopant atoms have ONE MORE VALENCE ELECTRON than the group 14 semiconductors (i.e. from group 15 e.g As).

> The extra electron provided by each dopant atom lies in a discrete level below the conduction band called the DONOR BAND.

> Electrons cannot move within these discrete levels but are readily promoted to the conduction band and allow conduction with NEGATIVE CHARGE CARRIERS (ELECTRONS).

Question 34

What are p-type semiconductors?

Answer 34

> dopant atoms have ONE FEWER VALENCE ELECTRONS than the semiconductor (i.e. from group 13 e.g. Ga or B).

> Energy levels associated with the electron deficient Ga-Si bond are not part of the valence band they form discrete ACCEPTOR LEVELS.

> Electrons are readily promoted to the acceptor levels allowing positive ‘holes’ to serve as charge carriers in the valence band.

Question 35

Why are semiconductors so expensive when Si = 2nd most abundant element?

Answer 35

> Semi-conductor devices ar emade from single crystals of > 99.99999999% pure silicon.

Question 36

How is semiconductor-grade silicon synthesised?

Answer 36

1) Refining Process: Silica (silicon dioxide) is reduced into silicon and carbon monoxide by heating in a furnace containing carbon:

SiO2 + 2C → Si + 2CO 98%

2) Chemical Vapour Deposition (CVD): (Sieman’s Process)

Si + 2HCl → (300°C) SiHCl3 + H2 + Impurities

SiHCl3 + H2 → (1000°C) Si + 3HCl

Question 37

Describe the process of Czochralski crystal growth.

Answer 37

> A crucible is filled with silicon and housed in an air tight chamber. Under argon, heat is applied to the crucible.

> During this initial melting phase, a dopant impurity may be added to the crucible so that the final “pulled” crystal will have the necessary electrical characteristics.

> A seed crystal is attached to a removable shaft and inserted into the top of the molten EGS.

> By slowly turning the seed crystal (~60 rpm) and raising it out of the molten silicon (~1 inch per hour), a large single crystal can be grown that exhibits the same crystalline structure as the original seed crystal.

Question 38

What is the pn junction?

Answer 38

> Many semiconductor devices are based on the pn junction (solid state equivalent of a diode allowing current to flow in only one direction).

> Fermi level for n-type region is higher than for p-type region ∴ electrons flow spontaneously from n-type to p-type.

Migration of:
Electrons: n → p
Holes: p → n

> Migration occurs until equilibrium is established when the junction potential V, builds up. The junction potential prevents further net transfer of charges.

> At equilibrium, a very small current of electrons flows in both directions due to the additional effect of intrinsic semiconductor behaviour.

NOTE: chemical potential is constant across the junction.

Question 39

What is the depletion region?

Answer 39

> The depletion region (aka the space charge region) lies at the junction of the n-type and p-type semiconductors and is deficient in majority charge carriers.

Question 40

Describe the current flow at equilibrium at the p-n junction.

Answer 40

> The majority charge carries diffuse from regions of high concentration to regions of low concentration generating the depletion region. This is the diffusion current (Idiff).

> Minority charge carriers (electrons on the p-type and holes on the n-type) drift due to the junction potential.

Electrons drift p →n
Holes drift n →p

> This is the drift current (Idrift)

i.e. Diffusion current and drift current flow in opposite direction.

Question 41

What is meant by “biasing the pn junction”?

Answer 41

> Applying an electrical potential across the junction.

Question 42

What happens when the pn junction is under reverse bias?

Answer 42

> p → -ve, n →+ve

> Junction potential is increased and reduces Idiff.

> I drift is unchanged.

> A very small current flow results.

Question 43

What happens when the pn junction is under forward bias?

Answer 43

> p → +ve, n → -ve

> Junction potential is decreased and Idiff increases.

> I drift is unchanged.

> A large current flow results.

Question 44

What are photovoltaics and why are semiconductors useful in solar cells?

Answer 44

> Photovoltaics/ Solar cells: Photovoltaic energy conversion relies on the solar flux of photons (which have distinctive energy distributions).

> Bandgaps for semiconductors Si, GaAs, InP make them suitable for use in solar cells. (e.g. Si → 1.1 eV → 1127 nm).

> Absorption in the near IR (700-1200) excites electrons without too much loss of efficiency to heat.

Question 45

What is the mechanism of action for photovoltaics?

Answer 45

1) Absorption of photons generating electron-hole pairs on both sides of the pn junction.
2) Electrons drift across the pn junction to a lower energy level. Holes drift in the opposite direction. An increase in the drift current results. New electron hole pairs continue to be formed while light falls on the solar cell.
3) As electrons continue to drift, a negative charge builds up in the emitter, a corresponding positive charge builds up in the base. The pn junction has separated the electrons from the holes and transformed the generation current between the bands into an electric current (drift current) across the pn junction.
4) If an electrical circuit is made between the emitter and base, current will flow. The current continues to flow while the solar cell is illuminated.

NOTE: The conventional current direction is defined as the opposite direction to electron flow.

Question 46

Outline the construction of a photovoltaic solar cell.

Answer 46

> Made in the form of widely-spaced thin metal strips (usually called fingers) that supply current to a larger bus bar.

> Thin layers minimise recombination (good).

> Top contact MUST allow light to pass through. The cell is covered with a thin layer of dielectric (insulator) material. This anti-reflection coating (ARC) minimises light reflection from the top surface.

Question 47

Why is amorphous silicon needed for solar panels and how is it made?

Answer 47

> Solar panels require large area thin films → practical limit to the size of a single crystal of Si.

> Solution is to use a form of amorphous silicon (𝛂-Si). It is made by chemical vapour deposition (CVD) which involves decomposition of gas phase reactants to produce a solid that is deposited on a substrate as a thin layer (600-800°C).

> Precursor for 𝛂-Si) is SiH4 which has been decomposed using plasma - CVD. (~300°C). Accidental contamination from H2 produced 𝛂-Si:H (hydrogen doped silicon).

Question 48

Describe the structure and band structure of 𝛂-Si.

Answer 48

> Amorphous silicon is made up of random networks of tetrahedral Si atoms i.e. not crystalline therefore no long range order.

> In normal 𝛂-Si, some Si atoms are only 3-coordinate ∴contain an unpaired, non-bonded electron known as a “dangling bond” which means there are localised states within the bandgap.

Question 49

Why is 𝛂-Si a poor intrinsic semiconductor?

Answer 49

> n doping: Electrons in conduction band are trapped by discrete levels in the band gap.

> p-doping: Holes in valence band are trapped by discrete levels in band gap.

i.e. doping does not have the desired effect and 𝛂-Si is a poor intrinsic semiconductor.

Question 50

Describe the structure and band structure of 𝛂-Si:H.

Answer 50

> There are NO dangling bonds in 𝛂-Si:H - all unpaired electrons participate in covalent bonding to H atoms.
∴𝛂-Si:H has a “clean” bandgap and can be doped by adding small amounts of B2H6 (p-type doping) or PH3 (n-type doping)during plasma -CVD).

> 𝛂-Si:H can form thin, flexible, non-crystalline films suitable for photovoltaics but degrades in sunlight and has a lower conversion factor (number of electrons harvested per photon).

Question 51

How do single crystal photovoltaic cells compare to amorphous silicon photovolataics?

Answer 51

> Single Crystal:
→Bandgap = 1.1.ev (1127nm)
→Poorly absorbing 
→ Thick film (need more material)
→high efficiency (up to 25%)
→Reliable
→Expensive

>Amorphous Silicon:
→Band gap = 1.65 eV (752 nm)
→strongly absorbing
→thin film
→low efficiency (10%)
→slowly degrades upon exposure to light
→cheap

Question 52

What is a photodetector?

Answer 52

> Photodectector = pn-junction run at reverse bias, in this mode current response (output) is proportional to intensity of light (on pn junction).

Question 53

What does photodetector performance depend on?

Answer 53

(i) Quantum efficiency (number of electron hole pairs generated per photon).
(ii) Current carrier transit time (absorbing photon → electron moves across junction).

Question 54

What are light emitting diodes (LEDS)?

Answer 54

> LEDS are in appliances such as digital clocks, remote controls, traffic lights and in tv/comp displays.

> They are based on simple pn junctions but the semiconductors used are not silicon, they are made from gallium arsenide.

> Unlike in photovoltaics and photodetectors, electron-hole recombination is encouraged. The process of light emission is known as electroluminescence.

Question 55

What advantages are there to LEDS?

Answer 55

> Mobility of conduction band electrons in GaAs is 2.5 times greater than in Si → possibility of faster integrated circuits (can use higher frequency).

Question 56

Describe the structure of GaAs.

Answer 56

> GaAs is a diamond-like binary compound with the same structure as zinc blende (ZnS).

> ABC layering of Ga3+ with 3/4 tetrahedral holes occupied.

> GaAs= direct band gap semiconductor (high quantum efficiency).

Question 57

What are direct and indirect and gaps?

Answer 57

> Direct band gap: promotion of an electron from the valence band to the conduction band or its reverse (electron-hole recombination) occurs only with absorption (or emission) of a photon.

> Indirect band gap: promotion of an electron from the valence band to the conduction band through a change in energy provided by a photon and a change in momentum provided also by a phonon. (lattice vibration).

Question 58

What are the selection rules for direct band gaps?

Answer 58

> Selection rule for electronic transition producing a net transition dipole moment.

Δl = ±1 (orbital angular momentum)
Δk= ±0 (electron momentum)

> NO change in momentum is required for the electron to be transferred between bands.

Question 59

Why do indirect band gap materials such as Si have a lower quantum efficiency than direct band gap materials?

Answer 59

> The lower probability of promotion of an electron from the valence band to the conduction band means that indirect band gap materials have a lower quantum efficiency.

> Electron-hole recombination that occurs through emission of a photon and a phonon is also a low-probability process; often a non-radiative recombination occurs via defects in the material .

Question 60

How is GaAs synthesised?

Answer 60

> The problem with making GaAs is that it doesnt occur in nature.

> Ga m.p. = 30°C, b.p=2204°C
As m.p=817°C, Tsub= 615°C

TEMPERATURES ARE NOT IDEAL

> An adaption of the Czochralski method is used.

Question 61

How is the wavelength of emitted photons by GaAs tuned?

Answer 61

> Can change the wavelength of emitted photons by tuning the band gap which is achieved by adding P (GaPxAs(1-x) - ratios may not be integer values).

> GaP0.4As0.6 is an efficient red LED.

> For x>0.45 an indirect band gap material results, hence low quantum efficiency for electroluminescence. Doping P with N can overcome this problem.

Question 62

In LEDs, recombination must take place in which 2 ways?

Answer 62

1) increasing the doping concentration of the substrate, so that additional free minority charge carriers (electrons) move to the top, recombine and emit light at the surface.

2) By increasing the diffusion length L=(Dτ_0.5, D=diffusion coefficient , τ= carrier life time.
However, beyond a critical distance between the zone of radiative recombination and the diode surface, the emitted photons will be reabsorbed primarily.

Question 63

How does LED emission depend on temperature?

Answer 63

> with increasing temperature:
→emission maximum shifts to lower energy (bathochromic shift)
→emission broadens
→quantum efficiency decreases

Question 64

Outline the advantages and disadvantages of LEDs.

Answer 64

Advantages:
→ more light per watt (1W= 1 Js-1)
→light at desired wavelength, no filter needed
→light up quickly
→very small
→Hg free, no vacuum
→dimmable without colour change
→on/off cycle resistant
→shock resistant
→long lifetime
→no sudden failure
→spectral purity used in diode LASERs

Disadvantages:
→still very high price per lumen
→temperature sensitive light yield and lifetime
→relies on correct current
→not usable when highly collimated beam is required (no “point” source)
→blue LEDs may produce harmful UV radiation
→needs transformer and AC/DC converter
→light not coherent (as opposed to LASERs).