Can Compression Turn an Insulator into a Metal?

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SUMMARY

Compression can transform an insulator into a metal by reducing the energy gap between the conduction and valence bands, thereby increasing conductivity. This phenomenon occurs as the material's band structure changes under pressure, allowing electrons to move more freely. Additionally, processes such as exposure to electric fields or light can excite electrons from the valence band to the conduction band. In extrinsic semiconductors, the presence of donor or acceptor levels within the band gap facilitates this excitation with lower energy requirements.

PREREQUISITES
  • Understanding of band theory in solid-state physics
  • Familiarity with semiconductor physics, particularly extrinsic semiconductors
  • Knowledge of photon energy calculations using the equation E=hf
  • Basic principles of the Ideal Gas Law and its assumptions
NEXT STEPS
  • Research the effects of pressure on the band structure of materials
  • Study the role of donor and acceptor levels in extrinsic semiconductors
  • Learn about the conditions under which materials transition from insulating to metallic states
  • Explore the Ideal Gas Law and its applications in thermodynamics
USEFUL FOR

Materials scientists, solid-state physicists, and students preparing for exams in semiconductor physics and thermodynamics will benefit from this discussion.

darkcrystal
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Hello,

I am in the middle of revising topics for the materials section of myend of year exam. For some reason my department give out past papers but do not give out answers.
Any answers to any of the follwing questions would be very helpful. I don't think too much detail is required just a basic knowledge.

1. Explain why in principle sufficient compreesion can turn an insulator into a metal?

2. In addition to heating describe another process which can excite an electron from the valence band to the conduction band?

3. Calculate the minimum energy a phton must have to produce an electron hole pait in an extrinsic semiconductor of band gap 0.6eV.

4. State the assumptions associated with the model used to develop the Ideal Gas law.

5. State what physical conditions fulfil the assumptions described in q4.

6. A gas container has an initial volume of 22.4 litres, the pressure is one atm and the temp is 273 deg K. If 0.3 mole of gas is released and the volume remains inchaned calculate the new pressure?

Thanks in advance
Dan
 
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darkcrystal said:
For some reason my department give out past papers but do not give out answers.

I had a department like that. It can be very annoying, although on the other hand it does force you to think a lot more.

Perhaps it would be more useful if you tell us how you would answer these questions first before we give you the answers?

Matt
 
attemtped answers

ok i think i have some answers

1. This is talking about the band structure of the materials. An insulator will start with a large energy gap between its conduction and valence bands. This means that electrons need to be given very large amounts of energy to get across the gap. As the material is compressed this gap gets smaller, its conductivity increases and it acts like a metal.

2. can also be done with electric fields , light and bombardment with particles

the light falling on the semiconductor needs to be of high enough frequency to be absorbed by the semiconductor giving electrons enough energy to jump the gap, leaving free electron / hole pairings. Extrinsicly doped materials require less energy as the band gaps in these are reduced do to impurities.

3. Egap = hf h=6.6x10^-34/1.6x10^-19 = 4.125x10^-15eV

f= 0.6/4.125x10^-15 = 1.45x10^14Hz

c=f x wavelength

wavelength = 3x10^8/1.45x10^14 = 2.1x10^-6m

hmm then can't think of another formula to use instead of e=hf


4. considered to be a point mass
collisions between particles are elastic
no long reange interactions
Ke equivalent to observed quantity of energy


5. its mass is very nearly zero, thus its volume is considered to be zero
no attractive or repulsive forces involved during collions thus Ke of molecules remains constant

not sure of something to put for the long range interactions statement

6. pV=nRT

at start
v=22.4
T=273
p=1
R=8.31

at finish

v=22.4
t=273
p=?
r= 8.31

i assumed you could find initial number of moles by subbing in values of initial conditions
1x22.4 = n x 8.31 x 273

n = 22.4/ 8.31 x 273 = 9.87x10^-3 moles

so when the 0.3 moles it taken away it gives a negative number

any help appreciated
 
I've only commented on the ones which I think need commenting on - the rest of your answers look fine.

darkcrystal said:
1. This is talking about the band structure of the materials. An insulator will start with a large energy gap between its conduction and valence bands. This means that electrons need to be given very large amounts of energy to get across the gap. As the material is compressed this gap gets smaller, its conductivity increases and it acts like a metal.

I'm not sure about this question myself, but I think you need to explain why the conduction band gets smaller when you compress the material. I think it will have something to do with the fact that as you compress it the unit cell in the material gets smaller which leads to a smaller band gap.

3. Egap = hf h=6.6x10^-34/1.6x10^-19 = 4.125x10^-15eV

f= 0.6/4.125x10^-15 = 1.45x10^14Hz

c=f x wavelength

wavelength = 3x10^8/1.45x10^14 = 2.1x10^-6m

hmm then can't think of another formula to use instead of e=hf

I would have simply said the photon needs an energy of 0.6eV. Or am I missing something obvious here?

5. its mass is very nearly zero, thus its volume is considered to be zero
no attractive or repulsive forces involved during collions thus Ke of molecules remains constant

I would have said it must be dilute and no long range interactions. Of course, if there are long range interactions you can try to counter this by making your gas even more dilute!

6. pV=nRT

at start
v=22.4
T=273
p=1
R=8.31

at finish

v=22.4
t=273
p=?
r= 8.31

i assumed you could find initial number of moles by subbing in values of initial conditions
1x22.4 = n x 8.31 x 273

n = 22.4/ 8.31 x 273 = 9.87x10^-3 moles

so when the 0.3 moles it taken away it gives a negative number

SI Units! Remember that volume must be in m^3 (1 litre is 10^-3 m^3) and pressure must be in Pascals (1atm is 10^5 Pa). If you include this you should get a more sensible answer.

Matt
 
thanks for the quick reply there matt.

I definitely see your point with question 3 being 0.6eV but that question was worth 7 marks. Hmm i don't really understand that.

Thanks for the responses to the other questions, much appreciated.
 
In semiconductors, when stress, i.e. pressure is applied to the crystal, the valence bands separate. In multi-quantum well structures and in quantum dots, the added stress separates the valence band structure into 6 distinct energy bands. I suspect that adding pressure to an insulator will just make it a better insulator. I need to look at some of my reference books to make sure. As for the semi-comductors, what I said above is true.
 
darkcrystal said:
thanks for the quick reply there matt.

I definitely see your point with question 3 being 0.6eV but that question was worth 7 marks. Hmm i don't really understand that.

Thanks for the responses to the other questions, much appreciated.

Also,it says that you have an "extrinsic semiconductor". So you would have donor/acceptor levels in the band gap, typically within a tenth of an eV from the conduction/valence band edge. So you don't really need a 0.6 eV photon to create an electron-hole pair. But without further information, it is impossible to accurately determine the minimum energy of such a photon.


Quick note : At STP, one mole of any ideal gas occupies 22.4 litres of volume.
 
Dr Transport said:
...I suspect that adding pressure to an insulator will just make it a better insulator...

No. Not true. Applying pressure to an insulator causes the valence and conduction band to spread out (just like you explained) causing them to eventually overlap at some critical pressure. Nitrogen, Oxygen and even Hydrogen have been made metallic at high pressures.
 
Gokul43201 said:
Also,it says that you have an "extrinsic semiconductor". So you would have donor/acceptor levels in the band gap, typically within a tenth of an eV from the conduction/valence band edge. So you don't really need a 0.6 eV photon to create an electron-hole pair. But without further information, it is impossible to accurately determine the minimum energy of such a photon.

It's been a while since I studied any solid state but won't all the electrons/holes in the donor/acceptor states already have been thermally excited to create electron-hole pairs? So if you want to create any more you have to excite something from the valence band again?

:confused:

Hmmm, I guess I should just go find my copy of Ashcroft and Mermin...

Matt
 
  • #10
baffledMatt said:
It's been a while since I studied any solid state but won't "all" the electrons/holes in the donor/acceptor states already have been thermally excited to create electron-hole pairs? So if you want to create any more you have to excite something from the valence band again?

:confused:

Hmmm, I guess I should just go find my copy of Ashcroft and Mermin...

Matt

It's never "all" the electrons/holes. The probability of excitation goes like exp(-E/kT)...so even if E is very small compared to kT (size of thermal fluctuations) there will be some unexcited charge carriers in the dopant levels.
 
  • #11
Gokul43201 said:
It's never "all" the electrons/holes. The probability of excitation goes like exp(-E/kT)...so even if E is very small compared to kT (size of thermal fluctuations) there will be some unexcited charge carriers in the dopant levels.

doh, of course! In fact, nomatter how high the temperature gets we will always expect to have some electrons in the donor states (basic stat. mech. which I really should know! :blush: )

Matt
 
  • #12
I forgot about metallic hydrogen...been outta grad school in the real world too long
 

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