HSC Physics Question Thread

[RuiAce]

Hey guys!

I was wondering how the equation w=qv was derived for work done in an electric field? Thanks

Actually, whilst HPL makes a valid point...

...the whole definition of voltage is that V=W/q, i.e. the potential difference across a circuit is the work required to move each unit charge across it.

[Neutron]

Hey guys! On the topic of p-type semiconductors, why is the acceptor level above the valence band if electrons are moving DOWN from the outer valence electron shells to fill up the holes of the covalent bonds within the doped silicon lattice?

[jamonwindeyer]

Hey guys! On the topic of p-type semiconductors, why is the acceptor level above the valence band if electrons are moving DOWN from the outer valence electron shells to fill up the holes of the covalent bonds within the doped silicon lattice?

Hey Neutron! Answering this in detail is tough, and is covered properly at the tertiary level (I did Solid State Physics at the back end of my first year)    Let's have a look!

To answer, first consider an N-Type semiconductor. N-Type semiconductors are formed by doping with a Group 5 Impurity atom, which introduces additional electrons into the lattice. The energy level of these donor electrons is higher than the energy level of the others, and they are easily excited into the conduction band. Thus, electrons act as majority charge carriers.

Right, now lets consider p-type semiconductors. These are formed by doping with a Group 3 impurity atom, which introduces additional electron vacancies or holes into the lattice. It is very easy for electrons to be excited from their place in the valence band to fill this vacancy, leaving holes in the valence band. These holes are then the majority charge carriers.

We note that the energy levels involved here can be referred to as activation energies. For N-Type, this is the energy required to excite an electron donation from the impurity. For P-Type, this is the energy required to excite an electron acceptance from the impurity.

So, my understanding would be that electrons do not jump DOWN from the P-type donor. The holes do. The electrons are actually excited UP to that energy level you are referring to 

[Happy Physics Land]

Hey guys! On the topic of p-type semiconductors, why is the acceptor level above the valence band if electrons are moving DOWN from the outer valence electron shells to fill up the holes of the covalent bonds within the doped silicon lattice?

Yeah l think jamon's explanation was quite sufficient. I will just drop in my 2 cents here.

The acceptor level is really determined by the band structure of the acceptor atom. So for example, in the case of gallium, the acceptor level would appear above the valence band because the acceptor level would be the same as the highest level in the valence band of gallium. The reason why this acceptor level is helpful is that it significantly reduces the forbidden energy gap for electrons from valence band to become mobile (this would only take about 0.05eV). This in turn leaves excess holes in the valence band, allowing for current to be conducted through the principal charge carriers of positively charged holes.

Valence band is essentially the lowest band in the valence shell, and if the acceptor level is below the valence band, that kinda means that the electrons have now moved into a lower energy shell. But we know that only the valence electrons in semiconductors can act as charge carriers. So if we think about it logically, whats the point of doping if all we are doing is moving an electron into a lower energy shell where it cant act as a charge carrier?

Another way to think about it: in an intrinsic semiconductor, the electron move up into conduction band in order to become mobile charge carriers and leave holes behind in the valence band. So if this is the case, then shouldn't the same be applied to extrinsic semiconductors too? Where the electrons would also move up in band to become mobile charge carriers?

Basically my first point about the acceptor level being the valence band of the acceptor atom is the more scientific explanation of why acceptor level is above the valence band for host semiconductor. The next two points I think are good ways to logically think about the existence of acceptor level in such manner.

[FallonXay]

Heyy, I'm having trouble understanding the relation between the blackbody radiation curve and Planck's formula E=hf (as well as light being quantized). How does this concept work? Thanks in advance! 

[Happy Physics Land]

Heyy, I'm having trouble understanding the relation between the blackbody radiation curve and Planck's formula E=hf (as well as light being quantized). How does this concept work? Thanks in advance! 

Hey FallonXay!

E=hf is Planck's second postulate, which means that quantitatively, energy possessed by each quantum is directly proportional to the frequency of incident electromagnetic radiation. A quantum is essentially a packet of energy, and this energy is quantised, meaning that they only exist in discrete levels of well-defined energy. So for example, if say I need 10 J of energy to jump onto a stair, I cant gain 9J otherwise I would fall when I almost reach the stairs, nor can I gain 11 J because I would also fall because I have taken a step higher than the staircase.

One aspect of blackbody radiation curve is the ultraviolet catastrophe, which means that intensity at short wavelength, high frequency UV, gamma and X-rays are displayed to be 0. This issue cannot be understood by classic theory of physics, because according to classical theory, higher frequency would cause atoms to oscillate with greater energy and hence according to classical theory the intensity of UV radiation should reach infinity because wavelength is approaching 0, meaning that frequency is very high. Experimental data however oppose this theory. In fact, if infinity amount of UV radiations are emitted by blackbodies, then humans cant really exist because we all know that UV rays are harmful to human bodies.

What Planck is suggesting through E=hf is that, for high frequency UV, X-ray and gamma rays, there will be high amounts of energy emitted in discrete packets called quanta (plural for quantum). These quanta are only released when the atom undergoes a change in discrete energy level (imagine yourself stepping down from the second stair on the staircase to the first stair). However, because UV has extremely high frequency, using E=hf, we know that an extremely high amount of energy are emitted. This means, that there must be a significant change in energy level in an atom of the blackbody in order to emit quanta that contain such high energy values. But because we dont have such a wide range of energy levels in an atom, this would be impossible to achieve. Therefore there are no UV, X-ray and gamma rays being emitted.

Another feature you would have noticed in a blackbody radiation curve is the peak wavelength radiation. So according to Planck's postulate, the intensity of radiation emitted (or you can understand it as the number of quanta emitted) is proportional to the number of atoms undergoing a change in energy level. That means, for certain wavelength/frequency radiations, there is a higher probability for certain changes in energy levels to take place. Therefore these more probable changes of energy would result in more intense emission of the corresponding wavelength of radiation.

These are the main features you would need to know. The peak wavelength radiation part seems irrelevant to E=hf, but if you recall what I said before, E=hf implies that quanta are emitted when there is a change in energy level. So thats how E=hf relates to peak wavelength radiation. Hope it made sense!

Best Regards
Happy Physics Land

[jamonwindeyer]

Heyy, I'm having trouble understanding the relation between the blackbody radiation curve and Planck's formula E=hf (as well as light being quantized). How does this concept work? Thanks in advance! 

Love your work HPL!  To add just a couple of points to what is a fantastic explanation already:

- The characteristic frequency is what we call the peak frequency (and wavelength), and it is dependent on the temperature of the object, which relates indirectly to Planck's formula E=hf, but not in any way that you need to know for this course.
- Since this frequency depends on temperature, black body curves vary drastically between objects. The rule that governs their shape is called Planck's Law, which you don't need to know specifically. The only important thing is that you recognise that the curve shape will change depending on the object.
- As HPL stated, it is found that the intensity of high frequency radiation is quite low for black bodies, but this is only true for objects like our sun or similar. There do exist black body radiation curves that have a characteristic frequency in the gamma range, perhaps from a gamma burst from a supernova. Again, not totally important, but again the focus is that the curves will change shape drastically depending on the object!

Pretty much the theme of these additions is this: The shape of the curve has to do with the object itself, and thus changes rapidly. E=hf is just our mechanism for deriving why that curve looks the way it does, instead of the Ultraviolet Catastrophe Curve (which instead stems from Rayleigh-Jean's Law).

Between this post and HPL's, you have way more detail than you need for this part of the course, hope it helps

[FallonXay]

Ahh ok.

 

But because we dont have such a wide range of energy levels in an atom, this would be impossible to achieve.

However, I don't really understand this sentence. What do you mean by not having enough energy levels in the atom? How would you determine how many energy levels an atom has?

(Also, thanks Jamonwindeyer and Happy Physics Land for your detailed answers!  )

[jamonwindeyer]

However, I don't really understand this sentence. What do you mean by not having enough energy levels in the atom? How would you determine how many energy levels an atom has?

(Also, thanks Jamonwindeyer and Happy Physics Land for your detailed answers!  )

You don't need to for HSC Physics! All good, electron energy levels in atoms are not overly important in this course.

All you need to understand is that the energy in an emitted quanta is determined based on some change in the atomic structure of the black body. An electron is demoted, a reaction takes place, something like that. This releases a photon with an energy exactly the same as the energy change in the atom, whatever that may be.

Gamma photons (for example) have a massive amount of energy. Massive energy changes are less common, and thus, we usually don't get many gamma photons emitted from a black body. Sometimes we do! But it is rare  Does that help at all? Anything still a little confusing?

(HPL or another chemist may wish to answer your questions specifically, but again, not necessary to understand in this course!) 

[Happy Physics Land]

Ahh ok.

 However, I don't really understand this sentence. What do you mean by not having enough energy levels in the atom? How would you determine how many energy levels an atom has?

(Also, thanks Jamonwindeyer and Happy Physics Land for your detailed answers!  )

Oh yeah hehehe no worries! What Im really saying here is that, for example you have sodium and sodium has the electron configuration of 1s²2s²2p⁶3s¹. Now for UV radiation to be emitted, according to E=hf, there would need to be an infinitely high amount of energy being released in the form of quanta. Each quanta is released by a change in energy level. This means that we need a very significant change in energy level in the atom to emit that much energy, and this change in energy level might be, say, from 2s shell into 6f shell. But since 6f shell isnt anywhere to be found in sodium, this change in energy level cannot exist! So what Im implying here is, if energy emitted becomes infinity, then there will be infinite change in energy shell levels, and hence this is not gonna happen with any atoms simply because they cant supply this many shells.

[Happy Physics Land]

I will just make an extra post here to hit 200 comments, YAY IM A RESPECTED MEMBER NOW !!!

[FallonXay]

oh ok, that makes sense

I will just make an extra post here to hit 200 comments, YAY IM A RESPECTED MEMBER NOW !!!

Also Congratulations!!! and thanks for your contribution

[Happy Physics Land]

oh ok, that makes sense
Also Congratulations!!! and thanks for your contribution

Ahahahaha how kind of you! Thanks mate!

[RuiAce]

I will just make an extra post here to hit 200 comments, YAY IM A RESPECTED MEMBER NOW !!!

M8.

[jamonwindeyer]

M8.

M(9-1)