I have a idea but I am not too sure, halides here is 3 degree, if I remove cl ion an Carbocation will produce, which cannot become planner because of steric requirement. Is it right. @JohnRennie
Momentum is frame dependent and there is always a frame in which the momentum is zero.
The best definition is from special relativity, but that's a bit advanced. In SR the rest mass is the relativistic magnitude (called the norm) of a 4D vector called the four-momentum.
The norm of a four vector is not frame dependent, so the rest mass is the same in every frame.
This lead to me more confusion, if I take you above equation correct and assume mass to be frame independent which says momentum is directly proportional velocity
It mean If I have given momentum of light, by de broglie wavelength, and I now the speed of light I can get mass of a photon. @JohnRennie
The equation $m = p/v$ is non-relativistic so it can only be used at speeds much less than the speed of light. Photons always travel at the speed of light so any attempt to use $m = p/v$ is pointless. The relativistic equation is:
$$ E^2 = p^2c^2 + m^2c^4 $$
where $p$ is the relativistic momentum. For all particles $p = h/\lambda$ where $\lambda$ is the de Broglie wavelength.
For photons the de Broglie wavelength is just the wavelength.
For massive particles we use:
$$ p = \frac{mv}{\sqrt{1 - v^2/c^2}} $$
This is the relativistic equation for the momentum of a massive object.
I think all of this is a bit of a diversion and not something you need to worry about for JEE level questions. You won't study relativity until you start a physics degree.
Thanks, sir as curious teenager I can, t control, it is habit to learn something new, I hope you will understand, sir do you know what is bubble chamber data. @JohnRennie
@JohnRennie suddenly switch is opened...current stopped so there will be a lot of electrons in the p side...I mean the holes will be filled up right sir?
@user8718165 remember that for every electron that flows across the junction into the P type side an electron flows out of the P side into the wire connected to the diode. So the total number of conduction electrons in the P side doesn't change during the flow.
What does happen is that when current is flowing the depletion zone is removed. When the current stops the depletion zone is recreated by electrons diffusing across the junction.
In effect when an electron flows from the N side across the barrier into the P side all the electrons in the P side move along one atom and the electron nearest the wire moves from the P side into the wire.
It's like water flowing in a pipe. When water flows into one end of the pipe an equal amount of water must flow out of the other end of the pipe.
@JohnRennie sir first of all, the electrons should diffuse from the n type across the junction and come all the way upto the p-type/wire contact to go into the wire...is it sir? because there aren't any free electrons at the p-type/wire edge to go into the wire
@JohnRennie I'm saying that sir...please help me if that's incorrect sir :-(
You might be able to find someone to buy them in India because computer stuff is harder to get in India than in the UK. In the UK they would be worthless because no-one would want them.
@yuvrajsingh have you got a flatscreen monitor for your Windows 10 PC?
@JohnRennie sir, if a battery is connected to a simple p-type semiconductor, after a while all the holes will accumulate where the (-) lead of the battery is connected to the p-type...right?
Because when you separate charges it creates a potential difference.
As the holes start to move towards the -ve wire connected to the diode this creates a potential difference that pulls the holes back in the opposite direction.
The end result is that we get a depletion zone around the junction but in the rest of the diode the concentration of holes is at the normal level. And the depletion zone is very thin.
Holes form when an electron in the valence band jumps into a gap state, where it is fixed and can't flow. This creates a negative gap state and a positive hole in the valene band.
If no surfactant is present then there is nothing to stablise the oil water interface, so if two oil drops meet they will merge to form a single larger drop.
But without gravity the oil drops won't move around so they won't meet.
I guess there could be an Ostwald ripening process, but that will be very slow.
So suppose you take two drops of slightly different sizes and connect them with a tube so the oil can flow along them. Then the oil will flow from the high pressure small drop to the low pressure large drop i.e. the small drop will shrink to nothing and the large drop will grow.
But oil is very slightly soluble in water. So oil can dissolve out of the small drop into the water, diffuse to the large drop and come out of solution enter it.
The end result is that oil molecules can travel between the drops through the water, just as if there was a tube between but much, much slower.
It's slow because the solubility of oil in water is very small, and diffusion through water is very slow.
In P doped semiconductors the doping creates gaps states just above the top of the valence band. The energy difference between the gap states and top of the valence band is comparable to $kT$ so electrons get thermally excited into the gaps states leaving holes behind.
I suspect you'd get an avalanche effect. As you increased the voltage there would come a point where one electron moved, and it would collide with other electrons and excite them. Then they would in turn collide with other electrons and the number of mobile electrons would grow exponentially.
But the point is that the motion of the electrons would heat the semiconductor so it wouldn't be at 0K any more.
@JohnRennie yeah sir...my qn is based on this topic sir
@JohnRennie if there is no E-field in the wire...why does the current flow sir through the wire...is it just due to the e-field in across the resistor? I'm confused :-/
@yuvrajsingh there's nothing wrong with your answer but I suspect it doesn't address what the Aditya Siroutiya is actually asking. I suspect he's asking about the physical mechanism that causes resistance. Bob D's answer makes the analogy with friction, which is actually pretty good. In effect the electrons flowing through a resistor experience a force analogous to friction and it's the work done against this force that causes the potential drop across the resistor.
What causes a potential drop in a resistor or load? Why a wire having neglible resistance have same potential across it?
