TrueTears question thread

[TrueTears]

Also for stopping voltage () against frequency graph, and we are required to work out planck's constant. The gradient gives

But do we assume the graph we are given is the characteristics of ONE electron, ie the stopping voltage to stop ONE electron. Hence right? [because if it's not for 1 electron then would be unknown since the question doesn't state it]

[mark_alec]

The gradient of V against f on a graph gives you h/e, where e is the electron charge. The photoelectric effect alone does not allow one to find e or h in isolation, you need to rely on another experiment (such as the Millikin oil drop experiment) to find both constants.

[kamil9876]

But do we assume the graph we are given is the characteristics of ONE electron, ie the stopping voltage to stop ONE electron. Hence right? [because if it's not for 1 electron then would be unknown since the question doesn't state it]

THat is unnecesary. Since you are after Voltage, there is no talk of Energy here. Voltage is a rate, Energy/Charge so it's a measure of how much energy each charge carries. If you are given JUST VOLTAGE and no mention of energy, then you shouldn't worry about the number of electrons involved. E.g: Say the voltage is 2Joules/electron. If you have 4 electrons, you have 8Joules/4electrons. The rate doesn't change as long as it's the same material.

let n be the number of electrons. Let q_e be the charge of ONE electron, W be the work function of ONE electron and E be the energy of ONE electron:




but since we have n electrons and so:


voila, n cancels out and you get the same equation as if you had only one electron.

on a side note: shouldn't be treated as a variable but a constant(charge on ONE electron, not on some arbitrary group of electrons). In fact I was going to use instead of (show respect to Euler) but I decided to stay consistent.

[TrueTears]

Thanks

The light from a candle can be best described as :

A. coherent, arising from the vibrations of electrons.
B. incoherent, arising only from the transition of electrons in excited energy levels falling to lower energy levels.
C. coherent, arising only from the transition of electrons in excited energy levels falling to lower energy levels.
D. incoherent, arising from the vibrations of electrons.

Answer is D.

My books don't talk about coherent/incoherent or about how the vibrations of electrons are related with light. [I understand all the emission/absorption spectrum which is related with choices B and C, but don't understand anything about A or D -.-"]

Can someone please explain exactly what those 2 things are?

Thanks.

[mark_alec]

A coherent light source is one in which all the wavefronts are in sync. The typical example is a laser. An incoherent light source is one in which the wavefronts head in all direction in all phases. The typical example is a light bulb.

[TrueTears]

A coherent light source is one in which all the wavefronts are in sync. The typical example is a laser. An incoherent light source is one in which the wavefronts head in all direction in all phases. The typical example is a light bulb.

Ah cool I see, so since a candle shoots light in all directions it would be a incoherent light source right?

And what about the business with vibrations with electrons?

[Over9000]

Sorry to use ur thread privilages TT but I wish to ask, is it possible for photons to travel faster than the speed of light somehow, like if the photons got an extra amount of energy somehow coz that wud be interesting.

[mark_alec]

Ah cool I see, so since a candle shoots light in all directions it would be a incoherent light source right?

Yes, a candle is an incoherent light source. A coherent light source also produces light only one wavelength.

And what about the business with vibrations with electrons?

All you need to know is that due to 'thermal vibrations', the energy is produced in a continuous band, not as discrete wavelengths.

[mark_alec]

Sorry to use ur thread privilages TT but I wish to ask, is it possible for photons to travel faster than the speed of light somehow, like if the photons got an extra amount of energy somehow coz that wud be interesting.

It is not possible.

Firstly, you need to think of how you would give a photon extra energy. It doesn't have an electric charge, so you can't perform work on it in an electric field. But it does have a mass. In actual fact, gravity does effect light, affecting its energy in a similar way to a classical mass. Look up gravitational redshift if you are interested.

[TrueTears]

Ah thank you so much mark.

Also with thermal vibrations is that just adding heat to the electron causing it to vibrate and generate energy by shooting out photons?

But isn't this the same as emission spectrum, where you heat a metal and the electron absorb a discrete amount of energy and releases a photon of a discrete amount of energy.

So what's the difference between thermal vibration [how does it work?] and emission spectrum?

Thanks!

[mark_alec]

With the emission spectrum, the electrons are excited to a higher energy level, and when they return to a lower state, they emit a photon of a characteristic wavelength. You get emission spectra for vapours that have energy given to them, not solids (though there may be some exceptions.)

The mechanism for the continuous spectrum I suspect is due to a charged particle (the electron) being accelerated due to the thermal energy being given to the solid. The electrons in a metal inhabit almost continuous energy bands, which could also be part of it. To be honest, I don't know the exact mechanism by which 'thermal' radiation is produced, except that it can be modelled by oscillators moving in a quadratic potential that can only have an energy equal to . You'll have to look at statistical physics and the work of Planck to get a better answer (or wait until third year uni.)

[TrueTears]

Ahhh I think I get it now.

So is this interpretation correct?

The candle is an incoherent light source because the wavefronts head in all directions and they are in all kind of phases. A continuous spectrum is produced due to thermal vibrations. This means the electrons are accelerated due to the thermal energy of the flame of the candle and hence emit a continuous spectrum.

[And just as a side thing, thermal vibration is different from emission due to the fact that thermal vibration produces a continuous spectrum where as emission spectrum produces light of only certain wavelengths]

Thanks so much mark.

[mark_alec]

The candle is an incoherent light source because the wavefronts head in all directions and they are in all kind of phases. A continuous spectrum is produced due to thermal vibrations. This means the electrons are accelerated due to the thermal energy of the flame of the candle and hence emit a continuous spectrum.

Scrap the bit about the light going in different directions, what is important is that it is out of phase. And I wouldn't bother writing the second sentence about the continuous spectrum, as I'm not sure of its validity.

[TrueTears]

Ah okay, thanks.

[TrueTears]

Q1, These 2 formulas are from Nelson's textbook, there are just a few things I am unsure of. The diagram is attached

1st formula:   [taking the central bright band as n = 0]

In a 2 slit Young's experiment:

is the angle that the nodal line makes with the centre line.

The formula has in front of which implies P must be a dark band.

What I'm wondering is, is this formula is a general formula for all dark bands for a 2 slit experiment? Ie, if I wanted to find the angle the 3rd nodal line makes with the centre line then would become .

But then, what if I wanted to find the angle an ANTI-nodal line makes with the centre line, would I change the formula to ? [Since the path difference for bright bands has co-efficient of "n" in front of .] Ie, If I wanted to find the angle the 2nd antinodal makes with the centre line then it would be (Since the centre bright band is n = 0). Or does the formula only work for nodal lines and hence dark bands?

Now, would this formula also work for single slit experiment? So instead of and being 2 separate slits, they are just the "corners" of a single slit. If it does work for a single slit, then would "" just be the width of the single slit? Would it also work for finding out the angle an antinodal line makes with the centre line or does it have to be nodal lines?

2nd formula:

: distance of nodal point from centre line
: distance between the 2 slits.
: distance of nodal point to the centre of the 2 slits.

In a 2 slit experiment: Again, does this formula only work for nodal lines (hence dark bands)? Because the formula contains which implies the band must be a dark band and as a result L is the length of the nodal line. So could you also use it to find out the distance of a bright band to the centre line. For example, would just be "" and would just be the distance of the bright band to the centre of the 2 slits. Or is that totally wrong?

In a single slit experiment: Could this formula work for a single slit experiment? So instead of and being 2 separate slits, they are just the "corners" of a single slit. If it does work then is "" the width of the single slit? And could you use it to find the distance of a bright band to the centre line? If so, "" would be the length of the antinodal line right? Or does it only work for dark bands?