TrueTears question thread

[TrueTears]

Just a confirmation.

The voltage in photoelectric cell can be retarding but it can also help accelerate the electron.

So when voltage (electric energy) is ADDED to the electron's kinetic energy that means the the speed of the electron increase right?

So on a photocurrent - voltage graph, assuming adding voltage to the kinetic energy is positive, then the positive side of the graph would just be constant horizontal line right? Because adding voltage to the kinetic energy doesn't increase the maximum amount of photocurrent it just increases its speed right? [only a change light intensity increases/decreases the max photocurrent yeah?]

But when voltage opposes the motion of the electron it converts the kinetic energy of the electron into electric energy so as this continues eventually the electron will have no kinetic energy hence no current, which is why the negative part of the photocurrent - voltage graph is decreasing until the current reaches 0 and at this point is the , stopping voltage?

Thankie

[TrueTears]

No wait...

Current is how many electrons pass through a point in one second, it makes sense to me that if you increase the speed of electrons current should increase? Hence it shouldn't be a horizontal line? shouldn't it be a line with positive gradient?

[Mao]

Not quite.

For photoelectric cells, the current depends on the light intensity. Remember that each photon knocks out one electron.

Work is done by the photon on the electron to knock it off the metal surface. The amount of work required varies (depending on location and how tightly bound the electron is to the metal cations). In this case, the threshold frequency gives the minimum work required (to liberate the least tightly bound electrons).

When a greater frequency is used, more electrons can be liberated as there are more energy available to liberate more tightly bound electrons. However, the least bound electrons still require the minimum work, whilst the higher frequency light provides more photon energy, hence these electrons have the greatest 'residual' kinetic energy. The less bound electrons have more kinetic energy than tighter bound electrons (as tighter bound electrons require more energy to liberate).

When a negative voltage is applied, work is done against electrons to reduce their motion. For electrons with not enough kinetic energy, they get knocked back onto the metal surface, but the more energetic ones can still escape. Since some are knocked back, the current decreases. As you apply more reverse voltage, more electrons get knocked back (because fewer electrons have sufficient energy to escape), hence current decreases. Until the reverse voltage reaches and exceeds the max KE, at this point, not even the most energetic can escape, hence no current flows.

Greater velocity does not necessitate greater current. It's difficult to explain, but the general gist is because distance between electrons is not fixed.

[TrueTears]

Thank you Mao!!! Beyond thanks actually...

[TrueTears]

Another question.

If a photon doesn't have mass, then how come it has momentum?

Is there a intuitive way of thinking about this? Or would I need beyond vce physics?

[dcc]

Another question.

If a photon doesn't have mass, then how come it has momentum?

Is there a intuitive way of thinking about this? Or would I need beyond vce physics?

Intuition and quantum mechanics rarely join forces.

[kamil9876]

haha my relativistic dynamics are piss poor:

But i can say that p=mv is not momentum when v is near c, let alone equal to it:

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/relmom.html

Notice on the bottom how it says:



and so p=E/c at the limit. and use

Actually, it even says on the bottom:

with the limiting case applying for the momentum of a photon.

[kamil9876]

Another question.

If a photon doesn't have mass, then how come it has momentum?

Is there a intuitive way of thinking about this? Or would I need beyond vce physics?

Intuition and quantum mechanics rarely join forces.

Haha was just about to say

edit: quote fail

[TrueTears]

lololol okay thanks.

[TrueTears]

Also what do you call interference when for example 2 waves undergo constructive interference but the final wave is not the maximum constructive interference, ie, it still gives an amplitude higher than the 2 waves but not the max amplitude.

Is it still called constructive interference? And for the interference that gives maximum amplitude is that called maximum constructive interference?

Is this the same for destructive?

Thanks.

[Mao]

'inteference' isn't exactly 'inteference', it's formally called 'superposition', via a process the same nature as 'addition of ordinates' you learn in maths.

but yes, it'll still be constructive.

[TrueTears]

Yeah, I just stick to the "interference" term coz I'm used to it haha

Thanks for confirmation!

[TrueTears]

Also with the formula where w is the width of the single slit.

What if we are given a question like, what is the ratio of if the point on the wall/plate  is a dark band occurring at and the central maximum is .

The ratio does not have or in front of . So does that mean we don't need to take into consideration what band (bright or dark) the wavelength hits on the wall/plate?

We just simply divide the wavelength of the slit by the width of the slit to work out extent of diffraction?

Thank you!

[kamil9876]

If you look at formula (1) of my proof of the first equation:



sub in n=1 and u get ur formula. The angle is the angle that the line connecting slit to point on screen makes with horizontal, in this case, the point on the screen being the first dark band. As gets bigger that means the first dark band is further away so diffraction occured to a greater extent. Hence the ratio is just a measure of the extent of diffraction. Because we only subbed in n=1, you'r right, only dark bands. If what you told me is true about no general formula being in the new course, then they couldn't ask you a question like this in the first place since it requires the general formula. But if you wanted to, use the general formula.

[TrueTears]

wait... I thought the extent of diffraction is independent of what point the wave lands on. I mean the wave can land on any point (dark or bright) the ratio is just a general statement to determine the OVERALL extent of diffraction right?

So we don't even need to care about whether there's dark/bright bands. All we need is the wavelength and width of slit. right?

Thanks.