Bozo's queries

[HarveyD]

ah yeah, i kinda see now

which option would you select for this:

[Lasercookie]

ah yeah, i kinda see now

which option would you select for this:

My initial assumption is blue - but I'm not too sure.

For a given metal and frequency of incident radiation, the rate at which photoelectrons are ejected is directly proportional to the intensity of the incident light.

The frequencies of the two light are different. I think this would mean that you can't really directly compare the two intensities. I don't know of any other way of figuring out the intensity.

Since there's only two options, I'd pick blue - I think that's the "most correct" option.

edit: didn't notice unable to determine.

[HarveyD]

yeah think tsfx may be wrong?
cause they had unable to determine lol

[Lasercookie]

yeah think tsfx may be wrong?
cause they had unable to determine lol

"Unable to determine" is correct as far as I know, as intensity is proportional to photocurrent "for a given metal and incident frequency". 

I didn't realise that "unable to determine" was an option :/

Hmm... these TSFX questions seem interesting - I might have a go at the exams later.

[iroflmfao]

TSFX exams are kinda off topic imo.
They kinda teach you everything at their masterclasses. Even if its not on the study design.

[Lasercookie]

Seeing that we've turned "Bozo's queries" into "Everyone's queries", might as well continue posting here (I hope you don't mind Bozo).

I've been intending to ask about this question for a while now, I kept forgetting to:



It's from STAV 2011, Q11 from the Light and Matter section.

I sort of understand the reasoning in the solutions, but the explanation feels kind of dense to me.

[xZero]

current = amount of charge flowing per second right? find the charge then divide that by the electrical charge of an electron to get the number of electrons/second and that should be the minimum number of photons per second

[Lasercookie]

current = amount of charge flowing per second right? find the charge then divide that by the electrical charge of an electron to get the number of electrons/second and that should be the minimum number of photons per second

Yep, that gives the correct answer. Thanks xZero

[iroflmfao]

Is AC voltage RMS ?

[Lasercookie]

Is AC voltage RMS ?

RMS voltage is the "effective voltage" - this also happens to be the amount that the voltage provides if it were equivalent to smooth D.C.

If the type of voltage isn't stated (it will be most of the time), then you assume it is RMS. This is what occurs in every day life as well. It'll say that the voltage in our houses is 240V AC. This is an RMS value.

[Lasercookie]

Well that's annoying, the images didn't work and we can no longer edit posts

Here they are (sorry for double posting)

[Aurelian]

Seeing that we've turned "Bozo's queries" into "Everyone's queries", might as well continue posting here (I hope you don't mind Bozo).

I've been intending to ask about this question for a while now, I kept forgetting to:

(Image removed from quote.)

It's from STAV 2011, Q11 from the Light and Matter section.

I sort of understand the reasoning in the solutions, but the explanation feels kind of dense to me.

Basically all you need to realise is that there is a one-to-one interaction between the incident photons and the ejected photoelectrons. As a result of this, the number of photons contained in the incident light beam per second is *the same* as the number of ejected photoelectrons per second, which you can figure out from the current and the charge on one electron =)

[iroflmfao]

Thanks LASER

Whats the unit of Work function LOL

[Lasercookie]

Thanks LASER

Whats the unit of Work function LOL

Work = Energy, so we'd be given it in either Joules or eV.
Most of the time we are given it in eV - makes things simpler.

The SI unit for work is Joules (derived from Newton metres e.g. W=Fx).

[HarveyD]

Any ideas on how to do this :/