OCR Physics Unit 2 - G482 - (June Exams Preparation)

Reply 360

zcanf

Original post by ReginaPhalange29

Is G481 worth 30%?

Posted from TSR Mobile

Yes.

(edited 10 years ago)

Reply 361

Layontheland

1

Original post by OllieGCSEs

Planck's constant experiment:
-Have a circuit with an LED of known wavelength and a variable p.d., and place a voltmeter across the LED.
-Increase the voltage from 0v slowly until the light is just on, and note the p.d across the LED (NB: this voltage is known as the threshold voltage, and is the minimum voltage required to pass a current through an LED)
-Replace the LED with another one of a different wavelength of light emitted and repeat the process
-As we know that E(energy) is proportional to the frequency of the light, we can say that E=hf, which is the same as E=hc/L (where h is the Planck constant, c is the speed of light and L is the wavelength)
-Therefore if we plot a graph of E against c/L, the gradient will be h :smile:

-Note that E is equal to the charge of an electron (1.6*10^-19)*the threshold voltage

I have a question too, which I'd greatly appreciate a response for: could someone please explain how spectral lines are evidence for the existence of energy levels in ionised atoms?

If you read my question and know the answer but cba to explain it, it might be worth realising that explaining the answer would develop your understanding :wink:

But if you don't know the answer, quote me and bump the question!

Posted from TSR Mobile

Couldn't you say plot a graph of Voltage against 1/Wavelength
so the gradient is equal to hc/e than just calculate the value of h?

Reply 362

Layontheland

1

Original post by PokeyZ

Can you lose marks for being too precise? In the mark schemes they almost always round the numbers which is fine, but when the question has multiple parts my answer ends up far off from the mark scheme answers.

I tend to give my answers to the sig. figure they have given the values too.
I've been told by my teacher to do 3. sig figure if unsure.

Reply 363

JamesJJT

Original post by Layontheland

I tend to give my answers to the sig. figure they have given the values too.
I've been told by my teacher to do 3. sig figure if unsure.

Official mark scheme notes state you will only lose 1 mark in the entire exam for sig fig errors.

Generally the answers round to 2, but 3 is the safest bet unless the question states otherwise.

Hope this helps!

Reply 364

Legal drugdealer

This is going to be killerrrrrrrr

Reply 365

jooh_23

basically you see those questions about what happens to the light intensity/difference between fringes when you change distance/colour of light and stuff can someone help!!

thx

Reply 366

OllieGCSEs

10

Original post by Layontheland

Couldn't you say plot a graph of Voltage against 1/Wavelength
so the gradient is equal to hc/e than just calculate the value of h?

Of course you could

Posted from TSR Mobile

Reply 367

viyas07

can somebody please please explain things we neeed to know about absorption spectrum, emission spectrum,line spectrum, band spectrum and everything or anything thats related to them and how to work out anything related to spectrum questions!!!!!

Posted from TSR Mobile

Reply 368

JamesJJT

Original post by jooh_23

basically you see those questions about what happens to the light intensity/difference between fringes when you change distance/colour of light and stuff can someone help!!

thx

Colours are: Red, Orange, Yellow, Green, Blue Violet
WAVELENGTHS DECREASE ------------->

As wavelengths decrease, The fringe separation will decrease. (Wavelength = ax/D where a and D don't change.) Violet diffracts the least, so it's always on the inside of the fringe (Closest to the central white line) and red is obviously furthest out each time.

Brightness of the fringes will gradually decrease the further the wave diffracts (the greater the angle) usually.

Reply 369

Dizzie31

Original post by JamesJJT

Colours are: Red, Orange, Yellow, Green, Blue Violet
WAVELENGTHS DECREASE ------------->

As wavelengths decrease, The fringe separation will decrease. (Wavelength = ax/D where a and D don't change.) Violet diffracts the least, so it's always on the inside of the fringe (Closest to the central white line) and red is obviously furthest out each time.

Brightness of the fringes will gradually decrease the further the wave diffracts (the greater the angle) usually.

by the way to remember the colours its the visible spectrum guys, rainbow colours!

Reply 370

JamesJJT

Original post by Dizzie31

by the way to remember the colours its the visible spectrum guys, rainbow colours!

Yes thank you!

Forgot to mention what I said was for white light only. Monochromatic light will not have spectral fringes.

Reply 371

ninetteharris

2

Original post by OllieGCSEs

Planck's constant experiment:
-Have a circuit with an LED of known wavelength and a variable p.d., and place a voltmeter across the LED.
-Increase the voltage from 0v slowly until the light is just on, and note the p.d across the LED (NB: this voltage is known as the threshold voltage, and is the minimum voltage required to pass a current through an LED)
-Replace the LED with another one of a different wavelength of light emitted and repeat the process
-As we know that E(energy) is proportional to the frequency of the light, we can say that E=hf, which is the same as E=hc/L (where h is the Planck constant, c is the speed of light and L is the wavelength)
-Therefore if we plot a graph of E against c/L, the gradient will be h :smile:

-Note that E is equal to the charge of an electron (1.6*10^-19)*the threshold voltage

I have a question too, which I'd greatly appreciate a response for: could someone please explain how spectral lines are evidence for the existence of energy levels in ionised atoms?

If you read my question and know the answer but cba to explain it, it might be worth realising that explaining the answer would develop your understanding :wink:

But if you don't know the answer, quote me and bump the question!

Posted from TSR Mobile

emission spectrum is when an atom is unstable and will release energy to try and become stable, it will realise energy in a form of a photon. the spectrum will look like, black background with the bright lines; the lines show the wavelength of the photons being released.

the absorption spectrum is when the atom, such as the zinc plate in the photoelectric effect, absorb a photon, the photon has a wavelength, and the the spectrum looks like lots of colours, and some black lines, the black lines show which wavelength has been absorbed. from that you can work out the energy.

i hope this helps, i really bad at explaining stuff :/

Reply 372

ninetteharris

2

http://www.astro.bas.bg/~petrov/herter00_files/lec05_06.gif

energy levels in a atom

Reply 373

Jonma

Can someone explain 2 ) (b) (ii) on June 2009 please? Why do you take one E.M.F away from the other instead of adding them?

Reply 374

R07

Original post by OllieGCSEs

I have a question too, which I'd greatly appreciate a response for: could someone please explain how spectral lines are evidence for the existence of energy levels in ionised atoms?

If you read my question and know the answer but cba to explain it, it might be worth realising that explaining the answer would develop your understanding :wink:

But if you don't know the answer, quote me and bump the question!

i think it is something like the spectral lines basically represent the radiation/energy emitted by the atoms when their excited electrons jump from higher to lower energy level right..!! so probably this shows that there are different energy levels in the atoms due to which we come across the spectrum. im not sure but it has to be related to some of this..!!

Reply 375

Layontheland

1

Do we actually state the words "the potential divider argument" or explain it.
If we have to explain it, can someone explain how we would phrase it?

Reply 376

zcanf

On an IV graph, can I do 1/Gradient for resistance or just do V/I?

Reply 377

Dizzie31

Original post by zcanf

On an IV graph, can I do 1/Gradient for resistance or just do V/I?

V / I is fine
although 1/ gradient is the same answer anyway..

Reply 378

blackstarz

1

Im a bit confused, according to I=anev if you increase the current there are more collisions so the temperature increases and the resistance increases but according to I=v/R it is the opposite? For certain questions how do you know whether you're meant to use the v=ir relationship or the as I increases temperature increases so R increases relationship? Thanks very much for any help and also how does the Voltage link in to all of this? Thanks !

Reply 379

Dizzie31

On the questions that increase the intensity, such as in the june 2009 paper q4b(ii)
Do you have to work out the amplitude? if so how
on the mark scheme it just says - sine wave of same frequency with any increased amplitude

OCR Physics Unit 2 - G482 - (June Exams Preparation)

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