AQA Physics: PHYA1 Tuesday 19th May 2015 (AM) and PHYA2 Thursday 04th June 2015 (PM)x

Online notes and papers: http://www.physicsandmathstutor.com/physics-revision/aqa-unit-1/ (He also has other units, this is just unit 1)
Websites: I highly recommend cyberphysics.

Reply 41

Osmium

1

Original post by PotterPhysics

Hi everyone, just wondering how most of you approach the 6-8 mark questions? There was this question on measuring the resistivity of a putty (question 3b here: https://d257ba8b9971d35b509279e471f1d8167ae3e7f0.googledrive.com/host/0B1ZiqBksUHNYZkdjdXdsZmZjX3M/From-AQA/1.3.1-2%20Current%20&%20Resistance.pdf ) and I basically had no idea how I would set up the circuit. We did some 6 markers in class and the teacher just said to memorize a few 6 markers in the hope that something similar would turn up, but for this one I didn't memorize anything similar so I had no idea what to do. Did you just have to know the resistivity experiment? I doubt that anyone could just work out that you had to use a "jockey" or whatever that thing is.

Any help at all is appreciated, especially from people who are good at these questions.

Thanks!

Edit: Linked the wrong question. (Now edited.)

I usually think of an equation that applies to the scenario in the question. In this case, it would be the resistivity equation. I would then think of what I could vary and measure from the variables in the equation. In this experiment, I could vary length and measure/calculate resistance by measuring current. Usually it is required to plot a graph, so in this case I would plot R(dependent) against L(independent) and match the equation to y=mx+c. From there it is possible to calculate resistivity from the gradient. Along the way I could add some information about the quantity of measurements, accuracy and other information depending on the question.

Reply 42

particlestudent

13

How is everyone feeling for this exam. Just done June 2014 Unit 1 and I got 89%.

I hope the electricity questions are not too complicated and I am hoping they give like 6 free marks on oscilloscope questions, like drawing the curve and finding the Vrms, Irms and maximum power.

Also can someone explain why the power dissipated in a circuit is at it's maximum point when R=r?

Reply 43

username1306902

9

can someone people tell me everything we need to know about line emission spectra

I'm in!

Original post by jarjarmonkey

can someone people tell me everything we need to know about line emission spectra

I'm assuming you mean both types of line spectrum?

Line Emission spectrum:
A hot (excited) gas emits photons as its electrons de-excite. The emitted photons are then passed through a prism. Only light of very discrete wavelengths is observed, and this relates to the changes in energy levels as the electrons de-excite. An example of such a spectrum is in the link below.

This is part of the process of visible light being produced by a flourescent tube; an electric current flows through mercury vapour at low pressure - the electrons collide with the mercury atoms and cause its electrons to excite. As the electrons de-excite, they emit photons in the UV range - these are, of course, not visible to the human eye. So the flourescent coating of the tube absorbs these photons, and itself gets excited, but instead of emitting UV light, it emits light in the visible range. (you need to know this example :smile:

)

Line Absorption spectrum:
When light is incident on a cool gas, certain frequencies (and therefore wavelengths) of the light don't pass through to the other side when observed through a prism. [This] The absorbed wavelengths of light are observed as discrete black lines.

[You don't really need to know this bit but it's helpful for you to understand the concept]

If you take the line emission and absorption spectra for the same element and combine them, you get a continuous spectrum.

http://ircamera.as.arizona.edu/astr_250/images/spectra.jpg (diagram)

Good luck with your exams!

Reply 46

username1306902

9

Original post by the_german_boy

I'm assuming you mean both types of line spectrum?

Line Emission spectrum:
A hot (excited) gas emits photons as its electrons de-excite. The emitted photons are then passed through a prism. Only light of very discrete wavelengths is observed, and this relates to the changes in energy levels as the electrons de-excite. An example of such a spectrum is in the link below.

This is part of the process of visible light being produced by a flourescent tube; an electric current flows through mercury vapour at low pressure - the electrons collide with the mercury atoms and cause its electrons to excite. As the electrons de-excite, they emit photons in the UV range - these are, of course, not visible to the human eye. So the flourescent coating of the tube absorbs these photons, and itself gets excited, but instead of emitting UV light, it emits light in the visible range. (you need to know this example :smile:

)

Line Absorption spectrum:
When light is incident on a cool gas, certain frequencies (and therefore wavelengths) of the light don't pass through to the other side when observed through a prism. [This] The absorbed wavelengths of light are observed as discrete black lines.

[You don't really need to know this bit but it's helpful for you to understand the concept]

If you take the line emission and absorption spectra for the same element and combine them, you get a continuous spectrum.

http://ircamera.as.arizona.edu/astr_250/images/spectra.jpg (diagram)

Good luck with your exams!

thanks man

Reply 47

Whizbox

7

Anyone know where to find some 6 mark questions to have a go at without having to dig through all the past papers?

Reply 48

Slenderman

2

Is this the discussion thread for AQA A Physics: Unit 1 - Particles, Quantum Phenomena and Electricity ?

Reply 49

rory58824

2

Original post by Whizbox

Anyone know where to find some 6 mark questions to have a go at without having to dig through all the past papers?

http://www.physicsandmathstutor.com <<< plenty of questions there all compiled for you :smile:

Original post by Slenderman

Is this the discussion thread for AQA A Physics: Unit 1 - Particles, Quantum Phenomena and Electricity ?

Yep it is, as well as Unit 2 - Mechanics, Materials and Waves

Anyone want to ask each other questions?

Reply 50

Slenderman

2

Has anyone got a list of all the definitions needed for Unit 1? For example threshold frequency, work function, restivity. etc. I need it please, I got the textbook but loads of them are so irrelevant. So does anyone have a nice list of them?

Reply 51

Cadherin

OP

14

Original post by Slenderman

Has anyone got a list of all the definitions needed for Unit 1? For example threshold frequency, work function, restivity. etc. I need it please, I got the textbook but loads of them are so irrelevant. So does anyone have a nice list of them?

Threshold frequency - the minimum frequency of an incident photon required to overcome the work function of the metal and emit an electron from the surface of a metal.
Work function - the minimum energy of an incident photon to emit an electron from the surface of a metal.
Resistivity - define using worded equation (although this came up last year so it is unlikely to come up again this year).

Most definitions you can define using the formula sheet equations, however.

Reply 52

Slenderman

2

Okay thanks.

Uhm I have a question off a paper.

http://filestore.aqa.org.uk/subjects/AQA-PHYA1-QP-JUN13.PDF

question 6.

How do i do part 2 onwards? If there is internal resistance within the power supply will that mean the emf will drop. Will current remain 4.2A? confused .. :frown:

Reply 53

crazy2015

2

How are you guys revising???

Reply 54

C0balt

16

Original post by Slenderman

Okay thanks.

Uhm I have a question off a paper.

http://filestore.aqa.org.uk/subjects/AQA-PHYA1-QP-JUN13.PDF

question 6.

How do i do part 2 onwards? If there is internal resistance within the power supply will that mean the emf will drop. Will current remain 4.2A? confused .. :frown:

You get the PD across 2.0ohm resistor by subtracting volts lost across internal resistance from emf.
Current through 2.0 is calculated by I=V/R.
Current is shared in parallel circuit so you can subtract the current through 2.0 from 4.2A to get current through R.
As R=V/I and PD is the same as the 2.0, you can now calculate the resistance of R.

Posted from TSR Mobile

Reply 55

Spelly456

2

I was doing the question: What distinguishes leptons from other particles? And the answer is they feel the weak interaction. Surely quarks also feel the weak interaction during beta decay? Any help?

Reply 56

C0balt

16

Original post by Spelly456

I was doing the question: What distinguishes leptons from other particles? And the answer is they feel the weak interaction. Surely quarks also feel the weak interaction during beta decay? Any help?

Yeah
That question is funny

I would go with safer does not feel strong force or they are fundamental
Posted from TSR Mobile

(edited 8 years ago)

Reply 57

Slenderman

2

Original post by C0balt

You get the PD across 2.0ohm resistor by subtracting volts lost across internal resistance from emf.
Current through 2.0 is calculated by I=V/R.
Current is shared in parallel circuit so you can subtract the current through 2.0 from 4.2A to get current through R.
As R=V/I and PD is the same as the 2.0, you can now calculate the resistance of R.

Posted from TSR Mobile

Thank you!!

I got another question though, aha.

Okay so lets say we got an electron in the ground state, n=1, to move that electron to the n=2 it must need 10eV.

Lets say a photon/bombarding electron comes along with 9.9ev of energy. It doesn't move.
Now lets say this photon/e- has exactly 10eV. Am I right to presume it will be excited and jump up to n=2 with no Kinetic Energy.
Now lets say it has 11eV, so it goes to n=2 with 1ev.

Is this all correct?

Also in my notes I have de Brogiles wavelength formula h/mv and i have written next to it 1/2mv^2

I did some rearrangement and i got

de broglies wavelength = h divied by square root 2mEk

What is this? Why have these equations merged? Where do I use it?

Sorry for these questions :frown:

Reply 58

C0balt

16

Original post by Slenderman

Thank you!!

I got another question though, aha.

Okay so lets say we got an electron in the ground state, n=1, to move that electron to the n=2 it must need 10eV.

Lets say a photon/bombarding electron comes along with 9.9ev of energy. It doesn't move.
Now lets say this photon/e- has exactly 10eV. Am I right to presume it will be excited and jump up to n=2 with no Kinetic Energy.
Now lets say it has 11eV, so it goes to n=2 with 1ev.

Is this all correct?

Also in my notes I have de Brogiles wavelength formula h/mv and i have written next to it 1/2mv^2

I did some rearrangement and i got

de broglies wavelength = h divied by square root 2mEk

What is this? Why have these equations merged? Where do I use it?

Sorry for these questions :frown:

Not really

Let's say an electron in an atom requires 10eV to move up one energy level. When an electron is incident to an atom with kinetic energy of 8eV, the orbital electron will not move in energy level. If an electron with 10eV is incident, the orbital electron will move up in energy level and the incident electron will have no kinetic energy left. If an electron with 12eV is incident, the orbital electron will move up in energy level and the INCIDENT electron will still have 2eV.

When it's photon, all energy of photon is absorbed. So the energy of photon must be exactly the same as the energy required for excitation. If the energy of the photon is more than energy required, there will be no excitation and if the energy if the photon is less than energy required there will obviously be no excitation.

De Broglie's wavelength equation cannot be written like 1/2 mv^2. I would imagine you had it written down due to this question? ImageUploadedByStudent Room1431607021.124655.jpg