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# AQA Physics Unit 1 PHYA1 20th May 2013 Watch

1. (Original post by Jimmy20002012)
In the specification it says that we should know how to calculate he specific charge of an ion, how do we do this, say we had lot 2 electrons from an atom forming a +2 ion?

Posted from TSR Mobile
Specific charge = charge/ mass.
Here the charge on a 2+ ion is 2 x (1.6x10^-19). The magnitude of a charge on a proton is equal to that of an electron. If an atom is deficient by 2 electrons, its charge is 2 x the magnitude of that charge. You then obviously divide this by the mass of the nucleus, and put C kg^-1 after it
2. (Original post by Steroidsman123)
Specific charge = charge/ mass.
Here the charge on a 2+ ion is 2 x (1.6x10^-19). The magnitude of a charge on a proton is equal to that of an electron. If an atom is deficient by 2 electrons, its charge is 2 x the magnitude of that charge. You then obviously divide this by the mass of the nucleus, and put C kg^-1 after it
do you use atomic mass units 1.661x10^-27 or the respective mass of a proton or neutron 1.67x10^-27?
3. (Original post by Jimmy20002012)
In the specification it says that we should know how to calculate he specific charge of an ion, how do we do this, say we had lot 2 electrons from an atom forming a +2 ion?

Posted from TSR Mobile
Specific charge is calculated using charge divided by mass.
For ions, I believe it's the net charge/mass, so for your example, if 2 electrons are lost, I believe it's 2(1.6*10^-19)/mass.

Here's how I worked this out:
4. (Original post by Bixel)
They're made when the electron goes from it's excited state back down to a lower energy level, emitting photons in the process. These photons have a specific wavelength depending on the difference in energy levels it goes between - that specific wavelength will be a specific colour on the line spectrum. When all of these different electrons fall back down to lower energy levels, a line spectrum is produced!

Sorry that the explanation was rough but that's the general idea!
Thanks!
5. (Original post by Goods)
do you use atomic mass units 1.661x10^-27 or the respective mass of a proton or neutron 1.67x10^-27?
Err...for the mass I'd just do the number of nucleons times 1.67 x10^-27, but I'm no physics genius I always seem to get calculations questions right but when it comes to internal resistance questions and 6 marks I always really struggle
Well actually yeah respective values as you said
6. (Original post by Jimmy20002012)
In the specification it says that we should know how to calculate he specific charge of an ion, how do we do this, say we had lot 2 electrons from an atom forming a +2 ion?

Posted from TSR Mobile
erm I think they mean charge-mass ratio... what JJ Thompson had figured out in his cathode ray experiment: Q/m (or e/m) ...
So I assume number of electrons would be given, so you do this :

C x number of electrons / mass

I'm not sure what you use for mass, possibly: atomic mass x atomic mass unit (1.661 x 10-27)

But could someone just confirm that last bit please
7. (Original post by posthumus)
erm I think they mean charge-mass ratio... what JJ Thompson had figured out in his cathode ray experiment: Q/m (or e/m) ...
So I assume number of electrons would be given, so you do this :

C x number of electrons / mass

I'm not sure what you use for mass, possibly: atomic mass x atomic mass unit (1.661 x 10-27)

But could someone just confirm that last bit please
nobody knows! they put it on the data sheet too troll...
8. Does anyone know how to do question 4 on page 69 of the official AQA Physics book?
9. (Original post by posthumus)
erm I think they mean charge-mass ratio... what JJ Thompson had figured out in his cathode ray experiment: Q/m (or e/m) ...
So I assume number of electrons would be given, so you do this :

C x number of electrons / mass

I'm not sure what you use for mass, possibly: atomic mass x atomic mass unit (1.661 x 10-27)

But could someone just confirm that last bit please
Now I think about it it doesnt really matter if you use atomic mass unit of the respective neutron and proton masses. The respective masses would be more accurate but to 2 or 3 significant figures I don't think it would really matter would it?
10. The markschemes always uses 1.67 x 10-27 multiplied by the number of nucleons

Posted from TSR Mobile
11. Quick question, if we round an answer to sig. fig. do we use the rounded answer in the follow through calculations?
12. (Original post by SpartASH)
Does anyone know how to do question 4 on page 69 of the official AQA Physics book?
The one with the diodes and resistors?

Bear in mind that a diode in the forward direction has a pd of 0.6V...
13. (Original post by mattj94)
Hi, I recently set up a revision website. So far AQA Unit 1 is the only module I have for Physics but I hope to add unit 2 notes soon, it would be great if you could check it out! http://www.mattsrevision.com/particl...d-electricity/
thank you so much ...... they really helped me to understand electricity.
14. (Original post by Goods)
nobody knows! they put it on the data sheet too troll...
It's 1/12th of the mass of an atom of carbon-12 (if you do chemistry, it should make a bit more sense why it's carbon-12).
15. (Original post by Goods)
nobody knows! they put it on the data sheet too troll...
Yh I've never seen such questions before

(Original post by Steroidsman123)
Now I think about it it doesnt really matter if you use atomic mass unit of the respective neutron and proton masses. The respective masses would be more accurate but to 2 or 3 significant figures I don't think it would really matter would it?
I see what you mean... though charge-mass ratio is typically given as " C kg^-1 " & it's SI units I think... so I think they would want you to convert atomic mass to kilograms

Also charge density of atom... includes all electrons in the atoms divided by the mass of the atom (which is basically mass of the nucleus) ?? I saw someone say, divide charge on the atom by the mass
Also it's crucial to multiply the number of electrons by C
16. (Original post by NedStark)
Quick question, if we round an answer to sig. fig. do we use the rounded answer in the follow through calculations?
No, what you do is you write out 5 or 6 sig figs of the full value, and write this next to it (c). That means that you've used this value in further calculations. You would then write your rounded answer to appropriate sig. figs. and then use your exact value in any follow up questions.
E.g. [Just a random example]
part i)
I = V/ R
10/3 = 3.333... (c)
I ≈ 3.3A (2sf)

ii) Q = It
Q = (3.33...) x 5.0
Q = 16.6666... (c) [For any further calculations]
Q ≈ 16.7C

In the mark schemes, it usually says something like this at the start: "The use of significant figures is tested once on each paper in a designated question or partquestion. The numerical answer on the designated question should be given to the same
number of significant figures as there are in the data given in the question or to one more than
this number. All other numerical answers should not be considered in terms of significant figures. "
17. (Original post by Goods)
nobody knows! they put it on the data sheet too troll...
A data sheet troll is the inclusion of the sun's mass.
When will that ever be useful...
18. (Original post by NabRoh)
A data sheet troll is the inclusion of the sun's mass.
When will that ever be useful...
Astrophysics presumably
19. (Original post by NabRoh)
A data sheet troll is the inclusion of the sun's mass.
When will that ever be useful...
I think the Astronomical Data along with some of the constants are used in A2
But still, I'm not sure why it's in our booklets...
20. (Original post by SpartASH)
Does anyone know how to do question 4 on page 69 of the official AQA Physics book?
Some one else might explain it better but here goes:

The current in the diagram is going clockwise, so diode X is forward biased and diode Y is reverse bias.
Diode X lets only 0.6v through and since it's in parralel with Q (10k resistor), then Q also has 0.6v.
Therefore Y(diode) must have 2.4v, and once again it's in parralel so P has 2.4v.

That's the pd's calculated for all 4. Next is the currents:

Since Y is reverse bias, it doesn't let current through so for that parralel combination, all current passes P. Therefore current through Y is 0A.
Use V=IR to calculate the current through P. This gets 0.48mA (i think)

0.48mA must also go through the combination of Q+X. Use V=IR to calculate the current through Q which is 0.42mA, the remaining 0.06mA goes through diode X.

Sorry if that isn't clear, it's basically me typing my thought process.

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