Higher Physics 2018-19 Watch

nat5dawg
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#41
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(Original post by AstralDwarf)
Can't figure out 2018 Specimen Paper Multiple Choice Q18 for the life of me.

Help appreciated!
Firstly find out how much energy has been converted to the photon - so basically just find the difference between the two energy levels. (remember electrons only orbit at discrete energy levels, so to emit the photon it must be that exact energy between those 2 levels)

Then, use E=hf (using the energy you just found out), to find out the frequency of the photon emitted.

FInally use V=f (lambda) to work out the wavelength.

Hope this helps a bit.
Last edited by nat5dawg; 2 weeks ago
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tomctutor
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(Original post by AstralDwarf)
Can't figure out 2018 Specimen Paper Multiple Choice Q18 for the life of me.

Help appreciated!
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E=hf gives f= (21.8-5.40)\times10^{-19}/6.63\times10^{-34}=2.474\times10^{15}Hz
\lambda=c/f = 3.00\times10^{8}/2.474\times10^{15}=1.213\times10  ^{-7}m

Ans: C
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Aaron_0610
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(Original post by AstralDwarf)
Can't figure out 2018 Specimen Paper Multiple Choice Q18 for the life of me.

Help appreciated!
Okay,

You have an energy transition and the arrow tells you what to do. You have to start by finding the energy of the transition so you always calculate it in the direction of the arrow, I always think about it as the top of the arrow take away the bottom. In this case:

-5.40x10^-19 - (-21.8x10^-19) = 1.64x10^-18 J

You now do E=hf to find your frequency.

E = hf, where h = planc’s constant
1.64x10^-18 = 6.63x10^-34 f
f = 2.47x10^15 Hz

(I know this is so much for a multiple choice question)
Now you use v = fλ to find your wavelength

v = fλ, where v = speed of light
3x10^8 = 2.47x10^15 λ
λ = 1.21x10^-7 m

So the answer is C.

Hope this helped! 😀
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a.nm_j
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any advice from crash highers to a fellow crash higher?
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Aaron_0610
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(Original post by a.nm_j)
any advice from crash highers to a fellow crash higher?
I'm a 2-year resit, from Nat 5, so I can't give you crash higher advice although considering I got 28% in the exam last year, I might be able to help.

Do past papers, if you can't do a question read how to do it, make notes on types you can't do. Go over main points, that they may ask in form of a long answer knowledge question. Understand it.

If you don't have the time over the next week just keep doing past papers.

This website provides (http://mrmackenzie.co.uk/higher-revision/), past papers back to 1994. Fun fact, SQA love to reuse multiple choice questions, if you try the majority of those multiple choice questions and memorise the answers/what to do, I can assure you you will bank 15/25 at least in the Multiple Choice, from either knowing how to do the questions or the exact same ones coming up. This means you go into Paper 2 with 20% odd of your marks already in the bag, plus the assignment.
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tomctutor
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(Original post by a.nm_j)
any advice from crash highers to a fellow crash higher?
On top of Aaron_0610 advice above, learn what the
fornulae listed in the Relationships are, what they relate to and some examples where they have been used: \mathrm{Formula\rightarrow substitute\rightarrow state \ answer} so becomes a maths exam!
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Taramunro2000
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Can anyone do question 8 part b) of the 2017 higher paper? We know the answer is 'out of the page' but how exactly do you figure that out? Thanks in advance

The electron beam is then passed into a “slalom magnet” beam guide.
The function of the beam guide is to direct the electrons towards a metal
target.
Inside the beam guides R and S, two different magnetic fields act on the
electrons.
Electrons strike the metal target to produce high energy photons of
radiation.

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turtle.8
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(Original post by Taramunro2000)
Can anyone do question 8 part b) of the 2017 higher paper? We know the answer is 'out of the page' but how exactly do you figure that out? Thanks in advance

The electron beam is then passed into a “slalom magnet” beam guide.
The function of the beam guide is to direct the electrons towards a metal
target.
Inside the beam guides R and S, two different magnetic fields act on the
electrons.
Electrons strike the metal target to produce high energy photons of
radiation.

Name:  Qustion 8 b.jpg
Views: 16
Size:  21.7 KB
I struggled with this question a few days ago but I've figured it out. So you use the right hand rule because electrons are negative. Your thumb (motion) is pointing upwards towards the top of the page because that is the motion that the electron is travelling as it goes in the motion of the arrow between R and S. Your middle finger (current) is pointing to the right because that is the direction of which the electron is starting from. Therefore your pointing finger (magnetic field) is pointing out of the page. Hope this makes sense and you understand now!
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rraduzky17
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I just want to ask. If there is a graph do you always need to start the scale at 0. I was doing the 2017 paper and I used a scale that started from 2 instead of 0.

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XZS
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Nah but do a break at the beginning of the axis to indicate that you’re no starting from 0
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FlightVelocity
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Start at zero. Better to be safe.
(Original post by rraduzky17)
I just want to ask. If there is a graph do you always need to start the scale at 0. I was doing the 2017 paper and I used a scale that started from 2 instead of 0.

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Mary527
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Can someone help with 1996 questions 12 the answer is D
http://mrmackenzie.co.uk/higher-revision/
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Mary527
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(Original post by Mary527)
Can someone help with 1996 questions 12 the answer is D
http://mrmackenzie.co.uk/higher-revision/
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sbneelu
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Okay so the first thing I did was to simplify the three resistors in parallel into one resistor. Since they all have the same resistance, you can just divide the resistance of each one by the number of resistors to get the total resistance. So since each one has a resistance of 3 kΩ the equivalent single resistor has a resistance of 1 kΩ. You can also do this the long way using 1/Rt = 1/R1 + 1/R2 + 1/R3 and get the same thing.

So you've got an EMF of 12V and a potential divider with resistors of resistances 3 kΩ and 1 kΩ, so the 12V gets split 3:1, so the p.d. across the 3 kΩ resistor (R1) is 9V and the p.d. across the 1kΩ resistor would be 3V.

So if the three resistors in parallel were a single resistor, the p.d. across the resistor would be 1 kΩ, so the p.d. across the two marked points is 1 kΩ. Since the voltage across each branch of a parallel circuit is the same, each of the resistors in connected parallel has a p.d. of 3 kΩ across it.

So the p.d. of R1 is 9V and the p.d.s across R1, R2 and R3 are all 3V, so the answer is D.
(Original post by Mary527)
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AstralDwarf
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In a few questions I've noticed that instead of using equations you just find the gradient of the graph. Never did anything like that in class. How are you meant to know whether to use the gradient or not?
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nat5dawg
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(Original post by AstralDwarf)
In a few questions I've noticed that instead of using equations you just find the gradient of the graph. Never did anything like that in class. How are you meant to know whether to use the gradient or not?
Usually you can just rearrange an equation into the equation of a line (y=mx+c)

So like the EMF equation E = V + Ir (in the eqn sheet)

Just rearrange that into y=mx+c where y is V (voltage) and x is the current

so you get V = -Ir + E
change that to y=mx+c to then finally get V= -r I + E

so EMF (E) is the y intercept and the gradient is -r (so take the negative of whatever gradient you got to get positive r)
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tomctutor
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Its the nature of physics, in any Physical Law its modeled in y=mx+c, e.g.
s=\frac{1}{2}at^2 for displacement:
would measure s plotted against t^2 then the gradient would give you
\frac{1}{2}a
You use YOUR best straight line if one is not shown.
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ZBAD$
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Hi all, could someone help me with the 2012 revised higher multiple choice questions 16 and 17. For 16 why is the answer not a proton? Also for 17 why is the emf 9 and not 6
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GladstoneBrookes
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(Original post by ZBAD$)
Hi all, could someone help me with the 2012 revised higher multiple choice questions 16 and 17. For 16 why is the answer not a proton? Also for 17 why is the emf 9 and not 6
For 16, the answer is not proton as protons and electrons do not have the same charge to mass ratio, so a proton would follow a path in the opposite direction to the electron, but with a larger radius. The answer is antimatter as an anti-electron would have the same mass and magnitude of charge, so follows the same shape of path, but since the charge is opposite to the electron (positive instead of negative), it moves in the opposite direction. Think about the right-hand rule here of it helps.
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GladstoneBrookes
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(Original post by ZBAD$)
Hi all, could someone help me with the 2012 revised higher multiple choice questions 16 and 17. For 16 why is the answer not a proton? Also for 17 why is the emf 9 and not 6
And for 17, the emf is the y-intercept of the graph of voltage against current, which (since the gradient is -3) would be 9V.
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