AQA Physics PHYA5 - Thursday 18th June 2015 [Exam Discussion Thread]

What does it mean by certain radioactive isotopes are easier to screen? Where can I find information on this in the nelson thornes book

Reply 1003

Original post by AR_95

What does it mean by certain radioactive isotopes are easier to screen? Where can I find information on this in the nelson thornes book

I assume it's referring to the various penetrating powers (of alpha, beta +-, gamma) of the radioactive emissions. For example, alpha would be easier to screen than gamma because alpha has a lower penetrating power.

Reply 1004

CD223

OP

10

Original post by AR_95

What does it mean by certain radioactive isotopes are easier to screen? Where can I find information on this in the nelson thornes book

Original post by PotterPhysics

I assume it's referring to the various penetrating powers (of alpha, beta +-, gamma) of the radioactive emissions. For example, alpha would be easier to screen than gamma because alpha has a lower penetrating power.

Yeah screening is an alternative to "shielding" :smile:

Hence why lead aprons are often worn etc.

Posted from TSR Mobile

Reply 1005

Original post by CD223

Yeah screening is an alternative to "shielding" :smile:

Hence why lead aprons are often worn etc.

Posted from TSR Mobile

That makes sense.

In the cgp book it says that a nuclear reactor is surrounded by a thick concrete casing which acts as a shield. Nuclear fission produces gamma radiation. On another page, it says that gamma radiation can be blocked by several meters of concrete. Does that mean nuclear reactors are cased in several meters of concrete?! (Edit: What I was thinking is it would be more practical (due to size considerations, etc.) to use a lead casing, as then it would only be a few cm thick.)

(edited 8 years ago)

Reply 1006

Original post by CD223

Does anyone know when you do and don't you take into account the mass of a particle/atom when considering binding energy? Is it when they're on their own? Ie: 1 electron/neutron/proton?

Posted from TSR Mobile

You can't have a binding energy between 1 thing. I considered a hadron comprised of quarks but they can't be classed as binding energy due to gluons and whatnot, anyway, tangent.
The binding energy is the energy required to break up a nucleus into its consistent protons and neutrons. Hence, the mass difference times c² is the binding energy. We always account for the mass for a particle. It's important to note for these questions you use the different masses for the proton and neutron unless otherwise stated (as the 3rd decimal place is important). If a particle is solitary binding energy=0 as it cannot be broken down further :smile:

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Reply 1007

Funny enough I was talking about the topic yesterday but...

How am I supposed to work out the mass of the left hand side ffs

EDIT :
Nevermind I didn't realise it was in the data sheet

(edited 8 years ago)

Reply 1008

Original post by AR_95

Funny enough I was talking about the topic yesterday but...

How am I supposed to work out the mass of the left hand side ffs

EDIT :
Nevermind I didn't realise it was in the data sheet

Energy is released from the nucleus so we only need to consider nuclear mass. Nuclear mass of left hand side is just 4 times the mass of a hydrogen nucleus, that is, 4 times the mass of a proton, i.e. 4u. This is because a lone proton doesn't have any mass defect.

Regarding edit: Is my answer incorrect?

Oh wait, yes, sorry. The proton rest mass isn't exactly 1u, we have to take it up to 5 decimal places since that is what the He one is given to. Silly oversight :redface:

(edited 8 years ago)

Reply 1009

Original post by PotterPhysics

Energy is released from the nucleus so we only need to consider nuclear mass. Nuclear mass of left hand side is just 4 times the mass of a hydrogen nucleus, that is, 4 times the mass of a proton, i.e. 4u. This is because a lone proton doesn't have any mass defect.

Regarding edit: Is my answer incorrect?

4u yes but you're not supposed to consider it as 4u. I realised it was 4 protons beforehand, but I couldn't find anywhere that the mass of proton = 1.00728U which is usually given in these types of questions. After a while and constant staring at the formula sheet, I found that the mass (1.00728u) for proton is actually in the list of all the constants (it's just under proton rest mass). The same is given for neutron and electron which is quite frankly wonderful

Reply 1010

Original post by AR_95

Funny enough I was talking about the topic yesterday but...

How am I supposed to work out the mass of the left hand side ffs

Assume that the electron neutrino has zero mass.
u is a constant. It would be wise to convert to kg for simplicity.
u=1.661x10^-27
Mass after=(4.00150u=6.646915x10-27kg)+2(9.11x10-31)
Mass before=4(1.673x10-27)
change in mass=4.3263x10-29
E=mc^2
E=3.89367x10-12J
Converting to electron Volts is to divide by the magnitude charge of the electron (1.60x10-19)
In MeV=24.34MeV

(edited 8 years ago)

Reply 1011

It took so long for me to type I didn't see you'd solved :tongue:

Reply 1012

Btw I saw and have done most of the past papers from the "OLD PPQ's" link, however I did see some people mentioning 2002/2003 questions. where are these from?

Reply 1013

Original post by Amanzz

Assume that the electron neutrino has zero mass.
u is a constant. It would be wise to convert to kg for simplicity.
u=1.661x10^-27
Mass after=(4.00150u=6.646915x10-27kg)+2(9.11x10-31)
Mass before=4(1.673x10-27)
change in mass=4.3263x10-29
E=mc^2
E=3.89367x10-12J
Converting to electron Volts is to divide by the magnitude charge of the electron (1.60x10-19)
In MeV=24.34MeV

I always knew how to do the question. My only problem (a small one) was finding the value of Mp (mass of proton in U) which I found was given on the formula sheet

Your way of doing it brings you close to the final answer but it wouldn't of been accepted in the mark scheme, which specifies 24.7 MeV

The way I did it/mark scheme suggested was

(4x1.00728u) - [ (4.00150u) + ( 2 x 5.4x10^-4u)

Which gave a remainder of 0.02652 u
Simply multiplying it by 931.3 gives 24.69 IE 24.7

I think you'd get 1 or 2 marks for the right method though

(edited 8 years ago)

Reply 1014

Do we use 931.3MeVu^-1 (as given in teh cgp book) or 931.5MeVu^-1 (as given in the aqa data booklet)? I'm thinking the latter but everyone uses 931.3...

Reply 1015

Also guys I just want reassurance. Heard a dreadful rumour from my teacher last year that they weren't giving you an equations sheet in the exam this year? Is that even remotely true?

Reply 1016

Original post by PotterPhysics

Do we use 931.3MeVu^-1 (as given in teh cgp book) or 931.5MeVu^-1 (as given in the aqa data booklet)? I'm thinking the latter but everyone uses 931.3...

Oh wow, I didn't even realise this haha. The mark scheme for that question included 931.5 initially, but in brackets next to it said allow 931.3. In this question using both values you still end up rounding to 24.7 so it wouldn't have mattered. I think both are just as acceptable but I'll keep an eye out from now on, on what they prefer.

(edited 8 years ago)

Reply 1017