When to put positron or beta + decay in feynman diagram, & how does Lower R=higher P

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akereem100
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Question 1)e) why can't I put Beta + decay instead of positron :confused: did the ms just forget to mention to accept positron or is there a preferred time to say positron or beta+ and electron or beta -

Question 6)c)ii) I thought there will be more energy released as some it will be lost (simply don't understand this question HELP !

LINK
Question
http://filestore.aqa.org.uk/subjects...W-QP-JUN10.PDF
Markscheme
http://filestore.aqa.org.uk/subjects...W-MS-JUN10.PDF


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AQA Physics A unit 1 june 2010
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samb1234
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Well for 6c ii you've already shown that the lower the voltage the lower the resistance. Therefore, since p=v^2/R the assumption that p will =9 is wrong as R is now less than 4. Therefore the power will be greater than 9 watts. (They might also want you to say that the decreased resistance is due to temperature change I'm not sure I haven't looked at the mark scheme).

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tiddlytom
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Hi,

For your first question, beta+ is the process of getting a positron and an electron-neutrino, and beta- is an electron and anti-electron-neutrino - so they aren't quite synonymous.

For the second part, do you get that the wire will be cooler so have lower resistance? Using P=(V^2)/R, if R goes down, P goes up. Does this help?

Let me know if you want it explained more
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