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PHYA1 - Physics Unit 1 Exam - 24th May 2011

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1. State what the ionisation energy is.

The minimum amount of energy that is needed to excite the electron from one energy level to transition to the ground state.

What I put, feel free to disagree.
Original post by Dizzy in my Head
1. State what the ionisation energy is.

The minimum amount of energy that is needed to excite the electron from one energy level to transition to the ground state.

What I put, feel free to disagree.


erm i put the 'minimum energy required to liberate an electron completely from the atom'. which i believe to be correct.
Original post by maths134
thats right. Mind if I add it to my mark scheme ?


Haha, sure, knock yourself out - I've been putting a few answers up, but I kind of need help with the six mark question, I know just bits and pieces. :wink:
Reply 323
Avatar for Aish93
Aish93
what did you put for the 1st Q? I got nucleus :smile:
Original post by fGDu

Original post by fGDu
not sure 3. is correct

4. Which of these two interactions cannot take place The second interaction because leptopn number is not conserved

5. Why must the charge of X be 0? To conserve charge (0->-1+1)

6. Which is the most stable particle in this interaction? Proton


I thought it was the top equation that doesn't work.
There are no leptons involved, so it is the Strong Interaction. And in this type of interaction Strangeness has to be conserved i.e it CAN'T be + or - 1 on either side.

The bottom equation was something = muon + antineutrino, which is 1 + -1 lepton = 0.

Anyone else agree?
Original post by maths134
erm i put the 'minimum energy required to liberate an electron completely from the atom'. which i believe to be correct.


Damn, yeah - the energy needed to remove an electron from the atom - why am I so stupid? I'll fail for sure now if I made mistakes like this the whole way through!
What was the neutral particle that decayed. question oneeeee
Original post by TooEasy123
I thought it was the top equation that doesn't work.
There are no leptons involved, so it is the Strong Interaction. And in this type of interaction Strangeness has to be conserved i.e it CAN'T be + or - 1 on either side.

The bottom equation was something = muon + antineutrino, which is 1 + -1 lepton = 0.

Anyone else agree?


Yeah, I got the top equation, too, because there wasn't any neutrino to conserve the pion-ness.

And I put weak interaction, because I just did. Because I'm stupid like that. Anyone got a definite answer to that?
Reply 328
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fGDu
14
Original post by TooEasy123
I thought it was the top equation that doesn't work.
There are no leptons involved, so it is the Strong Interaction. And in this type of interaction Strangeness has to be conserved i.e it CAN'T be + or - 1 on either side.

The bottom equation was something = muon + antineutrino, which is 1 + -1 lepton = 0.

Anyone else agree?


A muon has no lepton number

EDIT im an idiot goodbye 1 more mark
(edited 12 years ago)
Reply 329
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Spas_13
Original post by Aish93

Original post by Aish93
what did you put for the 1st Q? I got nucleus :smile:


Thats wrong mate, unlucky. Good luck in the retake :wink:
Reply 330
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nkah
Would you get marks for the proof question if you did the product of resistors in parallel / sum of resistors in parallel = (3+3) x 3 / (3+3) + 3 = 18/9 = 2 ohms, plus the 3 ohm resistor in series gives 5 ohms?

This is included in the NT book and its basically using the 1/R formula in a different way.
Reply 331
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nkah
Original post by Dizzy in my Head
Yeah, I got the top equation, too, because there wasn't any neutrino to conserve the pion-ness.

And I put weak interaction, because I just did. Because I'm stupid like that. Anyone got a definite answer to that?


Isn't it the bottom equation because I used the insert sheet and saw both of the products had a lepton number of -1, meaning the thing that decayed (can't remember exactly) would have a lepton number of -2, to conserve lepton number, which it doesn't?
Original post by Lucy149
What was the peak to peak voltage of that ac source?


The Y-sensitivity was 0.5V per division (if I remember correctly), and the line length was 8 divisions (4 up, 4 down), therefore 0.5x8=4V.

Would you get marks for the proof question if you did the product of resistors in parallel / sum of resistors in parallel = (3+3) x 3 / (3+3) + 3 = 18/9 = 2 ohms, plus the 3 ohm resistor in series gives 5 ohms?

This is included in the NT book and its basically using the 1/R formula in a different way.

I think that was what you were supposed to do. That's what I did anyway.

I thought it was the top equation that doesn't work.
There are no leptons involved, so it is the Strong Interaction. And in this type of interaction Strangeness has to be conserved i.e it CAN'T be + or - 1 on either side.

The bottom equation was something = muon + antineutrino, which is 1 + -1 lepton = 0.

The bottom equation didn't work. It was an antimuon and an antineutrino, not a muon, which comes to L = -2. The muon was positively charged. Quite sneaky tbh.
For ionisation energy:
Minimum energy to remove an electron from an atom
In a gaseous state
Reply 334
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LewisS911
Original post by nkah
Isn't it the bottom equation because I used the insert sheet and saw both of the products had a lepton number of -1, meaning the thing that decayed (can't remember exactly) would have a lepton number of -2, to conserve lepton number, which it doesn't?


Yup, thats exactlly what i did. I only realized once i saw the forumla sheet
Original post by Spas_13
Thats wrong mate, unlucky. Good luck in the retake :wink:



It asked for two baryons, one of which is a neutron (though you needed to state two - but you still got one mark if you didn't), the other a proton.

The nucleus is a collection of neutrons and protons, and it wanted a specific answer.
(edited 12 years ago)
Reply 336
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chemkid
11
For the proof question i proved that the resistance of the resistors was 3, using my answer of 5 ohms from the previous question - would i still get marks for that?
i.e. i did 1/2r + 1/r = 1/rt
rt = 2r/3

2r/3 + r = 5
5r = 15
r = 3
Reply 337
Avatar for Lucy149
Lucy149
4
i dont think we have mentioned the one in the first question yet about whether particle X was a lepton baryony or meson
i put baryon but i wouldnt count on any of my answers
Original post by fGDu

Original post by fGDu
A muon has no lepton number

EDIT im an idiot goodbye 1 more mark


Yeah but it has a muon-lepton number of 1. and im pretty sure the other was was a muon-antineutrino. so thats -1 muon-lepton number.
Original post by Dizzy in my Head
Yeah, I got the top equation, too, because there wasn't any neutrino to conserve the pion-ness.

And I put weak interaction, because I just did. Because I'm stupid like that. Anyone got a definite answer to that?


erm i put the same as you. But i believe it was teh second due to lepton number needing to be conserved. Because it was an antineutrino (-1) and a muon*+ (also -1) so it wasnt possible

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