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Beta decay

I thought beta minus is where a neutron -> proton

So surely there would be too many neutrons and vice versa for beta +

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Reply 1
Also for fusion and fission do they both have a decrease in mass?

Which is the mass defect

E= mc^2 so energy is released?
Reply 2
Fission help.jpg

Can someone explain ci)
Reply 3
Original post by Super199
I thought beta minus is where a neutron -> proton

So surely there would be too many neutrons and vice versa for beta +


Beta minus is neutron -> proton, it's where there are too many neutrons

Beta plus is proton -> neutron, it's where there are too many protons


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Reply 4
Original post by Super199
Also for fusion and fission do they both have a decrease in mass?

Which is the mass defect

E= mc^2 so energy is released?


I believe fusion is an increase in mass while fission is a decrease, not sure though


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Reply 5
Original post by Super199
Fission help.jpg

Can someone explain ci)


Calculate the mass defect and use e=mc^2


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Reply 6
Original post by Super199
I thought beta minus is where a neutron -> proton

So surely there would be too many neutrons and vice versa for beta +


Just remember conservation of charge :smile:

If neutron -> proton, electron must be released to keep charge neutral (beta minus)

If proton -> neutron, positron must be released to keep charge positive (beta plus)

:biggrin:


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Reply 7
Original post by Kyx
Beta minus is neutron -> proton, it's where there are too many neutrons

Beta plus is proton -> neutron, it's where there are too many protons


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So the thing is wrong?
Reply 8
Original post by Kyx
Calculate the mass defect and use e=mc^2


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How though? Thats what I did for ii
Reply 9
Original post by Super199
So the thing is wrong?


Yes


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Reply 10


b ii) and c i) are essentially the same question but with different values for the data

b ii) is for a single atom, c i) is for 0.001 kg. you have to multiply the answer for b ii) by something


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(edited 6 years ago)
Reply 11
Original post by Super199
How though? Thats what I did for ii


Calculate how many uranium atoms are needed to make 0.001 kg. multiply that by your answer to b ii)


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Reply 12
Original post by Kyx
Calculate how many uranium atoms are needed to make 0.001 kg. multiply that by your answer to b ii)


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Nah surely not.
Reply 13
Original post by Super199
Nah surely not.


I think so :smile:
Reply 14
Original post by Kyx
I think so :smile:


What numbers have you used?
Reply 15
Original post by Super199
What numbers have you used?


I'll just do the question now :smile:
Reply 16
Original post by Super199
What numbers have you used?


I got the answer to part b ii) correct.

Then I converted 0.001 kg to u to get 6.02 x 10^23

I divided this by the mass (in u) of the original uranium atom to get 2.55 x 10^21

This multiplied by the answer to b ii) gives 7.65x10^10
Reply 17
Original post by Kyx
I got the answer to part b ii) correct.

Then I converted 0.001 kg to u to get 6.02 x 10^23

I divided this by the mass (in u) of the original uranium atom to get 2.55 x 10^21

This multiplied by the answer to b ii) gives 7.65x10^10


Wait what numbers did you use here?

6.02 * 10^23 and what else?
Reply 18
Original post by Super199
Wait what numbers did you use here?

6.02 * 10^23 and what else?


236.053
Reply 19
Original post by Kyx
236.053


So I now get the workings, but I'm actually struggling to understand what's going on.

Like do you mind explaining each step?

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