# Beta decay

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#1
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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#2
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?
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#3

Can someone explain ci)
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3 years ago
#4
(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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3 years ago
#5
(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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3 years ago
#6
(Original post by Super199)

Can someone explain ci)
Calculate the mass defect and use e=mc^2

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3 years ago
#7
(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 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) Posted from TSR Mobile
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#8
(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?
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#9
(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
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3 years ago
#10
(Original post by Super199)
So the thing is wrong?
Yes

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3 years ago
#11
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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3 years ago
#12
(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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#13
(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.
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3 years ago
#14
(Original post by Super199)
Nah surely not.
I think so 0
#15
(Original post by Kyx)
I think so What numbers have you used?
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3 years ago
#16
(Original post by Super199)
What numbers have you used?
I'll just do the question now 0
3 years ago
#17
(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
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#18
(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?
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3 years ago
#19
(Original post by Super199)
Wait what numbers did you use here?

6.02 * 10^23 and what else?
236.053
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#20
(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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