# Beta decay

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I thought beta minus is where a neutron -> proton

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

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

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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?

Which is the mass defect

E= mc^2 so energy is released?

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#4

(Original post by

I thought beta minus is where a neutron -> proton

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

**Super199**)I thought beta minus is where a neutron -> proton

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

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

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#5

(Original post by

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?

**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?

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#7

**Super199**)

I thought beta minus is where a neutron -> proton

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

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)

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(Original post by

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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**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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#11

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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#12

(Original post by

How though? Thats what I did for ii

**Super199**)How though? Thats what I did for ii

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(Original post by

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

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**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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#16

(Original post by

What numbers have you used?

**Super199**)What numbers have you used?

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#17

(Original post by

What numbers have you used?

**Super199**)What numbers have you used?

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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(Original post by

I got the answer to part b ii) correct.

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

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

**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

6.02 * 10^23 and what else?

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(Original post by

236.053

**Kyx**)236.053

Like do you mind explaining each step?

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