why is neutrino/anti-neutrino released in beta +/- decay?

Watch
TimetoSucceed
Badges: 2
Rep:
?
#1
Report Thread starter 9 years ago
#1
i understand what the decays are

beta minus, neutron turns to proton in nucleus, emits electron

beta plus, proton turns to neutron, emits positron

but why is the neutrino/anti-neutrino released

As neutron/proton has lepton number 0, and positron has lepton number -1, so a electron-neutrino is released to make it back to 0? so it is conserved?

same with beta minus, electron = 1, anti-electron neutrino = -1, so lepton number is still 0?

is that the reason? is there anything else? please help thanks.
0
reply
honoris
Badges: 0
Rep:
?
#2
Report 9 years ago
#2
Balance out charge
not this sorry
5
reply
Circadian_Rhythm
Badges: 14
Rep:
?
#3
Report 9 years ago
#3
I think I asked my teacher this, and I have an answer in my head, IT IS PROBABLY WRONG.

Conservation of angular momentum.
0
reply
CharleyChester
Badges: 0
Rep:
?
#4
Report 9 years ago
#4
Yep its so lepton number is conserved
1
reply
Breedlove
Badges: 0
Rep:
?
#5
Report 9 years ago
#5
Conservation of lepton number.

^ninja.
1
reply
honoris
Badges: 0
Rep:
?
#6
Report 9 years ago
#6
Oh yeeah, also remember strangeness and baryon number are conserved.
0
reply
TimetoSucceed
Badges: 2
Rep:
?
#7
Report Thread starter 9 years ago
#7
(Original post by honoris)
Balance out charge
isnt it already balanced?

p ----> n + e+

1----->0 + 1?

and dont neutrinos have 0 charge? =/
0
reply
TimetoSucceed
Badges: 2
Rep:
?
#8
Report Thread starter 9 years ago
#8
(Original post by honoris)
Oh yeeah, also remember strangeness and baryon number are conserved.
arent both of them already conserved even without the neutrino being there?

As leptons [electron positron neutrino] aren't made out of quarks? and both proton and neutron already have strangeness 0 and bayron no. 1?

is the neutrinos sole purpose to conserve lepton number?

sorry i just want to understand the main reasons for the neutrinos release :/
0
reply
honoris
Badges: 0
Rep:
?
#9
Report 9 years ago
#9
(Original post by TimetoSucceed)
isnt it already balanced?

p ----> n + e+

1----->0 + 1?

and dont neutrinos have 0 charge? =/
Yes I was incorrect I did it last year and forgot, I was thinking of positrons.
I did get 90% I promise!

They are needed to conserve lepton numbers, for example in muon decay.
0
reply
TimetoSucceed
Badges: 2
Rep:
?
#10
Report Thread starter 9 years ago
#10
(Original post by honoris)
Yes I was incorrect I did it last year and forgot, I was thinking of positrons.
I did get 90% I promise!

They are needed to conserve lepton numbers, for example in muon decay.
thank you, i have another problem

Question: reperesent B- decay using quarks

I got d----->p + B-

I then wrote + [anti-electron-neutrino] sign, so that lepton number is conserved.

But in the answers it had +Ve [electron neutrino], doesnt that cause one side to have 0 lepton number, and the other to have 2?

Or is it different when representing it with quarks? Thanks :/
0
reply
sollythewise
Badges: 1
Rep:
?
#11
Report 9 years ago
#11
(Original post by TimetoSucceed)
i understand what the decays are

beta minus, neutron turns to proton in nucleus, emits electron

beta plus, proton turns to neutron, emits positron

but why is the neutrino/anti-neutrino released

As neutron/proton has lepton number 0, and positron has lepton number -1, so a electron-neutrino is released to make it back to 0? so it is conserved?

same with beta minus, electron = 1, anti-electron neutrino = -1, so lepton number is still 0?

is that the reason? is there anything else? please help thanks.
to deal with extra mass/energy left over after decay.
0
reply
honoris
Badges: 0
Rep:
?
#12
Report 9 years ago
#12
(Original post by TimetoSucceed)
thank you, i have another problem

Question: reperesent B- decay using quarks

I got d----->p + B-

I then wrote + [anti-electron-neutrino] sign, so that lepton number is conserved.

But in the answers it had +Ve [electron neutrino], doesnt that cause one side to have 0 lepton number, and the other to have 2?

Or is it different when representing it with quarks? Thanks :/
For B- decay. A down quark changes to an up quark, I believe this would mean a neutron (udd) changes to a proton (uud). An exchange particle is needed. It would be W- as neutron to a proton, needs a -ve to caryy away charge. B- (ie. electron) released to conserve this charge. The electron has a lepton number +1. Thus an anti neutrino is released with a lepton number of -1. So everything is conserved.
0
reply
TimetoSucceed
Badges: 2
Rep:
?
#13
Report Thread starter 9 years ago
#13
(Original post by honoris)
For B- decay. A down quark changes to an up quark, I believe this would mean a neutron (udd) changes to a proton (uud). An exchange particle is needed. It would be W- as neutron to a proton, needs a -ve to caryy away charge. B- (ie. electron) released to conserve this charge. The electron has a lepton number -1. Thus a neutrino is released with a lepton number of +1. So everything is conserved.
doesnt an electron have a lepton number of +1?
0
reply
honoris
Badges: 0
Rep:
?
#14
Report 9 years ago
#14
(Original post by TimetoSucceed)
doesnt an electron have a lepton number of +1?
Okay so electron lepton number +1, therefore an antineutrino needs to be released as that will have a lepton number -1.
0
reply
TimetoSucceed
Badges: 2
Rep:
?
#15
Report Thread starter 9 years ago
#15
yeah bro thats why i asked because the answers have it as a neutrino, i figure the answers are wrong in this case. Thanks man.
0
reply
Cora Lindsay
Badges: 10
Rep:
?
#16
Report 9 years ago
#16
(Original post by TimetoSucceed)
i understand what the decays are

beta minus, neutron turns to proton in nucleus, emits electron

beta plus, proton turns to neutron, emits positron

but why is the neutrino/anti-neutrino released

As neutron/proton has lepton number 0, and positron has lepton number -1, so a electron-neutrino is released to make it back to 0? so it is conserved?

same with beta minus, electron = 1, anti-electron neutrino = -1, so lepton number is still 0?

is that the reason? is there anything else? please help thanks.
Conservation of energy too. Decay energy is fixed for any individual decay process but the beta energy is very rarely equal to the total decay energy. So the neutrino carries away the energy which isn't associated with the beta particle. This was big mystery in the early days of studying beta decay.
0
reply
Spandy
Badges: 3
Rep:
?
#17
Report 6 years ago
#17
I believe conservation of lepton number is a purely experimental law and has no actual theoretical proof, same as entropy in thermo, sort of...
Btw, why an electron antineutrino? Why not a muon or tau antineutrino?
Rgds
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Current uni students - are you thinking of dropping out of university?

Yes, I'm seriously considering dropping out (30)
18.75%
I'm not sure (3)
1.88%
No, I'm going to stick it out for now (53)
33.13%
I have already dropped out (3)
1.88%
I'm not a current university student (71)
44.38%

Watched Threads

View All