particle physics a-level help!

Watch this thread
gracehumph
Badges: 2
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#1
Report Thread starter 3 years ago
#1
the question is:

a proton and a pi- go to a K- and a pi+

state which laws are conserved, and which law isn't to show that this reaction is not observed.

I understand that strangeness is not conserved, and the charge is conserved, but the mark scheme says that baryon number is conserved but I don't understand why?
can someone please explain, thanks.

I know I am probably missing something really easy.
0
reply
BenCushing
Badges: 3
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#2
Report 3 years ago
#2
There are two types of baryons. hadrons and mesons. All hadrons that aren't antiparticles will have a baryon number of one as they are made up of 3 quarks. mesons are made up of a quark and anti-quark and so have no baryon number.
2
reply
GalaxysEdge
Badges: 11
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#3
Report 3 years ago
#3
(Original post by BenCushing)
There are two types of baryons. hadrons and mesons. All hadrons that aren't antiparticles will have a baryon number of one as they are made up of 3 quarks. mesons are made up of a quark and anti-quark and so have no baryon number.
This is incorrect. There are two types of Hadrons, which are Baryons and Mesons. Baryons have 3 quarks and Mesons have 2 quarks.
1
reply
GalaxysEdge
Badges: 11
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#4
Report 3 years ago
#4
(Original post by gracehumph)
the question is:

a proton and a pi- go to a K- and a pi+

state which laws are conserved, and which law isn't to show that this reaction is not observed.

I understand that strangeness is not conserved, and the charge is conserved, but the mark scheme says that baryon number is conserved but I don't understand why?
can someone please explain, thanks.

I know I am probably missing something really easy.
Hi
Are you sure that is the correct particle interaction given in the question, as it doesn't seem to be possible. Baryon number is conserved in all interactions. Only the proton in that equation has a baryon number of 1, therefore the left hand side does not match up with the right hand side.
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

Were exams easier or harder than you expected?

Easier (13)
25.49%
As I expected (14)
27.45%
Harder (21)
41.18%
Something else (tell us in the thread) (3)
5.88%

Watched Threads

View All