# 11.2 Particle conservation laws  (Page 3/6)

 Page 3 / 6

Check Your Understanding What is the lepton number of an electron-positron pair?

0

## Strangeness conservation

In the late 1940s and early 1950s, cosmic-ray experiments revealed the existence of particles that had never been observed on Earth. These particles were produced in collisions of pions with protons or neutrons in the atmosphere. Their production and decay were unusual. They were produced in the strong nuclear interactions of pions and nucleons, and were therefore inferred to be hadrons; however, their decay was mediated by the much more slowly acting weak nuclear interaction. Their lifetimes were on the order of ${10}^{-10}$ to ${10}^{-8}\phantom{\rule{0.2em}{0ex}}\text{s},$ whereas a typical lifetime for a particle that decays via the strong nuclear reaction is ${10}^{-23}\phantom{\rule{0.2em}{0ex}}\text{s}.$ These particles were also unusual because they were always produced in pairs in the pion-nucleon collisions. For these reasons, these newly discovered particles were described as strange . The production and subsequent decay of a pair of strange particles is illustrated in [link] and follows the reaction

${\pi }^{\text{−}}+\text{p}\to {\text{Λ}}^{0}+{\text{K}}^{0}.$

The lambda particle then decays through the weak nuclear interaction according to

${\text{Λ}}^{0}\to {\pi }^{\text{−}}+\text{p},$

and the kaon decays via the weak interaction

${\text{K}}^{0}\to {\pi }^{+}+{\pi }^{\text{−}}.$

To rationalize the behavior of these strange particles, particle physicists invented a particle property conserved in strong interactions but not in weak interactions. This property is called strangeness    and, as the name suggests, is associated with the presence of a strange quark. The strangeness of a particle is equal to the number of strange quarks of the particle. Strangeness conservation requires the total strangeness of a reaction or decay (summing the strangeness of all the particles) is the same before and after the interaction. Strangeness conservation is not absolute: It is conserved in strong interactions and electromagnetic interactions but not in weak interactions. The strangeness number for several common particles is given in [link] .

## Strangeness conservation

(a) Based on the conservation of strangeness, can the following reaction occur?

${\pi }^{\text{−}}+\text{p}\to {\text{K}}^{+}+{\text{K}}^{\text{−}}+\text{n}.$

(b) The following decay is mediated by the weak nuclear force:

${\text{K}}^{+}\to {\pi }^{+}+{\pi }^{0}.$

Does the decay conserve strangeness? If not, can the decay occur?

## Strategy

Determine the strangeness of the reactants and products and require that this value does not change in the reaction.

## Solution

1. The net strangeness of the reactants is $0+0=0,$ and the net strangeness of the products is $1+\left(-1\right)+0=0.$ Thus, the strong nuclear interaction between a pion and a proton is not forbidden by the law of conservation of strangeness. Notice that baryon number is also conserved in the reaction.
2. The net strangeness before and after this decay is 1 and 0, so the decay does not conserve strangeness. However, the decay may still be possible, because the law of conservation of strangeness does not apply to weak decays.

## Significance

Strangeness is conserved in the first reaction, but not in the second. Strangeness conservation constrains what reactions can and cannot occur in nature.

Check Your Understanding What is the strangeness number of a muon?

0

## Summary

• Elementary particle interactions are governed by particle conservation laws, which can be used to determine what particle reactions and decays are possible (or forbidden).
• The baryon number conservation law and the three lepton number conversation law are valid for all physical processes. However, conservation of strangeness is valid only for strong nuclear interactions and electromagnetic interactions.

## Conceptual questions

What are six particle conservation laws? Briefly describe them.

Conservation energy, momentum, and charge (familiar to classical and relativistic mechanics). Also, conservation of baryon number, lepton number, and strangeness—numbers that do not change before and after a collision or decay.

In general, how do we determine if a particle reaction or decay occurs?

Why might the detection of particle interaction that violates an established particle conservation law be considered a good thing for a scientist?

It means that the theory that requires the conservation law is not understood. The failure of a long-established theory often leads to a deeper understanding of nature.

## Problems

Which of the following decays cannot occur because the law of conservation of lepton number is violated?

$\begin{array}{cccc}\left(\text{a}\right)\phantom{\rule{0.2em}{0ex}}\text{n}\to \text{p}+{\text{e}}^{\text{−}}\hfill & & & \left(\text{e}\right)\phantom{\rule{0.2em}{0ex}}{\pi }^{\text{−}}\to {\text{e}}^{\text{−}}+{\stackrel{\text{−}}{\upsilon }}_{\text{e}}\hfill \\ \left(\text{b}\right)\phantom{\rule{0.2em}{0ex}}{\mu }^{+}\to {\text{e}}^{+}+{\upsilon }_{\text{e}}\hfill & & & \left(\text{f}\right)\phantom{\rule{0.2em}{0ex}}{\mu }^{\text{−}}\to {\text{e}}^{\text{−}}+{\stackrel{\text{−}}{\upsilon }}_{\text{e}}+{\upsilon }_{\mu }\hfill \\ \left(\text{c}\right)\phantom{\rule{0.2em}{0ex}}{\pi }^{+}\to {\text{e}}^{+}+{\upsilon }_{\text{e}}+{\stackrel{\text{−}}{\upsilon }}_{\mu }\hfill & & & \left(\text{g}\right)\phantom{\rule{0.2em}{0ex}}{\text{Λ}}^{0}\to {\pi }^{\text{−}}+\text{p}\hfill \\ \left(\text{d}\right)\phantom{\rule{0.2em}{0ex}}\text{p}\to \text{n}+{\text{e}}^{+}+{\upsilon }_{\text{e}}\hfill & & & \left(\text{h}\right)\phantom{\rule{0.2em}{0ex}}{\text{K}}^{+}\to {\mu }^{+}+{\upsilon }_{\mu }\hfill \end{array}$

a, b, and c

Which of the following reactions cannot because the law of conservation of strangeness is violated?

$\begin{array}{ccc}\left(\text{a}\right)\phantom{\rule{0.2em}{0ex}}\text{p}+\text{n}\to \text{p}+\text{p}+{\pi }^{\text{−}}\hfill & & \left(\text{e}\right)\phantom{\rule{0.2em}{0ex}}{\text{K}}^{\text{−}}+\text{p}\to {\text{Ξ}}^{0}+{\text{K}}^{+}+{\pi }^{\text{−}}\hfill \\ \left(\text{b}\right)\phantom{\rule{0.2em}{0ex}}\text{p}+\text{n}\to \text{p}+\text{p}+{\text{K}}^{\text{−}}\hfill & & \left(\text{f}\right)\phantom{\rule{0.2em}{0ex}}{\text{K}}^{\text{−}}+\text{p}\to {\text{Ξ}}^{0}+{\pi }^{\text{−}}+{\pi }^{\text{−}}\hfill \\ \left(\text{c}\right)\phantom{\rule{0.2em}{0ex}}{\text{K}}^{\text{−}}+\text{p}\to {\text{K}}^{\text{−}}+{\sum }^{+}\hfill & & \left(\text{g}\right)\phantom{\rule{0.2em}{0ex}}{\pi }^{+}+\text{p}\to {\Sigma }^{+}+{\text{K}}^{+}\hfill \\ \left(\text{d}\right)\phantom{\rule{0.2em}{0ex}}{\pi }^{\text{−}}+\text{p}\to {\text{K}}^{+}+{\sum }^{\text{−}}\hfill & & \left(\text{h}\right)\phantom{\rule{0.2em}{0ex}}{\pi }^{\text{−}}+\text{n}\to {\text{K}}^{\text{−}}+{\text{Λ}}^{0}\hfill \end{array}$

Identify one possible decay for each of the following antiparticles:

(a) $\stackrel{\text{−}}{n}$ , (b) $\stackrel{\text{—}}{{\text{Λ}}^{0}}$ , (c) ${\text{Ω}}^{+}$ , (d) ${\text{K}}^{\text{−}}$ , and (e) $\stackrel{\text{−}}{\Sigma }$ .

a. ${\stackrel{\text{−}}{p}}_{e}{}^{+}ve$ ; b. $\stackrel{\text{−}}{p}{\pi }^{+}$ or $\stackrel{\text{−}}{p}{\pi }^{0}$ ; c. $\stackrel{\text{—}}{{\text{Ξ}}^{\text{0}}}{\pi }^{0}$ or $\stackrel{\text{—}}{{\Lambda }^{\text{0}}}{\text{K}}^{+}$ ; d. ${\mu }^{-}{\stackrel{\text{−}}{v}}_{\mu }$ or ${\pi }^{\text{−}}{\pi }^{0}$ ; e. $\stackrel{\text{−}}{p}{\pi }^{0}$ or $\stackrel{\text{−}}{n}{\pi }^{\text{−}}$

Each of the following strong nuclear reactions is forbidden. Identify a conservation law that is violated for each one.

$\begin{array}{}\\ \text{(a)}\phantom{\rule{0.2em}{0ex}}\text{p}+\stackrel{\text{−}}{\text{p}}\to \text{p}+\text{n}+\stackrel{\text{−}}{\text{p}}\hfill \\ \text{(b)}\phantom{\rule{0.2em}{0ex}}\text{p}+\text{n}\to \text{p}+\stackrel{\text{−}}{\text{p}}+\text{n}+{\pi }^{+}\hfill \\ \text{(c)}\phantom{\rule{0.2em}{0ex}}{\pi }^{\text{−}}+\text{p}\to {\Sigma }^{+}+{\text{K}}^{\text{−}}\hfill \\ \text{(d)}\phantom{\rule{0.2em}{0ex}}{\text{K}}^{\text{−}}+\text{p}\to {\text{Λ}}^{0}+\text{n}\hfill \end{array}$

how does colour appear in thin films
in the wave equation y=Asin(kx-wt+¢) what does k and w stand for.
derivation of lateral shieft
Hi
Hi
hi
ALFRED
how are you?
hi
asif
hi
Imran
I'm fine
ALFRED
total binding energy of ionic crystal at equilibrium is
How does, ray of light coming form focus, behaves in concave mirror after refraction?
Sushant
What is motion
Anything which changes itself with respect to time or surrounding
Sushant
good
Chemist
and what's time? is time everywhere same
Chemist
No
Sushant
how can u say that
Chemist
do u know about black hole
Chemist
Not so more
Sushant
DHEERAJ
Sushant
But ask anything changes itself with respect to time or surrounding A Not any harmful radiation
DHEERAJ
explain cavendish experiment to determine the value of gravitational concept.
Cavendish Experiment to Measure Gravitational Constant. ... This experiment used a torsion balance device to attract lead balls together, measuring the torque on a wire and equating it to the gravitational force between the balls. Then by a complex derivation, the value of G was determined.
Triio
For the question about the scuba instructor's head above the pool, how did you arrive at this answer? What is the process?
as a free falling object increases speed what is happening to the acceleration
of course g is constant
Alwielland
acceleration also inc
Usman
which paper will be subjective and which one objective
jay
normal distributiin of errors report
Dennis
normal distribution of errors
Dennis
acceleration also increases
Jay
there are two correct answers depending on whether air resistance is considered. none of those answers have acceleration increasing.
Michael
Acceleration is the change in velocity over time, hence it's the derivative of the velocity with respect to time. So this case would depend on the velocity. More specifically the change in velocity in the system.
Big
photo electrons doesn't emmit when electrons are free to move on surface of metal why?
What would be the minimum work function of a metal have to be for visible light(400-700)nm to ejected photoelectrons?
give any fix value to wave length
Rafi
40 cm into change mm
40cm=40.0×10^-2m =400.0×10^-3m =400mm. that cap(^) I have used above is to the power.
Prema
i.e. 10to the power -2 in the first line and 10 to the power -3 in the the second line.
Prema
there is mistake in my first msg correction is 40cm=40.0×10^-2m =400.0×10^-3m =400mm. sorry for the mistake friends.
Prema
40cm=40.0×10^-2m =400.0×10^-3m =400mm.
Prema
this msg is out of mistake. sorry friends​.
Prema
what is physics?
why we have physics
because is the study of mater and natural world
John
because physics is nature. it explains the laws of nature. some laws already discovered. some laws yet to be discovered.
Yoblaze
physics is the study of non living things if we added it with biology it becomes biophysics and bio is the study of living things tell me please what is this?
tahreem
physics is the study of matter,energy and their interactions
Buvanes
all living things are matter
Buvanes
why rolling friction is less than sliding friction
tahreem
thanks buvanas
tahreem
is this a physics forum