Application of Conservation Laws (AQA AS Physics): Revision Note

The lambda nought particle Λ⁰ is made up from the quarks uds. Show, in terms of the conservation of charge, strangeness, baryon number and lepton number whether the following interaction is permitted

${\mathrm{\Lambda }}^0\to {\mathrm{\pi }}^{-}+{\mathrm{e}}^{+}+{\nu }_e$

[4]

Answer:

Step 1: Determine conservation of charge

$0=\left(-1\right)+\left(+1\right)+0$

• So charge is conserved [1 mark]

Step 2: Determine conservation of strangeness

• Λ⁰ has an s quark, so must have a strangeness of –1

$\left(-1\right)=0+0+0$

• So strangeness is not conserved [1 mark]

Step 3: Determine conservation of baryon number

• Λ⁰ is a baryon since it has 3 quarks, so must have a baryon number of +1

$\left(+1\right)=0+0+0$

• So baryon number is not conserved

Step 4: Determine conservation of lepton number

• Λ⁰ is a baryon, so must have a lepton number of 0

$0=0+\left(-1\right)+\left(+1\right)$

• So lepton number is conserved [1 mark]

Step 5: Conclusion

• Since the baryon number is not conserved, this interaction is not permitted [1 mark]

The equation for β^– decay is

$\mathrm{n}=\mathrm{p}+{\mathrm{e}}^{-}+{\overline{\nu }}_e$

Using the quark model of beta decay, prove that the charge is conserved in this equation.

[2]

Answer:

Step 1: List the know quantities

• Charge of u quark: $+\frac{2}{3}\mathrm{e}$

• Charge of d quark: $-\frac{1}{3}\mathrm{e}$

• Charge of electron: $-1\mathrm{e}$

• Charge of electron anti-neutrino: 0

Step 2: State the equation in terms of quarks

• In β^– decay, a down quark changes to an up quark and causes the neutron to change into a proton [1 mark]

$\mathrm{udd}=\mathrm{uud}+{\mathrm{e}}^{-}+{\overline{\nu }}_e$

Step 3: Calculate charge of LHS and RHS of equation

• In β^– decay, a down quark changes to an up quark, causes the neutron to change into a proton

$\mathrm{LHS}\mathrm{charge}=+\frac{2}{3}\mathrm{e}-\frac{1}{3}\mathrm{e}-\frac{1}{3}\mathrm{e}=0$

$\mathrm{RHS}\mathrm{charge}=\left(+\frac{2}{3}\mathrm{e}+\frac{2}{3}\mathrm{e}-\frac{1}{3}\mathrm{e}\right)+\left(-1\mathrm{e}\right)+0=0$

Step 4: Conclusion

• Since the charges are equal on both sides, it must be conserved in the β^– decay equation [1 mark]

Note that quantum numbers for any exotic particles will be given in the question. All the rest can be found on the datasheet or can be deduced using the information provided.