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Q6Q

Expert-verifiedFound in: Page 1362

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Does the proposed decay ${{\mathit{\Lambda}}}^{{\mathbf{0}}}{\mathbf{\to}}{\mathit{p}}{\mathbf{+}}{{\mathit{K}}}^{{\mathbf{-}}}$ conserve (a) electric ****charge, (b) spin angular momentum, and (c) strangeness? (d) If the ****original particle is stationary, is there enough energy to create the ****decay products?**

(a) The electric charge is conserved.

(b) The spin angular momentum is conserved.

(c) The Strangeness is conserved.

(d) There is not enough energy to create the decay products.

The proposed decay is given as

${\Lambda}^{0}\to p+{K}^{-}$

Where, ${\Lambda}^{0}$ is lambda-zero

*p* is proton

${K}^{-}$ is Kaon

From the property of particles, we can determine the total charge, angular momentum, and Strangeness of decay

Properties of Lambda, proton, and Kaon

Particle, symbol | Charge | spin | Strangeness |

Lambda, ${\Lambda}^{0}$ | 0 | 1/2 | -1 |

Proton, | +1 | 1/2 | 0 |

Kaon, ${K}^{-}$ | -1 | 0 | -1 |

The proposed decay is given as

${\Lambda}^{0}\to p+{K}^{-}$

The charge on the ${\Lambda}^{0}$ is zero so the left side of the equation has a total charge of zero.

*p * has +1 charge and ${K}^{-}$ has -1 charge, so the total change on the left side of the equation is zero.

Since both, sides of the equation have the same charge, so the charge is conserved.

The proposed decay is given as

${\Lambda}^{0}\to p+{K}^{-}$

The ${\Lambda}^{0}$ has ** 1/2** spin, so the left side of the equation has 1/2** **spin angular momentum

*p* has 1/2** **spin and ${K}^{-}$ has 0 spin, so the total spin angular momentum on the right side of the equation is 1/2** .**

Since both, sides of the equation have the same spin angular momentum, so the spin angular momentum is conserved.

The proposed decay is given as

${\Lambda}^{0}\to p+{K}^{-}$

The ${\Lambda}^{0}$ has a Strangeness number -1 , so the left side of the equation has -1 strangeness.

*p* has strangeness number 0 and ${K}^{-}$ has strangeness number -1 , so the total Strangeness on the right side of the equation is -1

Since both, sides of the equation have the same Strangeness, Therefore Strangeness is conserved.

The energy require for decay can be given as

${Q}_{decay}=$ Initial total mass-energy – final total mass-energy

$\begin{array}{rcl}Q& =& {m}_{\Lambda}{c}^{2}-\left({m}_{p}{c}^{2}+{m}_{{K}^{-1}}{c}^{2}\right)\\ Q& =& \left({m}_{\Lambda}-\left({m}_{p}+{m}_{{K}^{-1}}\right)\right){c}^{2}\end{array}$

The rest mass of lambda is,

${m}_{A}=1115.6\mathrm{MeV}/{\mathrm{c}}^{2}$

The rest mass of a proton

${m}_{p}=938.3\mathrm{MeV}/{\mathrm{c}}^{2}$

The rest mass of Kaon

${m}_{p}=493.7\mathrm{MeV}/{\mathrm{c}}^{2}$

Since the rest mass of lambda particles is less than the total rest mass of the

Proton and Kaon.

Therefore, its rest energy is less than the energy of the proton, and Keon

The initial energy is less than the final energy of the particles therefore there is not enough energy to create the decay products.

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