Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration

Q6Q

Expert-verified
Fundamentals Of Physics
Found in: Page 1362

Answers without the blur.

Just sign up for free and you're in.

Illustration

Short Answer

Does the proposed decay Λ0p+K- 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.

See the step by step solution

Step by Step Solution

Step 1: Identifying the data given in the question

The proposed decay is given as

Λ0p+K-

Where, Λ0 is lambda-zero

p is proton

K- is Kaon

Step 2: The concept used to solve the problem

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, Λ0

0

1/2

-1

Proton, p

+1

1/2

0

Kaon, K-

-1

0

-1

Step 3: (a) Finding whether the electric charge is conserved or not

The proposed decay is given as

Λ0p+K-

The charge on the Λ0 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.

Step 4: (b) Finding whether the spin angular momentum is conserved or not

The proposed decay is given as

Λ0p+K-

The Λ0 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.

Step 5: (c) Finding whether the Strangeness is conserved or not

The proposed decay is given as

Λ0p+K-

The Λ0 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.

Step 6: (d) Finding is there enough energy or not

The energy require for decay can be given as

Qdecay= Initial total mass-energy – final total mass-energy

Q=mΛc2-mpc2+mK-1c2Q=mΛ-mp+mK-1c2

The rest mass of lambda is,

mA=1115.6 MeV/c2

The rest mass of a proton

mp=938.3 MeV/c2

The rest mass of Kaon

mp=493.7 MeV/c2

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.

Recommended explanations on Physics Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.