EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
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Chapter 41, Problem 42P

(a)

To determine

The total energy of the decay products.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given: The given reaction is Σ0Λ0+γ .

Formula used:

Write the expression for total energy of the decay products.

  E=(mΣ0)c2 …… (1)

Here, E is total energy, mΣ0 is rest mass of particle and c is the speed of light.

Calculation:

Substitute 1193MeV/c2 for mΣ0 in equation (1).

  E=(1193MeV/c2)c2=1193MeV

Conclusion:

Thus, the total energy of the decay products is 1193MeV .

(b)

To determine

The approximate momentum of the photon.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given: The given reaction is Σ0Λ0+γ .

Formula used:

Write the expression for total energy of the decay products.

  E=(mΣ0)c2

Here, E is total energy, mΣ0 is rest mass of particle and c is the speed of light.

Write the expression for the rest energy of baryon.

  EΛ0=(mΛ0)c2

Here, mΛ0 is mass of baryon and EΛ0 is rest energy of baryon.

Write the expression for the rest energy of photon.

  Eγ=EEΛ0

Substitute (mΛ0)c2 for EΛ0 in above expression.

  Eγ=E(mΛ0)c2

Here, Eγ is rest energy of photon.

Write the expression for momentum of photon.

  pγ=Eγc

Substitute E(mΛ0)c2 for Eγ in above expression,

  pγ=E(mΛ0)c2c …… (2)

Here, pγ is momentum of photon.

Calculation:

Substitute 1193MeV for E and 1116MeV/c2 for mΛ in equation (2).

  pγ=1193MeV(1116MeV/c2)c2c=77MeV/c

Conclusion:

Thus, the approximate momentum of the photon is 77MeV/c .

(c)

To determine

The kinetic energy of baryon.

(c)

Expert Solution
Check Mark

Explanation of Solution

Given: The given reaction is Σ0Λ0+γ .

Formula used:

Write the expression for total energy of the decay products.

  E=(mΣ0)c2

Here, E is total energy, mΣ0 is rest mass of particle and c is the speed of light.

Write the expression for the rest energy of baryon.

  EΛ0=(mΛ0)c2

Here, mΛ0 is mass of baryon and EΛ0 is rest energy of baryon.

Write the expression for the rest energy of photon.

  Eγ=EEΛ0

Substitute (mΛ0)c2 for EΛ0 in above expression.

  Eγ=E(mΛ0)c2

Here, Eγ is rest energy of photon.

Write the expression for momentum of photon.

  pγ=Eγc

Substitute E(mΛ0)c2 for Eγ in above expression,

  pγ=E(mΛ0)c2c

Here, pγ is momentum of photon.

The momentum of baryon and photon is equal.

Write the expression for the kinetic energy of baryon.

  KΛ0=pΛ022mΛ0

Here, pΛ0 is momentum of baryon and KΛ0 is kinetic energy of baryon.

Substitute pγ for pΛ0 in above expression.

  KΛ0=pγ22mΛ0 …… (3)

Calculation:

Substitute 77MeV/c for pγ and 1116MeV/c2 for mΛ in equation (3).

  KΛ0=(77MeV/c)22(1116MeV/c2)=2.7MeV

Conclusion:

Thus, the kinetic energy of baryon is 2.7MeV .

(d)

To determine

The estimate energy of the photon; the estimate momentum of the photon.

(d)

Expert Solution
Check Mark

Explanation of Solution

Given: The given reaction is Σ0Λ0+γ .

Formula used:

Write the expression for total energy of the decay products.

  E=(mΣ0)c2

Here, E is total energy, mΣ0 is rest mass of particle and c is the speed of light.

Write the expression for the rest energy of baryon.

  EΛ0=(mΛ0)c2

Here, mΛ0 is mass of baryon and EΛ0 is rest energy of baryon.

Write the expression for the rest energy of photon.

  Eγ=EEΛ0

Substitute (mΛ0)c2 for EΛ0 in above expression.

  Eγ=E(mΛ0)c2

Here, Eγ is rest energy of photon.

Write the expression for momentum of photon.

  pγ=Eγc

Substitute E(mΛ0)c2 for Eγ in above expression,

  pγ=E(mΛ0)c2c

Here, pγ is momentum of photon.

The momentum of baryon and photon is equal.

Write the expression for the kinetic energy of baryon.

  KΛ0=pΛ022mΛ0

Here, pΛ0 is momentum of baryon and KΛ0 is kinetic energy of baryon.

Substitute pγ for pΛ0 in above expression.

  KΛ0=pγ22mΛ0

Write the expression for estimate energy of photon.

  Eγ=E(mΛ0)c2KΛ0 …… (4)

Here, Eγ is estimate energy of photon.

Write the expression for estimate momentum of photon.

  pγ=Eγc …… (5)

Here, pγ is estimate momentum of photon.

Calculation:

Substitute 1193MeV for E , 1116MeV/c2 for mΛ and 2.7MeV for KΛ0 in equation (4).

  Eγ=1193MeV(1116MeV/c2)c22.7MeV=74MeV

Substitute 74MeV for Eγ in equation (5).

  pγ=74MeVc

Conclusion:

The estimate energy of the photon is 74MeV ; the estimate momentum of the photon is 74MeV/c .

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