College Physics:
College Physics:
11th Edition
ISBN: 9781305965515
Author: SERWAY, Raymond A.
Publisher: Brooks/Cole Pub Co
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Chapter 26, Problem 36P

(a)

To determine

Equation for conservation of energy.

(a)

Expert Solution
Check Mark

Answer to Problem 36P

  The equation for conservation of energy is K.Ei+mic2=K.Ef+mfc2 .

Explanation of Solution

The formula used to calculate the initial rest energy is,

ERi=mic2

  • ERi is the initial rest energy
  • c is the speed of light
  • mi is the initial mass

The formula used to calculate the final relativistic energy is,

Ei=K.Ei+ERi=K.Ei+mic2

The formula used to calculate the final rest energy is,

ERf=mfc2

  • ERf is the final rest energy
  • mf is the final mass

The formula used to calculate the final relativistic energy is,

Ef=K.Ef+ERf=K.Ef+mfc2

The formula used to calculate the conservation energy is,

Ei=EfK.Ei+mic2=K.Ef+mfc2

Thus, the equation for conservation of energy is K.Ei+mic2=K.Ef+mfc2 .

Conclusion:

The equation for conservation of energy is K.Ei+mic2=K.Ef+mfc2 .

(b)

To determine

The total mass of initial particle.

(b)

Expert Solution
Check Mark

Answer to Problem 36P

  The total mass of initial particle is 236.052588u .

Explanation of Solution

Given info:

Mass of U92235 is 235.043923 u .

Mass of n01 is 1.008665u .

The formula used to calculate the mass of initial particle is,

mi=mU92235+mn01

  • mi is the mass of initial particle
  • mU92235 is the mass of u92235
  • mn01 is the mass of n01

Substitute 235.043923 u for mU92235 and 1.008665u for mn01 to find mi .

mi=235.043923 u+1.008665u=236.052588u

Thus, the total mass of initial particle is 236.052588u .

Conclusion:

The total mass of initial particle is 236.052588u .

 (c)

To determine

The total mass of final particle.

 (c)

Expert Solution
Check Mark

Answer to Problem 36P

  The total mass of final particle is 235.861612u .

Explanation of Solution

Given info:

Mass of L57148a is 147.932236 u .

Mass of B3587r is 86.9207119 u .

The formula used to calculate the mass of initial particle is,

mf=mL57148a+mB3587r+mn01

  • mf is the mass of initial particle
  • mL57148a is the mass of L57148a
  • mB3587r is the mass of B3587r

Substitute 147.932236 u for mL57148a , 86.9207119 u for mB3587r and 1.008665u for mn01 to find mf .

mf=147.932236 u+86.9207119 u+1.008665u=235.861612u

Thus, the total mass of final particle is 235.861612u .

Conclusion:

The total mass of final particle is 235.861612u .

(d)

To determine

The difference between the masses.

(d)

Expert Solution
Check Mark

Answer to Problem 36P

  The mass difference is 0.190976u .

Explanation of Solution

The formula used to calculate the mass difference is,

Δm=mimf

  • Δm is the mass is the mass difference

Substitute 236.052588u for mi and 235.861612u for mf to find Δm .

Δm=236.052588u235.861612u=0.190976u

Thus, the mass difference is 0.190976u .

Conclusion:

The mass difference is 0.190976u .

(e)

To determine

The final kinetic energy in MeV.

(e)

Expert Solution
Check Mark

Answer to Problem 36P

  The final kinetic energy in MeV is 177.893MeV .

Explanation of Solution

Consider the initial kinetic energy as zero, the kinetic energy is equal to the mass difference.

K.Ef=Δm=(0.190976u)(931.494MeV1u)=177.893MeV

Thus, the final kinetic energy in MeV is 177.893MeV .

Conclusion:

The final kinetic energy in MeV is 177.893MeV

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Chapter 26 Solutions

College Physics:

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