Physics for Scientists and Engineers with Modern Physics
Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN: 9781337553292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 43, Problem 58AP

(a)

To determine

Prove that K1Ktot=m2m1+m2 and K2Ktot=m1m1+m2.

(a)

Expert Solution
Check Mark

Answer to Problem 58AP

It is proved that K1Ktot=m2m1+m2 and K2Ktot=m1m1+m2.

Explanation of Solution

Two fragments must move in opposite direction for the momentum conservation. Write the equation for conservation of momentum.

  m1v1 =m2v2  

Here, m1,m2 are the masses of fragments and v1,v2 are velocities of fragments.

Rewrite the above relation in terms of v2.

    v2=(m1m2)v1

Write the equation for kinetic energy of mass m1.

  K1=12m1v12

Here, K1 is the kinetic energy of mass m1.

Write the equation for kinetic energy of mass m2.

  K2=12m2v22

Here, K2 is the kinetic energy of mass m2.

Rewrite the above expression by substituting (m1m2)v1 for v2.

  K2=12m2((m1m2)v1)2=(m1m2)12m1v12

Rewrite the above equation by substituting K1 for 12m1v12.

  K2=(m1m2)K1

Write the equation for total kinetic energy.

    Ktotal=K1+K2

Here, Ktotal is the total kinetic energy.

Conclusion:

Write the equation for the fraction of total kinetic possessed by mass m1.

  K1Ktotal=K1K1+K2

Rewrite the above expression by substituting (m1m2)K1 for K2.

  K1Ktotal=K1K1+(m1m2)K1=m2m1+m2

Write the equation for the fraction of total kinetic possessed by mass m2.

  K2Ktotal=1K1K1+K2

Rewrite the above expression by substituting m2m1+m2 for K1Ktotal.

  K2Ktotal=1m2m1+m2=m1m1+m2

Therefore, the It is proved that K1Ktot=m2m1+m2 and K2Ktot=m1m1+m2.

(b)

To determine

The disintegration energy for nuclear fission.

(b)

Expert Solution
Check Mark

Answer to Problem 58AP

The disintegration energy for nuclear fission is 177.4 MeV.

Explanation of Solution

Write the equation for disintegration energy.

    Q=(mUmBrmLa)c2

Here, Q is the disintegration energy, mU is the mass of 92236U, mBr is the mass of 3587Br, mLa is the mass of 57149La, and c is the speed of light in vacuum.

Conclusion:

Substitute 236.045562u for mU, 86.920711u for mBr, 148.934370u for mLa, and 931.5 MeV/u for c2 in the above equation to find Q.

  Q=(236.045562u86.920711u148.934370u)×(931.5 MeV/u)=177.4MeV

Therefore, the disintegration energy for nuclear fission is 177.4 MeV.

(c)

To determine

The splitting of disintegration energy between the two primary fragments.

(c)

Expert Solution
Check Mark

Answer to Problem 58AP

The splitting of disintegration energy for Br is 112.0 MeV and for La is 65.4 MeV.

Explanation of Solution

Write the equation to find the disintegration energy for Br.

  KBr=QmLamBr+mLa

Here, KBr is the disintegration energy for Br.

Write the equation to find the disintegration energy for La.

  KLa=QKBr

Conclusion:

Substitute 177.4MeV for Q, 149u for mLa, and 87u for mBr in the equation for KBr.

    KBr=(177.4MeV)149u87u+149u=112.0MeV

Substitute 177.4MeV for Q and 112.0MeV for KBr in the equation for KLa.

  KLa=177.4MeV112.0MeV=65.4MeV

Therefore, the splitting of disintegration energy for Br is 112.0 MeV and for La is 65.4 MeV.

(d)

To determine

The speed of fragments just after the fission.

(d)

Expert Solution
Check Mark

Answer to Problem 58AP

The speed of Br nuclie and La nuclei are 1.58×107m/s and 9.20×106m/s respectively.

Explanation of Solution

Write the equation to find the speed of Br nuclie.

  vBr=2KBrmBr

Here, vBr is the speed of Br nuclie.

Write the equation to find the speed of La nuclie.

  vBr=2KBrmBr

Here, vBr is the speed of Br nuclie.

  vLa=2KLamLa

Here, vLa is the speed of Br nuclie.

Conclusion:

Substitute 112.0 MeV for KBr and 87u for mBr in the equation to find vBr.

    vBr=2(112.0 MeV(106eV1MeV)(1J1.60×1019eV))87u(1kg1.66×1027u)=2.50×1014m2/s2=1.58×107m/s

Substitute 65.4MeV for KLa and 149u for mLa in the equation to find vBr.

    vBr=2(112.0 MeV(106eV1MeV)(1J1.60×1019eV))87u(1kg1.66×1027u)=84.64×1012m2/s2=9.20×106m/s

Therefore, the speed of Br nuclie and La nuclei are 1.58×107m/s and 9.20×106m/s respectively.

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