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Chapter 44, Problem 46CP

(a)

To determine

Show that the threshold kinetic energy is Kmin=[m32(m1+m2)2]c22m2.

(a)

Expert Solution
Check Mark

Answer to Problem 46CP

The threshold kinetic energy is Kmin=[m32(m1+m2)2]c22m2.

Explanation of Solution

Write the equation showing the conservation of energy.

    Emin+m2c2=(m3c2)2+(p3c)2                                                                         (I)

Here, Emin is the minimum energy required for the bombarding particle to induce the reaction, m2 is the mass of the stationary particle, m3 is the mass of the product, p3 is the momentum of the product and c is the speed of light.

Write the relativistic energy equation and substitute p1 for p3 and m1 for in equation (I) because p1=p3 due to conservation of momentum.

    (p3c)2=(p1c)2=Emin2(m1c2)                                                                                    (II)

Substitute equation (II) in (I).

    Emin+m2c2=(m3c2)2+Emin2(m1c2)

Conclusion:

Take square on both sides.

    Emin2+2Eminm2c2+(m2c2)2=(m3c2)2+Emin2(m1c2)Emin=(m32m12m22)c22m2                                   (III)

Write the equation for minimum kinetic energy.

    Kmin=Eminm1c2

Substitute equation (III) in above equation to find Kmin.

    Kmin=(m32m12m22)c22m2m1c2=(m32m12m22)c22m1m2c22m2=(m32m12m222m1m2)c22m2=[m32(m1+m2)2]c22m2

Thus, the threshold kinetic energy is Kmin=[m32(m1+m2)2]c22m2.

(b)

To determine

The threshold energy for the reaction p+p=p+p+p+p¯.

(b)

Expert Solution
Check Mark

Answer to Problem 46CP

The threshold energy for the reaction p+p=p+p+p+p¯ is 5.63GeV.

Explanation of Solution

Write the equation for the threshold energy.

    Kmin=[m32(m1+m2)2]c22m2

Substitute mp+mp+mp+mp¯ for m3, mp for m1 and mp for m2 in the above equation.

    Kmin=[(mp+mp+mp+mp¯)2(mp+mp)2]c22mp

Conclusion:

Substitute 938.3MeV/c2 for mp and 938.3MeV/c2 for mp¯ to find Kmin.

    Kmin=[(938.3MeV/c2+938.3MeV/c2+938.3MeV/c2+938.3MeV/c2)2(938.3MeV/c2+938.3MeV/c2)2]c22(938.3MeV/c2)=[4(938.3MeV/c2)22(938.3MeV/c2)2]c22(938.3MeV/c2)=(5630MeV)(103GeV1MeV)=5.63GeV

Thus, the threshold energy for the reaction p+p=p+p+p+p¯ is 5.63GeV.

(c)

To determine

The threshold energy for the reaction π+p=K0+Λ0.

(c)

Expert Solution
Check Mark

Answer to Problem 46CP

The threshold energy for the reaction π+p=K0+Λ0 is 768MeV.

Explanation of Solution

Write the equation for the threshold energy.

    Kmin=[m32(m1+m2)2]c22m2

Substitute mK0+mΛ0 for m3, mπ for m1 and mp for m2 in the above equation.

    Kmin=[(mK0+mΛ0)2(mπ+mp)2]c22mp

Conclusion:

Substitute 497.7MeV/c2 for mK0, 1115.6MeV/c2 for mΛ0, 139.6MeV/c2 for mπ and 938.3MeV/c2 for mp¯ to find Kmin.

    Kmin=[(497.7MeV/c2+1115.6MeV/c2)2(139.6MeV/c2+938.3MeV/c2)2]c22(938.3MeV/c2)=768MeV

Thus, the threshold energy for the reaction π+p=K0+Λ0 is 768MeV.

(d)

To determine

The threshold energy for the reaction p+p=p+p+π0.

(d)

Expert Solution
Check Mark

Answer to Problem 46CP

The threshold energy for the reaction p+p=p+p+π0 is 280MeV.

Explanation of Solution

Write the equation for the threshold energy.

    Kmin=[m32(m1+m2)2]c22m2

Substitute mp+mp+mπ0 for m3, mp for m1 and mp for m2 in the above equation.

    Kmin=[(mp+mp+mπ0)2(mp+mp)2]c22mp

Conclusion:

Substitute 135MeV/c2 for mπ0 and 938.3MeV/c2 for mp¯ to find Kmin.

    Kmin=[(938.3MeV/c2+938.3MeV/c2+135MeV/c2)2(938.3MeV/c2+938.3MeV/c2)2]c22(938.3MeV/c2)=[(2(938.3MeV/c2)+135MeV/c2)22(938.3MeV/c2)2]c22(938.3MeV/c2)=280MeV

Thus, the threshold energy for the reaction p+p=p+p+π0 is 280MeV.

(e)

To determine

The threshold energy for the reaction p+p¯=Z0.

(e)

Expert Solution
Check Mark

Answer to Problem 46CP

The threshold energy for the reaction p+p¯=Z0 is 4.43TeV.

Explanation of Solution

Write the equation for the threshold energy.

    Kmin=[m32(m1+m2)2]c22m2

Substitute mZ0 for m3, mp for m1 and mp¯ for m2 in the above equation.

    Kmin=[(mZ0)2(mp+mp¯)2]c22mp¯

Conclusion:

Substitute 91.2×103MeV/c2 for mZ0 and 938.3MeV/c2 for mp¯ and mp¯ to find Kmin.

    Kmin=[(91.2×103MeV/c2)2(938.3MeV/c2+938.3MeV/c2)2]c22(938.3MeV/c2)=[(91.2×103MeV/c2)22(938.3MeV/c2)2]c22(938.3MeV/c2)=(4.43×106MeV)(106TeV1MeV)=4.43TeV

Thus, the threshold energy for the reaction p+p¯=Z0 is 4.43TeV.

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