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Chapter 39, Problem 19P

(a)

To determine

The scattering angle of the proton and electron.

(a)

Expert Solution
Check Mark

Answer to Problem 19P

The scattering angle of the proton and electron is 43.0°.

Explanation of Solution

Write the expression for the conservation of momentum along horizontal direction.

  hλ=hλcosθ+pcosθ                                                                                            (I)

Here, λ is wavelength of the photon before collision, λ is the wavelength of the photon after collision, p is the momentum of electron and θ is the direction scattering for both electron and photon.

Write the expression for the conservation of momentum along the vertical direction.

  0=hλsinθpsinθ

Rewrite the above equation in terms of p.

  hλsinθ=psinθp=hλ

Substitute the above expression in (I) to rewrite.

  hλ=hλcosθ+(hλ)cosθ1λ=2λcosθcosθ=λ2λ                                                                                    (I)

Write the expression for the Compton scattering.

  λ=λ+hmc(1cosθ)

Here, m is the mass of the electron.

Substitute the above expression in (II) to rewrite.

  cosθ=λ+hmc(1cosθ)2λ2λcosθ=λ+hmchmccosθ(2λ+hmc)cosθ=λ+hmc                                                                    (III)

Write the expression for the wavelength of the photon before collision.

  λ=hcE0

Here, E0 is the energy of the photon.

Substitute the above equation in (III) to rewrite in terms of θ.

  (2(hcE0)+hmc)cosθ=hcE0+hmc1E0mc(2mc2+E0)cosθ=1E0mc(mc2+E0)cosθ=mc2+E02mc2+E0θ=cos1(mc2+E02mc2+E0)

Write the expression to calculate the rest mass energy of electron.

  E=mc2

Here, E is the rest mass energy of the photon.

Substitute the above equation in the expression for θ to rewrite.

  θ=cos1(E+E02E+E0)

The rest mass energy of the photon is 0.511MeV.

Substitute 0.511MeV for E and 0.880MeV for E0 in the above equation to calculate θ.

  θ=cos1(0.880MeV+0.511MeV2(0.511MeV)+0.880MeV)=cos1(1.3911.902)=43.0°

Conclusion:

Therefore, the scattering angle of the proton and electron is 43.0°.

(b)

To determine

The energy and momentum of the scattered photon.

(b)

Expert Solution
Check Mark

Answer to Problem 19P

The energy and momentum of the scattered photon is respectively 0.600MeV and 3.20×1022kgm/s.

Explanation of Solution

Write the expression to calculate the energy of the scattered photon.

    E=hcλ

Here, E is the energy of the scattered photon.

Substitute the expression for λ using the expression given in part (a).

    E=hcλ+hmc(1cosθ)

Rewrite the above expression using λ=hcE0 in terms of E.

    E=hchcE0+hmc(1cosθ)E=11E0+hmc2(1cosθ)

Rewrite the equation using E=mc2 to calculate E.

    E=11E0+1E(1cosθ)

Substitute 0.511MeV for E, 43.0° for θ and 0.880MeV for E0 in the above equation to calculate E.

    E=110.880MeV+10.511MeV(1cos43.0°)=1(1.14+0.526)MeV1=0.600MeV

Write the expression to calculate the momentum of the scattered electron.

    p=Ec

Here, p is the momentum of the scattered electron and c is the speed of light.

Substitute 0.600MeV for E and 3.00×108m/s for c in the above equation to calculate p.

    p=0.600MeV(106eV1MeV)(1.602×1019J1eV)3.00×108m/s=3.20×1022kgm/s

Conclusion:

Therefore, the energy and momentum of the scattered photon is respectively 0.600MeV and 3.20×1022kgm/s.

(c)

To determine

The kinetic energy and momentum of the scattered electron.

(c)

Expert Solution
Check Mark

Answer to Problem 19P

The energy and momentum of the scattered electron is respectively 0.280MeV and 3.20×1022kgm/s.

Explanation of Solution

Write the expression to calculate the kinetic energy of the scattered electron.

    K=E0E

Here, K is the kinetic energy of the scattered electron.

Substitute 0.600MeV for E and 0.880MeV for E0 in the above equation to calculate K.

    K=0.880MeV0.600MeV=0.280MeV

Refer the equation for p in part (a) to calculate the momentum of the electron.

Write the expression for the momentum of the electron.

    p=hλ

Substitute the expression for λ using the expression given in part (a) to rewrite the above equation.

    p=hλ+hmc(1cosθ)

Rewrite the above expression using λ=hcE0 in terms of p.

    p=hhcE0+hmc(1cosθ)p=1cE0+1mc(1cosθ)

Substitute 0.880MeV for E0, 3.00×108m/s for c, 43.0° for θ and 9.11×1031kg for m in the above equation to calculate p.

    p=13.00×108m/s0.880MeV(106eV1MeV)(1.602×1019J1eV)+19.11×1031kg(3.00×108m/s)(1cos43.0°)=10.0213×1023(kgm/s)1+0.00984×1023(kgm/s)1=3.21×1022kgm/s3.20×1022kgm/s

Conclusion:

Therefore, the energy and momentum of the scattered electron is respectively 0.280MeV and 3.20×1022kgm/s.

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