Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card)
Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card)
5th Edition
ISBN: 9781305586871
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 31, Problem 60P

(a)

To determine

The minimum photon energy required to produce reaction in reference frame.

(a)

Expert Solution
Check Mark

Answer to Problem 60P

The minimum photon energy required to produce reaction in reference frame is Eγ=127MeV.

Explanation of Solution

At threshold point, photon and a proton colliding head on to produce a proton and a pion at rest.

  p+γp+π0

Write the expression for the conservation of energy of the following reaction.

  mpc21u2/c2+Eγ=mpc2+mπc2        (I)

Here, mp is the mass of the proton, mπ is the mass of the pion, Eγ is the energy of the gamma ray, c is the velocity of light, and u is the velocity of the particle.

Write the expression for conservation of momentum.

  mpu1u2/c2Eγc=0        (II)

Rewrite the above equation for Eγ.

  Eγ=mpc21u2/c2uc        (III)

The energy of the each particle is,

  mpc2=938.3MeVmπc2=135.0MeV

Conclusion:

Combining the equations (I) and (III).

  mpc21u2/c2+mpc21u2/c2uc=mpc2+mπc2

Substitute 938.3MeV for mpc2 and 135.0MeV for mπc2 in the above relation to find u/c.

  938.3MeV1u2/c2+938.3MeV1u2/c2uc=938.3MeV+135.0MeV(938.3MeV)(1+u/c)(1u/c)(1+u/c)=1073.3MeVuc=0.134

Substitute 0.134 for u/c and 938.3MeV for mpc2 in the equation (III) to find Eγ.

  Eγ=938.3MeV1(0.134)2(0.134)=127MeV

Therefore, the minimum photon energy required to produce reaction in reference frame is Eγ=127MeV.

(b)

To determine

The wavelength of the photon.

(b)

Expert Solution
Check Mark

Answer to Problem 60P

The wavelength of the photon is 1.06mm.

Explanation of Solution

Write the expression from Wien’s displacement law.

  λmaxT=2.898mmK

Here, λmax is the maximum wavelength of the photon and T is the absolute temperature.

Rewrite the above expression for λmax.

  λmax=2.898mmKT        (IV)

Conclusion:

Substitute 2.73K for T in the equation (IV) to find λmax.

  λmax=2.898mmK2.73K=1.06mm

Therefore, the wavelength of the photon is 1.06mm.

(c)

To determine

The energy of the photon.

(c)

Expert Solution
Check Mark

Answer to Problem 60P

The energy of the photon is 1.17×103eV.

Explanation of Solution

Write the expression for energy of the photon.

  E=hcλ        (V)

Here, h is the plank’s constant, c is the velocity of light, and E is the energy of the photon.

Conclusion:

Substitute 6.626×1034Js for h, 3×108m/s for c, and 1.06mm for λ in the equation (V) to find E

  E=(6.626×1034Js)(3×108m/s)(1.06mm)(103m1mm)=19.878×1026Jm(6.242×1018eV1J)(1.06×103m)=1240eVnm1.06×103m=1.17×103eV

Therefore, the energy of the photon is 1.17×103eV.

(d)

To determine

The energy of the proton in the reference frame mentioned in part (a).

(d)

Expert Solution
Check Mark

Answer to Problem 60P

The energy of the proton in the reference frame mentioned in part (a) is 5.81×1019eV.

Explanation of Solution

Write the expression from Doppler effect.

  E=1+v/c1v/cE        (VI)

Here, E is the energy of the photon moving to left, E is the energy of the photon in unprimed frame, and v is the velocity of the source.

The speed of the proton is,

  uc=u/c+v/c1+uv/c2        (VII)

Here, u is the velocity of the proton in reference frame.

The energy of the proton is,

  E=mpc21u2/c2        (VIII)

Conclusion:

Substitute 1.27×108eV for E and 1.17×103eV for E in equation (VI) to find v/c.

(1.27×108eV)=1+v/c1v/c(1.17×103eV)vc=11.71×1022

Substitute 0.134 for u/c and 11.71×1022 for v/c in equation (VII) to find u/c.

  uc=0.134+11.71×10221+0.134(11.71×1022)=11.30×1022

Substitute 938.3MeV for mpc2 and 11.30×1022 for u/c in equation (VIII) to find E.

  E=(938.3MeV)1(11.30×1022)2=6.19×1010(938.3MeV)(106eV1MeV)=5.81×1019eV

Therefore, the energy of the proton in the reference frame mentioned in part (a) is 5.81×1019eV

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Chapter 31 Solutions

Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card)

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