EBK COLLEGE PHYSICS, VOLUME 2
EBK COLLEGE PHYSICS, VOLUME 2
11th Edition
ISBN: 9781337514644
Author: Vuille
Publisher: CENGAGE LEARNING - CONSIGNMENT
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Chapter 27, Problem 27P

(a)

To determine

The momentum of the electron.

(a)

Expert Solution
Check Mark

Answer to Problem 27P

The momentum of the electron is 1.18×1023kg.m/s .

Explanation of Solution

EBK COLLEGE PHYSICS, VOLUME 2, Chapter 27, Problem 27P

The figure shows the situation before and after the scattering process.

The Compton shift formula is,

λ'λ=hmec(1cos180°)=2hmec

    • λ0 is the wavelength of the x-ray
    • h is the Plank’s constant
    • c is the speed of the light
    • me is the mass of the electron

Substitute 0.110nm for λ , 6.63×1034J.s for h , 9.11×1031kg for me and 3.00×108m/s for c .

λ'=(0.110nm)(109m1nm)+2(6.63×1034J.s)(9.11×1031kg)(3.00×108m/s)=0.115nm

The momentum of the recoiling electron is,

pe=p+p'

    • p is the momentum of the incident photon
    • p' is the momentum of the scattered photon

pe=hλ+hλ'

Substitute 0.110nm for λ , 0.115nm for λ' , 6.63×1034J.s for h and 3.00×108m/s for c .

pe=(6.63×1034J.s)(1(0.110nm)(109m1nm)+1(0.115nm)(109m1nm))=1.18×1023kg.m/s

Conclusion:

Thus, the momentum of the electron is 1.18×1023kg.m/s .

(b)

To determine

The kinetic energy of the electron.

(b)

Expert Solution
Check Mark

Answer to Problem 27P

The kinetic energy of the electron is 478eV .

Explanation of Solution

The kinetic energy of the electron is,

KE=pe22me

Substitute 1.18×1023kg.m/s for pe and 9.11×1031kg for me .

KE=(1.18×1023kg.m/s)22(9.11×1031kg)(1eV1.60×1019J)=478eV

Conclusion:

Thus, the kinetic energy of the electron is 478eV

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