Modern Physics, 3rd Edition
Modern Physics, 3rd Edition
3rd Edition
ISBN: 9780534493394
Author: Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Publisher: Cengage Learning
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Question
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Chapter 3, Problem 26P

(a)

To determine

The wavelengths of photons scattered at angles 30.0°, 60.0°, 90.0° , 120.0°, 150.0°, 180.0° and 210°.

(a)

Expert Solution
Check Mark

Answer to Problem 26P

The wavelengths of photons scattered at angles 30.0°, 60.0°, 90.0° , 120.0°, 150.0°, 180.0° and 210° are 4.03×1011m, 4.12×1011m, 4.24×1011m, 4.36×1011m, 4.45×1011m 4.48×1011m and 4.45×1011m respectively.

Explanation of Solution

Write the expression to calculate the change in wavelength.

  Δλ=hmec(1cosθ)        (I)

Here, h is the Plank’s constant, me is the mass of electron, c is the speed of light, Δλ is the change in wavelength, θ is the scattering angle.

Write the expression to calculate the wavelengths of scattered photons.

  λ=λ+Δλ        (II)

Here, λ is the wavelengths of scattered photons, λ is the wavelength of X-rays before scattering.

Substitute equation (I) in (II) to calculate the wavelength of scattered photons.

  λ=λ+hmec(1cosθ)        (III)

Conclusion:

Substitute 0.040nm for λ, 6.626×1034Js for h, 9.11×1031kg for me, 2.998×108m/s for c and 30.0° for θ to find the wavelength of scattered photon at 30.0° in expression (III).

  λ=0.040nm(109m1.0nm)+6.626×1034Js(9.11×1031kg)(2.998×108m/s)(1cos(30.0°))=0.040×109m+0.3×1012m=4.03×1011m

Substitute 0.040nm for λ, 6.626×1034Js for h, 9.11×1031kg for me, 2.998×108m/s for c and 60.0° for θ to find the wavelength of scattered photon at 60.0° in expression (III).

  λ=0.040nm(109m1.0nm)+6.626×1034Js(9.11×1031kg)(2.998×108m/s)(1cos(60.0°))=0.040×109m+1.2×1012m=4.12×1011m

Substitute 0.040nm for λ, 6.626×1034Js for h, 9.11×1031kg for me, 2.998×108m/s for c and 90.0° for θ to find the wavelength of scattered photon at 90.0° in expression (III).

  λ=0.040nm(109m1.0nm)+6.626×1034Js(9.11×1031kg)(2.998×108m/s)(1cos(90.0°))=0.040×109m+2.4×1012m=4.24×1011m

Substitute 0.040nm for λ, 6.626×1034Js for h, 9.11×1031kg for me, 2.998×108m/s for c and 120.0° for θ to find the wavelength of scattered photon at 120.0° in expression (III).

  λ=0.040nm(109m1.0nm)+6.626×1034Js(9.11×1031kg)(2.998×108m/s)(1cos(120.0°))=0.040×109m+3.6×1012m=4.36×1011m

Substitute 0.040nm for λ, 6.626×1034Js for h, 9.11×1031kg for me, 2.998×108m/s for c and 150.0° for θ to find the wavelength of scattered photon at 150.0° in expression (III).

  λ=0.040nm(109m1.0nm)+6.626×1034Js(9.11×1031kg)(2.998×108m/s)(1cos(150.0°))=0.040×109m+4.5×1012m=4.45×1011m

Substitute 0.040nm for λ, 6.626×1034Js for h, 9.11×1031kg for me, 2.998×108m/s for c and 180.0° for θ to find the wavelength of scattered photon at 180.0° in expression (III).

  λ=0.040nm(109m1.0nm)+6.626×1034Js(9.11×1031kg)(2.998×108m/s)(1cos(180.0°))=0.040×109m+4.8×1012m=4.48×1011m

Substitute 0.040nm for λ, 6.626×1034Js for h, 9.11×1031kg for me, 2.998×108m/s for c

and 210.0° for θ to find the wavelength of scattered photon at 180.0° in expression (III).

    λ=0.040nm(109m1.0nm)+6.626×1034Js(9.11×1031kg)(2.998×108m/s)(1cos(210.0°))=0.040×109m+4.5×1012m=4.45×1011m

Therefore the wavelengths of photons scattered at angles 30.0°, 60.0°, 90.0° , 120.0°, 150.0°, 180.0° and 210° are 4.03×1011m, 4.12×1011m, 4.24×1011m, 4.36×1011m, 4.45×1011m 4.48×1011m and 4.45×1011m respectively.

(b)

To determine

The energy of electrons at scattering angles 30.0°, 60.0°, 90.0° , 120.0°, 150.0°, 180.0° and 210°.

(b)

Expert Solution
Check Mark

Answer to Problem 26P

The energy of electrons at scattering angles 30.0°, 60.0°, 90.0° , 120.0°, 150.0°, 180.0° and 210° are 231eV, 905eV, 1760eV, 2570eV, 3140eV, 3330eV and 3140eV respectively.

Explanation of Solution

The energy of the electrons will be the energy lost from the photons.

Write the expression to find the energy of electrons.

  Ke=hcλ0hcλ=hc(1λ1λ)        (I)

Here, Ke is the energy of electrons, λ is the wavelength before scattering, λ is the wavelength of scattered photon.

Conclusion:

Substitute 0.040×109m for λ, 6.626×1034Js for h, 2.998×108m/s for c and 0.0403×109m for λ to find the energy of scattered electron when scattering angle is 30.0° in expression (I).

  Ke=(6.626×1034Js)(2.998×108m/s)(10.040×109m10.0403×109m)(1.0J1.602×1019eV)=231eV

Substitute 0.040×109m for λ, 6.626×1034Js for h, 2.998×108m/s for c and 0.0412×109m for λ to find the energy of scattered electron when scattering angle at 60.0° in expression (I).

  Ke=(6.626×1034Js)(2.998×108m/s)(10.040×109m10.0412×109m)(1.0J1.602×1019eV)=905eV

Substitute 0.040×109m for λ, 6.626×1034Js for h, 2.998×108m/s for c and 0.0424×109m for λ to find the energy of scattered electron when scattering angle at 90.0° in expression (I).

  Ke=(6.626×1034Js)(2.998×108m/s)(10.040×109m10.0424×109m)(1.0J1.602×1019eV)=1760eV

Substitute 0.040×109m for λ, 6.626×1034Js for h, 2.998×108m/s for c and 0.0436×109m for λ to find the energy of scattered electron when scattering angle at 120.0° in expression (I).

  Ke=(6.626×1034Js)(2.998×108m/s)(10.040×109m10.0436×109m)(1.0J1.602×1019eV)=2570eV

Substitute 0.040×109m for λ, 6.626×1034Js for h, 2.998×108m/s for c and 0.0445×109m for λ to find the energy of scattered electron when scattering angle at 150.0° in expression (I).

  Ke=(6.626×1034Js)(2.998×108m/s)(10.040×109m10.0445×109m)(1.0J1.602×1019eV)=3140eV

Substitute 0.040×109m for λ, 6.626×1034Js for h, 2.998×108m/s for c and 0.0448×109m for λ to find the energy of scattered electron when scattering angle at 180.0° in expression (I).

    Ke=(6.626×1034Js)(2.998×108m/s)(10.040×109m10.0448×109m)(1.0J1.602×1019eV)=3330eV

Substitute 0.040×109m for λ, 6.626×1034Js for h, 2.998×108m/s for c and 0.0445×109m for λ to find the energy of scattered electron when scattering angle at 210.0° in expression (I).

  Ke=(6.626×1034Js)(2.998×108m/s)(10.040×109m10.0445×109m)(1.0J1.602×1019eV)=3140eV

Therefore, the energy of electrons at scattering angles 30.0°, 60.0°, 90.0° , 120.0°, 150.0°, 180.0° and 210° are 231eV, 905eV, 1760eV, 2570eV, 3140eV, 3330eV and 3140eV respectively.

(c)

To determine

The scattering angle with which the electron gets the greatest energy.

(c)

Expert Solution
Check Mark

Answer to Problem 26P

The scattering angle with which the electron gets the greatest energy is 180°.

Explanation of Solution

When the scattering angle is 180.0°, the photon strikes on the electron which is at rest with a head on collision. In the head on collision, the photon imparts its momentum to the electron.

After the head on collision the photon scattered straight back and the kinetic energy gained by the initially stationary electron will be the maximum. The kinetic energy gained by the electron at scattering angle 180° is 3330eV.

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