COLLEGE PHYSICS,VOLUME 1
COLLEGE PHYSICS,VOLUME 1
2nd Edition
ISBN: 9781319115104
Author: Freedman
Publisher: MAC HIGHER
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Chapter 26, Problem 55QAP
To determine

(a)

Wavelength of the Compton scattered photon and the kinetic energy of the scattered electron if the scattering angle is 0.000.

Expert Solution
Check Mark

Answer to Problem 55QAP

Wavelength of the Compton scattered photon is 0.140 nm and the kinetic energy of the scattered electron is 0 if the scattering angle is 0.000.

Explanation of Solution

Given:

Initial wavelength λi=0.140 nm and the scattering angle is 0.000

Formula used:

According to Compton scattering the change in wavelength of the incident photon is given by

  Δλ=(λfλi)=hmec(1cosθ) (1.1)

Now the energy of the scattered photon is

  Ef=hcλf (1.2)

Kinetic energy of the scattered electron is given by

  K=EiEf=hc(λfλiλfλi)=hc(Δλλfλi)(1.3)

Calculation:

For θ=0.000 using Eq. (1.1) we get,

  Δλ=hmec(1cos0°)=hmec(11)=0

So,

  λf=λi=0.140 nm

And using the value of Δλ=0 in Eq. (1.3), we get

K = 0.

Conclusion:

So the wavelength of the Compton scattered photon is 0.140 nm and the kinetic energy of the scattered electron is 0

To determine

(b)

Wavelength of the Compton scattered photon and the kinetic energy of the scattered electron if the scattering angle is 30.000

Expert Solution
Check Mark

Answer to Problem 55QAP

Wavelength of the Compton scattered photon is 0.140325 nm and the kinetic energy of the scattered electron is 20.5 eV

Explanation of Solution

Given:

Initial wavelength λi=0.140 nm and the scattering angle is 30.000

Formula used:

According to Compton scattering the change in wavelength of the incident photon is given by

  Δλ=(λfλi)=hmec(1cosθ)(1.1)

Now the energy of the scattered photon is

  Ef=hcλf (1.2)

Kinetic energy of the scattered electron is given by

  K=EiEf=hc(λfλiλfλi)=hc(Δλλfλi)(1.3)

Calculation:

For θ=300 using Eq. (1.1) we get

  Δλ=(λfλi)=2.43×1012(1cos300) m=0.325×103 nmλf=0.140+0.325×103 nm = 0.140325 nm

Now using the values of Δλ, λf and λi in Eq. (1.3) we get

  K=hc( Δλ λ f λ i )=6.63× 10 34×3× 108×0.325× 10 120.140325×0.140× 10 18 Jor, K = 329× 10 201.6× 10 19eV=20.5 eV

[Above we have used 1 nm = 10-9 m and 1eV=1.6×1019J ]

Conclusion:

So the wavelength of the Compton scattered photon is 0.140325 nm and the kinetic energy of the scattered electron is 20.5 eV.

To determine

(c)

Wavelength of the Compton scattered photon and the kinetic energy of the scattered electron if the scattering angle is 45.000.

Expert Solution
Check Mark

Answer to Problem 55QAP

Wavelength of the Compton scattered photon is 0.140712 nm and the kinetic energy of the scattered electron is 44.9 eV

Explanation of Solution

Given:

Initial wavelength λi=0.140 nm and the scattering angle is 45.000

Formula used:

According to Compton scattering the change in wavelength of the incident photon is given by

  Δλ=(λfλi)=hmec(1cosθ)(1.1)

Now the energy of the scattered photon is

  Ef=hcλf (1.2)

Kinetic energy of the scattered electron is given by,

  K=EiEf=hc(λfλiλfλi)=hc(Δλλfλi)(1.3)

Calculation:

For θ=45.000 using Eq. (1.1)

  Δλ=(λfλi)=2.43×1012(1cos450) mor, λf=0.140712 nm

Now using the values of Δλ, λf and λi in Eq. (1.3) and using the conversions as used earlier in part (b) above [1 nm = 10-9 m and 1eV=1.6×1019J ] we get

  K=44.9 eV

Conclusion:

Wavelength of the Compton scattered photon is 0.140712 nm and the kinetic energy of the scattered electron is 44.9 eV

To determine

(d)

Wavelength of the Compton scattered photon and the kinetic energy of the scattered electron if the scattering angle is 60.000

Expert Solution
Check Mark

Answer to Problem 55QAP

Wavelength of the Compton scattered photon is 0.141215 nm and the kinetic energy of the scattered electron is 75.32 eV if the scattering angle is 60.000.

Explanation of Solution

Given:

Initial wavelength λi=0.140 nm and the scattering angle is 60.000

Formula used:

According to Compton scattering the change in wavelength of the incident photon is given by

  Δλ=(λfλi)=hmec(1cosθ)(1.1)

Now the energy of the scattered photon is

  Ef=hcλf (1.2)

Kinetic energy of the scattered electron is given by,

  K=EiEf=hc(λfλiλfλi)=hc(Δλλfλi)(1.3)

Calculation:

For θ=60.000 using Eq. (1.1)

  Δλ=(λfλi)=2.43×1012(1cos600) m = 0.001215 nmor, λf=0.141215 nm

Now using the values of Δλ, λf and λi in Eq. (1.3) and using the conversions as used earlier in part (b) above [1 nm = 10-9 m and 1eV=1.6×1019J ] we get,

  K=75.32 eV

Conclusion:

Wavelength of the Compton scattered photon is 0.141215 nm and the kinetic energy of the scattered electron is 75.32 eV

To determine

(e)

Wavelength of the Compton scattered photon and the kinetic energy of the scattered electron if the scattering angle is 90.000

Expert Solution
Check Mark

Answer to Problem 55QAP

Wavelength of the Compton scattered photon is 0.14243 nm and the kinetic energy of the scattered electron is 151.32 eV

Explanation of Solution

Given:

Initial wavelength λi=0.140 nm and the scattering angle is 90.000

Formula used:

According to Compton scattering the change in wavelength of the incident photon is given by

  Δλ=(λfλi)=hmec(1cosθ) (1.1)

Now the energy of the scattered photon is

  Ef=hcλf (1.2)

Kinetic energy of the scattered electron is given by,

  K=EiEf=hc(λfλiλfλi)=hc(Δλλfλi)(1.3)

Calculation:

For θ = 900 using Eq. (1.1)

  Δλ=(λfλi)=2.43×1012(1cos900) m = 2.43×1012 mor, λf=0.14243 nm

Now using the values of Δλ, λf and λi in Eq. (1.3) and using the conversions as used earlier in part (b) above [1 nm = 10-9 m and 1eV=1.6×1019J ] we get

  K=151.32 eV

Conclusion:

Wavelength of the Compton scattered photon is 0.14243 nm and the kinetic energy of the scattered electron is 151.32 eV.

To determine

(f)

Wavelength of the Compton scattered photon and the kinetic energy of the scattered electron if the scattering angle is 180.000.

Expert Solution
Check Mark

Answer to Problem 55QAP

Wavelength of the Compton scattered photon is 0.14486 nm and the kinetic energy of the scattered electron is 297.43 eV

Explanation of Solution

Given:

Initial wavelength λi=0.140 nm and the scattering angle is 180.000

Formula used:

According to Compton scattering the change in wavelength of the incident photon is given by

  Δλ=(λfλi)=hmec(1cosθ)(1.1)

Now the energy of the scattered photon is

  Ef=hcλf (1.2)

Kinetic energy of the scattered electron is given by,

  K=EiEf=hc(λfλiλfλi)=hc(Δλλfλi)(1.3)

Calculation:

For θ = 1800 using Eq. (1.1)

  Δλ=(λfλi)=2.43×1012(1cos1800) m = 4.86×1012 mor, λf=0.14486 nm

Now using the values of Δλ, λf and λi in Eq. (1.3) and using the conversions as used earlier in part (b) above [1 nm = 10-9 m and 1eV=1.6×1019J ] we get

  K= 297.43 eV

Conclusion:

Wavelength of the Compton scattered photon is 0.14486 nm and the kinetic energy of the scattered electron is 297.43 eV

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Chapter 26 Solutions

COLLEGE PHYSICS,VOLUME 1

Ch. 26 - Prob. 11QAPCh. 26 - Prob. 12QAPCh. 26 - Prob. 13QAPCh. 26 - Prob. 14QAPCh. 26 - Prob. 15QAPCh. 26 - Prob. 16QAPCh. 26 - Prob. 17QAPCh. 26 - Prob. 18QAPCh. 26 - Prob. 19QAPCh. 26 - Prob. 20QAPCh. 26 - Prob. 21QAPCh. 26 - Prob. 22QAPCh. 26 - Prob. 23QAPCh. 26 - Prob. 24QAPCh. 26 - Prob. 25QAPCh. 26 - Prob. 26QAPCh. 26 - Prob. 27QAPCh. 26 - Prob. 28QAPCh. 26 - Prob. 29QAPCh. 26 - Prob. 30QAPCh. 26 - Prob. 31QAPCh. 26 - Prob. 32QAPCh. 26 - Prob. 33QAPCh. 26 - Prob. 34QAPCh. 26 - Prob. 35QAPCh. 26 - Prob. 36QAPCh. 26 - Prob. 37QAPCh. 26 - Prob. 38QAPCh. 26 - Prob. 39QAPCh. 26 - Prob. 40QAPCh. 26 - Prob. 41QAPCh. 26 - Prob. 42QAPCh. 26 - Prob. 43QAPCh. 26 - Prob. 44QAPCh. 26 - Prob. 45QAPCh. 26 - Prob. 46QAPCh. 26 - Prob. 47QAPCh. 26 - Prob. 48QAPCh. 26 - Prob. 49QAPCh. 26 - Prob. 50QAPCh. 26 - Prob. 51QAPCh. 26 - Prob. 52QAPCh. 26 - Prob. 53QAPCh. 26 - Prob. 54QAPCh. 26 - Prob. 55QAPCh. 26 - Prob. 56QAPCh. 26 - Prob. 57QAPCh. 26 - Prob. 58QAPCh. 26 - Prob. 59QAPCh. 26 - Prob. 60QAPCh. 26 - Prob. 61QAPCh. 26 - Prob. 62QAPCh. 26 - Prob. 63QAPCh. 26 - Prob. 64QAPCh. 26 - Prob. 65QAPCh. 26 - Prob. 66QAPCh. 26 - Prob. 67QAPCh. 26 - Prob. 68QAPCh. 26 - Prob. 69QAPCh. 26 - Prob. 70QAPCh. 26 - Prob. 71QAPCh. 26 - Prob. 72QAPCh. 26 - Prob. 73QAPCh. 26 - Prob. 74QAPCh. 26 - Prob. 75QAPCh. 26 - Prob. 76QAPCh. 26 - Prob. 77QAPCh. 26 - Prob. 78QAPCh. 26 - Prob. 79QAPCh. 26 - Prob. 80QAPCh. 26 - Prob. 81QAPCh. 26 - Prob. 82QAPCh. 26 - Prob. 83QAPCh. 26 - Prob. 84QAPCh. 26 - Prob. 85QAPCh. 26 - Prob. 86QAPCh. 26 - Prob. 87QAPCh. 26 - Prob. 88QAPCh. 26 - Prob. 89QAPCh. 26 - Prob. 90QAPCh. 26 - Prob. 91QAPCh. 26 - Prob. 92QAPCh. 26 - Prob. 93QAPCh. 26 - Prob. 94QAPCh. 26 - Prob. 95QAPCh. 26 - Prob. 96QAPCh. 26 - Prob. 97QAPCh. 26 - Prob. 98QAPCh. 26 - Prob. 99QAPCh. 26 - Prob. 100QAPCh. 26 - Prob. 101QAPCh. 26 - Prob. 102QAPCh. 26 - Prob. 103QAPCh. 26 - Prob. 104QAP
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