COLLEGE PHYSICS
COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Chapter 26, Problem 58QAP
To determine

The wavelength of x-ray source

Expert Solution & Answer
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Answer to Problem 58QAP

The wavelength of x-ray source is λi=1.618 nm = 16.18 Angstrom.

Explanation of Solution

Given info:

Scattering angle of the photon is θ=600 and the speed of electron is v=4.5×105m/s. The value of K is 92.13×1021 J

Formula used:

According to Compton scattering the change in wavelength of the incident photon is given by
  Δλ=(λfλi)=hmec(1cosθ)

Now the energy of the scattered photon is
Ef=hcλf

Kinetic energy of the scattered electron is given by
  K=EiEf=hc(λfλiλfλi)=hc(Δλλfλi)=12mv2

Calculation:

Putting θ=600 in Eq. (1.1) we get
  Δλ=(λfλi)=2.43×1012(1cos60°) m = 1.215×10-12m

Now putting the given value of K, and calculated Δλ in Eq. (1.3) we get
  K=hc( Δλ λ f λ i )=12×9.1×1031×(4.5×105)2=92.13×1021 Jor, λfλi×10-18 = 6.63× 10 34×3× 108×1.215× 10 -1292.13× 10 21or, λfλi = 2.62 nm2or, λf= 2.62 λinm

Now substituting the value of λf in the expression for Δλ=(λfλi) we get
  Δλ=( 2.62 λ i λi) nm=0.001215 nmor, 2.62λi2=0.001215λior, λi2+0.001215λi2.62=0

Solving this quadratic equation for λf we get
  λi=0.001215± (0.001215)2+4×1×(2.62)2×1 nm

Now as λi cannot be negative so considering only the positive root of λi we get
  λi=1.618 nm = 16.18 Angstrom

Conclusion:

So, the initial wavelength of the photon is λi=1.618 nm = 16.18 Angstrom

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Chapter 26 Solutions

COLLEGE PHYSICS

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