Explanation 1) Concept: We can derive the equation for Compton shift by using equation 38 - 8 , 38 - 9 , a n d 38 - 10 and by using trigonometric identity. 2) Formula: i) γ = 1 1 - v / c 2 ii) sin 2 ⁡ θ + cos 2 ⁡ θ = 1 3) Given: i) h λ = h λ ' cos ⁡ ϕ + γ m v cos ⁡ θ ii) h m λ ' sin ⁡ ϕ = v 1 - v / c 2 sin ⁡ θ iii) h λ = h λ ' + m c ( γ - 1 ) 4) Calculations: From equation 38 - 9 , we have h λ = h λ ' cos ⁡ ϕ + γ m v cos ⁡ θ We can rewrite this equation as h m λ - h m λ ' cos ⁡ ϕ = γ v cos ⁡ θ As γ = 1 1 - v / c 2 Thus, h m λ - h m λ ' cos ⁡ ϕ = v 1 - v / c 2 cos ⁡ θ … … … … … … … … … … … … … … ( 1 ) Now from equation 38 - 10 , we have 0 = h λ ' sin ⁡ ϕ - γ m v sin ⁡ θ We can rewrite this equation as h m λ ' sin ⁡ ϕ = γ m v sin ⁡ θ As γ = 1 1 - v / c 2 Thus, h m λ ' sin ⁡ ϕ = v 1 - v / c 2 sin ⁡ θ … … … … … … … … … … … … … ( 2 ) Now squaring and adding equations 1 and ( 2 ) , h m λ - h m λ ' cos ⁡ ϕ 2 + h m λ ' sin ⁡ ϕ 2 = v 2 1 - v / c 2 cos 2 ⁡ θ + v 2 1 - v / c 2 sin 2 ⁡ θ h m 2 1 λ - 1 λ ' cos ⁡ ϕ 2 + 1 λ ' sin ⁡ ϕ 2 = v 2 1 - v / c 2 cos 2 ⁡ θ + sin 2 ⁡ θ As sin 2 ⁡ θ + cos 2 ⁡ θ = 1 h m 2

Fundamentals Of Physics 11th Edition Loose-leaf Print Companion Volume 2 With Wileyplus Card Set

11th Edition

ISBN: 9781119463252

Author: David Halliday

Publisher: John Wiley and Sons

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Chapter 38, Problem 82P

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To derive:

∆λ=hmc(1-cos⁡ϕ)

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Chapter 38, Problem 81P

Chapter 38, Problem 83P

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Chapter 38 Solutions

Fundamentals Of Physics 11th Edition Loose-leaf Print Companion Volume 2 With Wileyplus Card Set

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To derive: ∆ λ = h m c ( 1 - cos ⁡ ϕ )

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To derive: ∆λ=hmc(1-cos⁡ϕ)

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Chapter 38 Solutions

To derive:

∆λ=hmc(1-cos⁡ϕ)