Explanation According to Compton shift formula, Δ λ = h m e × c ( 1 − cos θ ) (1) Here, h is the Planck’s constant. m e is the mass of electron. c is the speed of light . θ is the angle of deviation. Substitute 6 .63 × 10 − 34 J ⋅ s for h , 9.11 × 10 − 31 kg for m e , 180 ° for θ and 3 × 10 8 m / s for c in equation (1) to find ( Δ λ ) θ = 180 ° . ( Δ λ ) θ = 180 ° = ( 6 .63 × 10 − 34 J ⋅ m ) ( 9.11 × 10 − 31 kg ) × ( 3 × 10 8 m / s ) ( 1 − cos 180 ) = ( 6 .63 × 10 − 34 J ⋅ m ) ( 9.11 × 10 − 31 kg ) × ( 3 × 10 8 m / s ) ( 1 − ( − 1 ) ) = 4.85 × 10 − 12 m The wavelength shift of one-fourth maximum is, Δ λ = 1 4 × ( Δ λ ) θ = 180 ° (2) Substitute 4.85 × 10 − 12 m for ( Δ λ ) θ = 180 ° in equation (2), to get the value of Δ λ

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Chapter 30, Problem 54PCE

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The scattering angle that will produce a wavelength shift of one-fourth of the maximum.

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Chapter 30, Problem 53PCE

Chapter 30, Problem 55PCE

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Ch. 30.1 - Prob. 1EYU Ch. 30.2 - Prob. 2EYU Ch. 30.3 - Prob. 3EYU Ch. 30.4 - Prob. 4EYU Ch. 30.5 - Prob. 5EYU Ch. 30.6 - Prob. 6EYU Ch. 30.7 - Prob. 7EYU Ch. 30 - Prob. 1CQ Ch. 30 - Prob. 2CQ Ch. 30 - Prob. 3CQ

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The scattering angle that will produce a wavelength shift of one-fourth of the maximum.

Solution Summary: The author explains the scattering angle that will produce a wavelength shift of one-fourth of the maximum is 60°.

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The scattering angle that will produce a wavelength shift of one-fourth of the maximum.

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