Explanation Given info: Wavelength of the x ray is 0.0300 nm , order of interference is 1,angle at which intensity maximum for a crystal occurred is 35.8 ° and energy of the electron is 4.50 keV . Write the expression for wavelength of x ray for m = 1 . λ = d sin θ 1 (1) Here, λ is the wavelength of the x ray wave d is the separation between adjacent atoms in the surface plane. θ is the angle at which maxima occurs Write the wavelength of the electron wave for m = 1 . λ ′ = d sin θ ′ 1 (2) Here, λ ′ is the wavelength of the x ray wave d ′ is the separation between adjacent atoms in the surface plane. θ ′ is the angle at which maxima occurs Divide equation (1) by equation (2) to get sin θ ′ . λ λ ′ = d sin θ d sin θ ′ sin θ ′ = ( λ ′ λ ) sin θ (3) Write the expression for the wavelength of electron wave in terms of energy. λ ′ = h 2 m E ′ (4) Here, E ′ is the energy of the electron beam. Substitute h 2 m E ′ for λ ′ in equation (3) to get sin θ ′ . sin θ ′ = ( h λ 1 2 m E ′ ) sin θ Wavelength of the x ray is 0.0300 nm , order of interference is 1,angle at which intensity maximum for a crystal occurred is 35.8 ° Substitute 0.0300 nm for λ ′ , 35.8 ° for θ , 4.50 keV for E ′ , 9

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Chapter 39, Problem 39.77P

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The angle at which maximum occurs when a beam of 4.50 keV electrons is used.

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Chapter 39, Problem 39.76P

Chapter 39, Problem 39.78P

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Chapter 39 Solutions

MASTERINGPHYSICS W/ETEXT ACCESS CODE 6

Ch. 39.2 - Prob. 39.2TYU Ch. 39.3 - Prob. 39.3TYU Ch. 39.4 - Prob. 39.4TYU Ch. 39.5 - Prob. 39.5TYU Ch. 39.6 - Prob. 39.6TYU Ch. 39 - Prob. 39.1DQ Ch. 39 - Prob. 39.2DQ Ch. 39 - Prob. 39.3DQ Ch. 39 - When an electron beam goes through a very small...Ch. 39 - Prob. 39.5DQ

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The angle at which maximum occurs when a beam of 4.50 keV electrons is used.

Solution Summary: The author calculates the angle at which maximum occurs when a beam of 4.50sintheta keV electrons is used.

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The angle at which maximum occurs when a beam of 4.50 keV electrons is used.

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Chapter 39 Solutions