Explanation Given: The given Energy of a particle is E = f ( λ ) = m 0 2 c 4 + h 2 c 2 λ 2 (1) Calculation: Assume a = m 0 2 c 4 and b = h 2 c 2 Rewrite the equation (1) with the assumed variable. E = f ( λ ) = a + b λ 2 = a λ 2 + b λ 2 = a λ 2 + b λ Calculate the asymptotes. Substitute 0, ± ∞ for λ in the above equation by applying limit. lim x → 0 + f ( v ) = lim x → 0 + a λ 2 + b λ = a ( 0 2 ) + b 0 = ∞ Therefore, the equation λ = 0 is the vertical asymptote. lim x → ∞ f ( v ) = lim x → ∞ a λ 2 + b λ = lim x → ∞ a λ 2 + b λ λ λ lim x → ∞ f ( v ) = lim x → ∞ a + b / λ 2 1 = a + b / ∞ 2 1 = a + 0 1 = a Therefore, the equation E = a = m 0 c 2 is the horizontal asymptote Differentiate the equation a λ 2 + b λ with respect to λ by applying UV method. f ( v ) = a λ 2 + b λ f ( v ) = v u ' − u v ' v 2 f ' ( v ) = [ λ ( 1 2 ( a λ 2 + b ) − 1 2 ) ( 2 a λ ) ] − [ ( a λ 2 + b ) ( 1 ) ] λ 2 = a λ 2 ( a λ 2 + b ) − 1 2 − ( a λ 2 + b ) 1 2 λ 2 = ( a λ 2 + b ) − 1 2 [ a λ 2 − ( a λ 2 + b ) ] λ 2 = − b λ 2 a λ 2 + b < 0 Therefore, the function f is decreasing on ( 0 , ∞ )

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ISBN: 9780357019788

Author: Stewart

Publisher: CENGAGE L

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Chapter 4.5, Problem 56E

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To Sketch: The graph of energy of particle E as a function of wavelength λ

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Chapter 4.5, Problem 55E

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The graph of energy of particle E as a function of wavelength λ

Solution Summary: The author illustrates the graph of energy of particle E as a function of wavelength lambda.

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