Explanation Given data: De Broglie wavelength of the electron is 0.20 nm. Formula used: Write the expression for energy of atom and temperature. K = 3 2 ( k B ) T (1) Here, K is the energy of rubidium, k B is the temperature coefficient with value 1.38 × 10 − 23 J K , and T is the temperature. Write the expression for momentum and kinetic energy. K = p 2 2 m (2) Here, p is the momentum, and m is the mass of rubidium ( 85.5 u ) . Write the expression for de Broglie wavelength for moving particle. λ = h p (3) Here, h is the plank’s constant, which has the standard value as 6.63 × 10 − 34 Js . λ is the wavelength. Substitute 1.38 × 10 − 23 J K for k B and 1.0 × 10 − 7 K for T in equation (1), K = 3 2 ( 1.38 × 10 − 23 J K ) ( 1.0 × 10 − 7 K ) = 2.07 × 10 − 30 J Mass of rubidium in kg, m = 85.5 u { ∵ 1 u = 1.66 × 10 − 27 kg } = 85.5 ( 1.66 × 10 − 27 kg ) = 1.42 × 10 − 25 kg Substitute 2

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Chapter 28, Problem 34P

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The de Broglie wavelength of rubidium and number of times larger than the diameter of the atom.

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Chapter 28, Problem 33P

Chapter 28, Problem 35P

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The de Broglie wavelength of rubidium and number of times larger than the diameter of the atom.

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