a) Explanation Given data: E 1 = 0 eV E 2 = 4 eV E 3 = 6 eV Formula used: Consider the Wavelength. λ = h c E n + 1 − E n (1) Here, h is the planks constant, which has a standard value as 6.63 × 10 − 34 J ⋅ s , c is the speed of light , which has a standard value as 3 × 10 8 m s , E n is the energy of n level, E n + 1 is the energy of n + 1 level, and λ is the wavelength. Explanation: Wavelength of emission spectrum For n + 1 is 2 and n is 1, Substitute 6.63 × 10 − 34 J ⋅ s for planks constant, 3 × 10 8 m s for speed of light, 0 eV for energy of n level, and 4 eV for energy of n + 1 level in equation (1), λ = ( 6.63 × 10 − 34 J ⋅ s ) × ( 3 × 10 8 m s ) 4 eV − 0 eV = ( 6.63 × 10 − 34 J ⋅ s ) × ( 3 × 10 8 m s ) 4 eV Consider the conversion factor for J to eV. 1 J ⋅ s = 1 1.6 × 10 − 19 eV ⋅ s Convert the unit by using conversion factor. λ = ( 6.63 × 10 − 34 eV ⋅ s 1.6 × 10 − 19 ) × ( 3 × 10 8 m s ) 4 eV = ( 4.14 × 10 − 15 eV ⋅ s ) × ( 3 × 10 8 m s ) 4 eV = ( 4.14 × 10 − 15 eV ⋅ s ) × ( 3 × 10 8 m s ) 4 eV = 3 .105 × 10 − 7 m Simplify the equation. λ = 310.5 × 10 − 9 m ≅ 311 nm For n + 1 is 3 and n is 2, Substitute 6.63 × 10 − 34 J ⋅ s for planks constant, 3 × 10 8 m s for speed of light, 4 eV for energy of n level, and 6 eV for energy of n + 1 level in equation (1), λ = ( 6.63 × 10 − 34 J ⋅ s ) × ( 3 × 10 8 m s ) 6 eV − 4 eV = ( 6.63 × 10 − 34 J ⋅ s ) × ( 3 × 10 8 m s ) 2 eV Convert the unit by using J to eV conversion factor. λ = ( 6.63 × 10 − 34 eV ⋅ s 1.6 × 10 − 19 ) × ( 3 × 10 8 m s ) 2 eV = ( 4.14 × 10 − 15 eV ⋅ s ) × ( 3 × 10 8 m s ) 2 eV = ( 4.14 × 10 − 15 eV ⋅ s ) × ( 3 × 10 8 m s ) 2 eV = 6 .22 × 10 − 7 m Simplify the equation. λ = 622 × 10 − 9 m = 622 nm For n + 1 is 3 and n is 1, Substitute 6.63 × 10 − 34 J ⋅ s for planks constant, 3 × 10 8 m s for speed of light, 0 eV for energy of n level, and 6 eV for energy of n + 1 level in equation (1), λ = ( 6.63 × 10 − 34 J ⋅ s ) × ( 3 × 10 8 m s ) 6 eV − 0 eV = ( 6.63 × 10 − 34 J ⋅ s ) × ( 3 × 10 8 m s ) 6 eV Convert the unit by using J to eV conversion factor b) To determine The wavelength of emission spectrum of the atom’s. c) To determine The wavelength of absorption spectrum of the atom’s.

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Author: Knight

Publisher: PEARSON CO

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Question

Chapter 29, Problem 10P

To determine

To draw: The energy level diagram of atom’s.

To determine

The wavelength of emission spectrum of the atom’s.

To determine

The wavelength of absorption spectrum of the atom’s.

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Chapter 29, Problem 9P

Chapter 29, Problem 11P

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To draw : The energy level diagram of atom’s.

Solution Summary: The author explains how the energy level diagram of atoms has been drawn.

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To draw: The energy level diagram of atom’s.

The wavelength of emission spectrum of the atom’s.

The wavelength of absorption spectrum of the atom’s.

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Chapter 29 Solutions