a) Explanation Given data: E 1 = 0 eV E 2 = 1.5 eV E 3 = 4 eV Formula used: Wavelength is given by 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: 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 1.5eV for energy of n + 1 level in equation (1), λ = ( 6.63 × 10 − 34 J ⋅ s ) × ( 3 × 10 8 m s ) 1.5 eV − 0 eV = ( 6.63 × 10 − 34 J ⋅ s ) × ( 3 × 10 8 m s ) 1.5 eV Consider the conversion factor for J to eV. 1 J ⋅ s = 1 1.6 × 10 − 19 eV ⋅ s Convert the value by using the conversion factor. λ = ( 6.63 × 10 − 34 eV ⋅ s 1.6 × 10 − 19 ) × ( 3 × 10 8 m s ) 1.5 eV = ( 4.14 × 10 − 15 eV ⋅ s ) × ( 3 × 10 8 m s ) 1.5 eV = ( 4.14 × 10 − 15 eV ⋅ s ) × ( 3 × 10 8 m s ) 1.5 eV = 8 .28 × 10 − 7 m Simplify the equation. λ = 828 × 10 − 9 m ≅ 830 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, 1.5 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 − 1 b) To determine The wavelength of absorption spectrum of the atom’s.

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The Bohr model

Students often get confused between Rutherford’s model and Bohr model of an atom. They interpret the postulates of both these models to be the same, which is incorrect. Rutherford’s model differs from Bohr’s model in the following ways:

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Chapter 29, Problem 14P

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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 13P

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Chapter 29 Solutions

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The wavelength of emission spectrum of the atom’s.

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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