Explanation Given data: Wavelength is 103 nm Formula used: Consider the expression for 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. Consider the expression of energy. E = h c λ (2) Explanation: Substitute 6.63 × 10 − 34 J ⋅ s for h , 3 × 10 8 m s for c , and 103 nm for λ in equation (2), E = ( 6.63 × 10 − 34 J ⋅ s ) × ( 3 × 10 8 m s ) 103 nm = ( 6.63 × 10 − 34 J ⋅ s ) × ( 3 × 10 8 m s ) 103 × 10 − 9 m Consider the conversion factor for J to eV. 1 J ⋅ s = 1 1.6 × 10 − 19 eV ⋅ s Convert the energy into eV ⋅ s by using conversion factor. E = ( 6.63 × 10 − 34 eV ⋅ s 1.6 × 10 − 19 ) × ( 3 × 10 8 m s ) 103 × 10 − 9 m = ( 4.14 × 10 − 15 eV ⋅ s ) × ( 3 × 10 8 m s ) 103 × 10 − 9 m = 12.1 eV The photon is 12.1 eV is absorbed by hydrogen atom in ground state, the amount of energy left after absorption is, E f = E i + E (3) Here, E f is the final energy after absorption, E i is the energy of hydrogen in ground state, which has a standard value of − 13.6 eV , and E is the amount of energy absorbed by hydrogen atom. Substitute 12.1 eV for E , and − 13.6 eV for E i in equation (3), E f = − 13.6 eV + 12.1 eV = − 1 .5 eV Draw energy level diagram as shown in Figure 1. From the figure the value of n = 3 , As the atoms undergo transitions to the ground state the possible transitions are, n = 3 to 2 , n = 3 to1 , n = 2 to1 For n + 1 is 2 and n is 1. Substitute 6.63 × 10 − 34 J ⋅ s for h , 3 × 10 8 m s for c , − 3.4 eV for energy of n level, and − 13.6 eV for energy of n + 1 level in equation (1), λ = ( 6.63 × 10 − 34 J ⋅ s ) × ( 3 × 10 8 m s ) − 3.4 eV − ( − 13.6 eV ) = ( 6.63 × 10 − 34 J ⋅ s ) × ( 3 × 10 8 m s ) 10.2 eV Consider the conversion factor for J to eV. 1 J ⋅ s = 1 1.6 × 10 − 19 eV ⋅ s Convert the value by using conversion factor. λ = ( 6.63 × 10 − 34 eV ⋅ s 1.6 × 10 − 19 ) × ( 3 × 10 8 m s ) 10.2 eV = ( 4

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Author: Knight

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Question

Chapter 29, Problem 40P

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The wavelength seen in the emission spectrum of hydrogen gas.

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Chapter 29, Problem 39P

Chapter 29, Problem 41P

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Chapter 29 Solutions

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The wavelength seen in the emission spectrum of hydrogen gas.

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The wavelength seen in the emission spectrum of hydrogen gas.

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Chapter 29 Solutions