Explanation Given data: Length of the box is 0.85 nm. Initial energy level of the electron is 1. Formula used: Write the expression for minimum energy of confined electron. E n = h 2 8 m L 2 n 2 (1) Here, E n is the energy, h is the plank’s constant with value 6.63 × 10 − 34 J ⋅ s , m is the mass of the electron, n is the integral multiple, and L is the length of the box. Write the expression for photon’s energy and wavelength. Δ E = h c λ (2) Here, h is the plank’s constant with value 6.63 × 10 − 34 J ⋅ s , c is the speed of light with value 3.0 × 10 8 m s , and λ is the wavelength. Explanation: Energy is the difference between the final and initial energy level. Consider the expression. Δ E 1 → n = h 2 8 m L 2 ( n 2 − 1 ) (3) Combining equation (3) and (2). h c λ = h 2 8 m L 2 ( n 2 − 1 ) λ = 8 m c L 2 h ( n 2 − 1 ) Substitute 9.11 × 10 − 31 kg for m , 3.0 × 10 8 m s for c , ( 0.85 × 10 − 9 m ) 2 for L , 6.63 × 10 − 34 J ⋅ s for h , λ 1 → n = 8 ( 9.11 × 10 − 31 kg ) ( 3.0 × 10 8 m s ) ( 0.85 × 10 − 9 m ) 2 ( 6.63 × 10 − 34 J ⋅ s ) ( n 2 − 1 ) (4) For n = 2 , substitute 2 for n in equation (5), λ 1 → 2 = 8 ( 9.11 × 10 − 31 kg ) ( 3.0 × 10 8 m s ) ( 0

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Chapter 28, Problem 52P

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The three longest wavelengths absorbed by the molecule.

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Chapter 28, Problem 51P

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Chapter 28 Solutions

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Ch. 28 - Prob. 1CQ Ch. 28 - Prob. 2CQ Ch. 28 - Prob. 3CQ Ch. 28 - Prob. 4CQ Ch. 28 - Prob. 5CQ Ch. 28 - Prob. 6CQ Ch. 28 - Prob. 7CQ Ch. 28 - Prob. 8CQ Ch. 28 - Prob. 9CQ Ch. 28 - Prob. 10CQ

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The three longest wavelengths absorbed by the molecule.

Solution Summary: The author explains the three longest wavelengths absorbed by the molecule: 790, 300, and 160 nm. Energy is the difference between the final and initial energy levels.

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The three longest wavelengths absorbed by the molecule.

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Chapter 28 Solutions