Chapter1_solutionseditedBodd

Chapter1_solutionseditedBodd - Chapter 1 - Solutions 1. a....

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Chapter 1 - Solutions 1. a. Find the quantum number of the initial energy state n f = 6 (fifth excited state), λ = 5910 nm c = λν ν = R - 2 i 2 f n 1 n 1 = 3.29 × 10 15 Hz - 2 i 2 f n 1 n 1 λ c = 3.29 × 10 15 Hz - 2 i 2 f n 1 n 1 m 1 nm 10 nm 5910 s m 10 3.00 9 8 × × = 3.29 × 10 15 Hz - 2 i 2 n 1 6 1 0154 . 0 n 1 6 1 2 i 2 = - 0123 . 0 n 1 2 i = ; n i = 9, therefore the emitted photon corresponds to the third lowest-energy spectral line in the Humphreys series.
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b. λ A B C D Lymar Balmer (n f = 1) (n f = 2) Paschen Humphreys (n f = 3) (n f = 6) E atom spectral line A spectral line B spectral line C spectral line D A B C n = 1 n = 2 n = 3 n = 4 n = 5 n = 6 n = 9 n = 8 n = 7 D
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3. a. E nxnynz = + + 2 Z 2 Z 2 y 2 y 2 x 2 x 2 L n L n L n 8m h - - = + + × × 2 2 Z 2 2 y 2 2 x 31 2 12 2 34 (550) n (700) n (550) n kg) 10 (8)(9.11 pm) (10 Js) 10 (6.626 = 6.024 × 10 - 14 J + + 302500 n 490000 n 302500 n 2 Z 2 y 2 x E 211 = E 112 = 1.12 × 10 - 18 J
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Chapter1_solutionseditedBodd - Chapter 1 - Solutions 1. a....

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