Explanation E 3 0 − E 2 0 = h ν = 2 π ℏ c λ = E 1 ( 1 9 − 1 4 ) = − 5 36 E 1 ⇒ λ = − 36 5 2 π ℏ c E 1 E 1 = − 13.6 eV ℏ c = 1.97 × 10 − 11 MeV ⋅ cm λ = 36 5 ( 2 π ) ( 1.97 × 10 − 11 × 10 6 eV ⋅ cm ) ( 13.6 eV ) = 6.55 × 10 − 5 cm ( 1 nm 10 − 7 cm ) = 655 nm ν = c λ = 3.00 × 10 8 m / s 6.55 × 10 − 7 m = 4358 × 10 14 Hz E f s 1 = ( E n ) 2 2 m c 2 ( 3 − 4 n j + 1 2 ) For n = 2 : l = 0 or l = 1 , so j = 1 2 or 3 2 . Thus n = 2 splits into two levels: j = 1 / 2 _ : E 2 1 = ( E 2 ) 2 2 m c 2 ( 3 − 8 1 ) = − 5 2 ( E 2 ) 2 m c 2 = − 5 2 ( 1 4 ) 2 ( E 1 ) 2 m c 2 = − 5 32 ( 13.6 eV ) 2 ( 0.511 × 10 6 eV ) = − 5.66 × 10 − 5 eV j = 3 / 2 _ : E 2 1 = ( E 2 ) 2 2 m c 2 ( 3 − 8 2 ) = − 1 2 ( E 2 ) 2 m c 2 = − 1 32 ( 3.62 × 10 − 4 eV ) = − 1.13 × 10 − 5 eV For n = 3 : l = 0 , 1 or 2, so j = 1 2 , 3 2 or 5 2 . . Thus n = 3 splits into three levels: j = 1 / 2 _ : E 3 1 = ( E 3 ) 2 2 m c 2 ( 3 − 12 1 ) = − 9 2 ( E 3 ) 2 m c 2 = − 9 2 ( 1 9 ) 2 ( E 1 ) 2 m c 2 = − 1 18 ( 3.62 × 10 − 4 eV ) = − 2.01 × 10 − 5 eV j = 3 / 2 _ : E 3 1 = ( E 3 ) 2 2 m c 2 ( 3 − 12 2 ) = − 3 2 ( E 3 ) 2 m c 2 = − 1 54 ( 3.62 × 10 − 4 eV ) = − 0.67 × 10 − 5 eV j = 5 / 2 _ : E 3 1 = ( E 3 ) 2 2 m c 2 ( 3 − 12 3 ) = − 1 2 ( E 3 ) 2 m c 2 = − 1 162 ( 3

3rd Edition

ISBN: 9781107189638

Author: Griffiths, David J., Schroeter, Darrell F.

Publisher: Cambridge University Press

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Chapter 7.3, Problem 7.21P

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Fine structure slitting of Balmer line of hydrogen spectrum.

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Introduction To Quantum Mechanics

Ch. 7.1 - Prob. 7.1P Ch. 7.1 - Prob. 7.2P Ch. 7.1 - Prob. 7.3P Ch. 7.1 - Prob. 7.4P Ch. 7.1 - Prob. 7.5P Ch. 7.1 - Prob. 7.6P Ch. 7.2 - Prob. 7.8P Ch. 7.2 - Prob. 7.9P Ch. 7.2 - Prob. 7.10P Ch. 7.2 - Prob. 7.11P

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