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Unformatted text preview: 24.1. Model: Balmer’s formula predicts a series of spectral lines in the hydrogen spectrum. Solve: Substituting into the formula for the Balmer series, λ = − 91.18 nm 1 2 1 2 2 n ⇒ = − = λ 91.18 nm 410.3 nm 1 2 1 6 2 2 where n = 3, 4, 5, 6, … and where we have used n = 6. Likewise for n = 8 and n = 10, λ = 389.0 nm and λ = 379.9 nm. 24.2. Model: Balmer’s formula predicts a series of spectral lines in the hydrogen spectrum. Solve: Balmer’s formula is λ = − 91.18 nm 1 2 1 2 2 n n = 3, 4, 5, 6, … As , Thus, 91.18 nm 364.7 nm n n n → ∞ → = ( ) = →∞ 1 4 2 . λ . 24.3. Model: Balmer’s formula predicts a series of spectral lines in the hydrogen spectrum. Solve: Using Balmer’s formula, λ = = − ⇒ − = ⇒ = 389.0 nm 91.18 nm 1 2 1 1 4 1 0 2344 8 2 2 2 n n n . 24.4. Model: The angles of incidence for which diffraction from parallel planes occurs satisfy the Bragg condition. Solve: The Bragg condition is 2 d m m cos θ λ = , where m = 1, 2, 3, … For first and second order diffraction, 2 1 1 d cos θ λ = ( ) 2 2 2 d cos θ λ = ( ) Dividing these two equations, cos cos cos cos cos cos . θ θ θ θ 2 1 2 1 1 1 2 2 2 6 8 4 1 5 = ⇒ = ( ) = ° ( ) = ° − − 24.5. Model: The angles of incidence for which diffraction from parallel planes occurs satisfy the Bragg condition. Solve: The Bragg condition is 2 d m m cos θ λ = . For m = 1 and for two different wavelengths, 2 1 1 1 d cos θ λ = ( ) 2 1 1 1 d cos ′ = ( ) ′ θ λ Dividing these two equations, cos cos cos cos cos . . ′ = ′ ⇒ ′ ° = ⇒ ′ = ( ) = ° − θ θ λ λ θ θ 1 1 1 1 1 1 1 54 0 4408 63 8 0.15 nm 0.20 nm 24.6. Model: The angles corresponding to the various orders of diffraction satisfy the Bragg condition. Solve: The Bragg condition for m = 1 and m = 2 gives 2 1 2 2 1 2 d d cos cos θ λ θ λ = ( ) = ( ) Dividing these two equations, cos cos cos cos cos . θ θ θ 1 2 1 1 2 45 2 45 2 69 3 = = ° ⇒ = ° = ° − 24.7. Model: The angles corresponding to the various diffraction orders satisfy the Bragg condition. Solve: The Bragg condition is 2 d m m cos θ λ = , where m = 1, 2, 3, … The maximum possible value of m is the number of possible diffraction orders. The maximum value of cos θ m is 1. Thus, 2 2 2 4 2 d m m d = ⇒ = = ( ) ( ) = λ λ 0.180 nm 0.085 nm . We can observe up to the fourth diffraction order. 24.8. Model: Use the photon model of light. Solve: The energy of the photon is E h f h c photon Js m / s m J = = = × ( ) × × = × − − − λ 6 63 10 3 0 10 500 10 3 98 10 34 8 9 19 . . . Assess: The energy of a single photon in the visible light region is extremely small. 24.9. Model: Use the photon model of light....
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 Spring '08
 Medvedev
 Diffraction, Light, NM, Balmer

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