Chapter 25 - 25.1. Model: Balmers formula predicts a series...

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25.1. Model: Balmer’s formula predicts a series of spectral lines in the hydrogen spectrum. Solve: Substituting into the formula for the Balmer series, 22 91.18 nm 11 2 n λ = ⎛⎞ ⎜⎟ ⎝⎠ 91.18 nm 410.3 nm 26 ⇒= = where n = 3, 4, 5, 6, … and where we have used n = 6. Likewise for n = 8 and n = 10, 389.0 nm and = 379.9 nm. =
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25.2. Model: Balmer’s formula predicts a series of spectral lines in the hydrogen spectrum. Solve: Balmer’s formula is 22 91.18 nm 11 2 n λ = ⎛⎞ ⎜⎟ ⎝⎠ n = 3, 4, 5, 6, … () 2 As , 1 0. Thus, 4 91.18 nm 364.7 nm. n nn →∞ →∞ = =
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25.3. Model: Balmer’s formula predicts a series of spectral lines in the hydrogen spectrum. Solve: Using Balmer’s formula, 2 22 91.18 nm 1 1 389.0 nm 0.2344 8 11 4 2 n n n λ == = = ⎛⎞ ⎜⎟ ⎝⎠
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25.4. Model: The angles of incidence for which diffraction from parallel planes occurs satisfy the Bragg condition. Solve: The Bragg condition is 2c o s , m dm θ λ = where m = 1, 2, 3, … For first and second order diffraction, ( ) 1 o s 1 d = ( ) 2 o s 2 d = Dividing these two equations, () ( ) 11 2 21 1 cos o s 2 c o s c o s 2 c o s 6 8 4 1 cos θθ −− =⇒ = = °= °
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25.5. Model: The angles of incidence for which diffraction from parallel planes occurs satisfy the Bragg condition. Solve: The Bragg condition is 2c o s . m dm θ λ = For m = 1 and for two different wavelengths, ( ) 11 o s 1 d = ( ) o s 1 d = Dividing these two equations, () 1 1 1 cos cos 0.15 nm cos 0.4408 64 cos cos54 0.20 nm θλ ′′ =⇒ = ⇒= °
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25.6. 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 ( ) ( ) 12 2c o s 1 o s 2 dd θ λθ λ == Dividing these two equations, 1 2 11 cos cos45 cos45 cos cos 69.3 22 2 θθ °° ⎛⎞ = = ° ⎜⎟ ⎝⎠
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25.7. Model: The angles corresponding to the various diffraction orders satisfy the Bragg condition. Solve: The Bragg condition is 2c o s m dm 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 m is 1. Thus, ( ) () 2 0.180 nm 2 24 . 2 0.085 nm d dm m =⇒ == = We can observe up to the fourth diffraction order.
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25.8. Model: Use the photon model of light. Solve: The energy of the photon is () 8 34 19 photon 9 3.0 10 m/s 6.63 10 Js 3.98 10 J 500 10 m c Eh f h λ −− ⎛⎞ × == = × = × ⎜⎟ × ⎝⎠ Assess: The energy of a single photon in the visible light region is extremely small.
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25.9. Model: Use the photon model of light. Solve: The energy of the single photon is ( )( ) 34 8 19 photon 6 6.63 10 Js 3.0 10 m/s 1.99 10 J 1.0 10 m c Eh f h λ ×× ⎛⎞ == = = × ⎜⎟ × ⎝⎠ ( )( ) 23 19 5 mol A photon 6.023 10 1.99 10 J 1.2 10 J EN E ⇒= = × × = × Assess: Although the energy of a single photon is very small, a mole of photons has a significant amount of energy.
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25.10. Model: Use the photon model of light. Solve: The energy of a photon with wavelength λ 1 is 11 1 .
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This note was uploaded on 01/30/2011 for the course PHYS 131 taught by Professor E.salik during the Winter '10 term at Cal Poly Pomona.

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Chapter 25 - 25.1. Model: Balmers formula predicts a series...

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