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 410 3 nm 91 18 nm 1 2 n 1 2 2 91 18 nm 1 2 2 1 2 6 where n 3 4 5 6 and where we have used n 6 Likewise for n 8 and n 10 379 9 nm 389 0 nm and 25 2 Model Balmer s formula predicts a series of spectral lines in the hydrogen spectrum Solve Balmer s formula is n 3 4 5 6 91 18 nm 1 2 n 1 2 2 1 2 n 0 Thus n 4 91 18 nm 364 7 nm As n 25 3 Model Balmer s formula predicts a series of spectral lines in the hydrogen spectrum Solve Using Balmer s formula 389 0 nm 0 2344 n 8 1 4 1 2 n 91 18 nm 1 2 n 1 2 2 25 4 Model The angles of incidence for which diffraction from parallel planes occurs satisfy the Bragg condition Solve The Bragg condition is 2 cos where m 1 2 3 For first and second order diffraction m d m Dividing these two equations 2 cos d 1 1 2 cos d 2 2 cos cos 2 1 2 2 cos 1 2cos 1 cos 1 2cos68 41 25 5 Model The angles of incidence for which diffraction from parallel planes occurs satisfy the Bragg condition Solve The Bragg condition is 2 cos For m 1 and for two different wavelengths m d m 2 cos d 1 1 1 2 cos d 1 1 1 Dividing these two equations cos cos 1 1 1 1 cos 1 cos54 0 15 nm 0 20 nm 1 1 cos 0 4408 64 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 d 2 cos 1 1 d 2 cos 2 2 Dividing these two equations cos 1 cos 2 2 cos 45 2 1 cos 1 cos 45 2 69 3 25 7 Model The angles corresponding to the various diffraction orders satisfy the Bragg condition Solve The Bragg condition is 2 cos m number of possible diffraction orders The maximum value of cos m is 1 Thus where m 1 2 3 The maximum possible value of m is the m d 2 d m m 2 d 2 0 180 nm 0 085 nm 4 2 We can observe up to the fourth diffraction order 25 8 Model Use the photon model of light Solve The energy of the photon is E photon hf h 6 63 10 34 Js 3 98 10 19 J c 8 3 0 10 m s 500 10 m 9 Assess The energy of a single photon in the visible light region is extremely small 25 9 Model Use the photon model of light Solve The energy of the single photon is E photon hf h c 6 63 10 Js 3 0 10 m s 8 34 1 0 10 m 6 1 99 10 19 J E mol N E A photon 6 023 10 23 1 99 10 19 J 1 2 10 J 5 Assess Although the energy of a single photon is very small a mole of photons has a significant amount of energy 25 10 Model Use the photon model of light Solve The energy of a photon with wavelength 1 is to 2E1 E 1 hf 1 hc 1 Similarly E hc 2 2 Since E2 is equal hc 2 hc 1 2 2 1 2 600 nm 2 300 nm Assess A photon with 300 nm has twice the energy of a photon with 600 nm This is an expected result because energy is inversely proportional to the wavelength 25 11 Model Use the photon model of light Solve The energy of the x ray photon is E hf h 6 63 10 34 Js c 8 3 0 10 m s 1 0 10 m 9 2 0 10 16 J Assess This is a very small amount of energy but it is larger than the energy of a photon in the visible wavelength region 25 12 Solve Your mass is say m 70 kg and your velocity is 1 m s Thus your momentum is p mv 70 kg 1 m s 70 kg m s Your de Broglie wavelength is h p 6 63 10 34 Js 70 kg m s 9 10 36 m 25 13 Solve Broglie wavelength is a The baseball s momentum is p mv 0 200 kg 30 m s 6 0 kg m s The baseball s de h p 6 63 10 34 Js 6 0 kg m s 1 1 10 34 m v h m 6 63 10 34 Js 0 200 kg 0 20 10 m 9 1 7 10 23 m s b Using h p h mv we have 25 14 Visualize We ll employ Equations 25 8 terms of kinetic energy Solve First solve Equation 25 9 for p p mE 2 h p and 25 9 2 E p m 2 to express the wavelength in h p h mE 2 6 63 10 34 J s 2 9 11 10 31 kg 2 4 10 19 J 1 0 nm Assess The energy given is about 1 5 eV which is a reasonable amount of energy The resulting wavelength is a few to a few dozen times the size of an atom 25 15 Solve a For an electron the momentum p mv 9 11 10 kg v The de Broglie wavelength is 31 0 20 10 m 9 0 20 10 m 9 v 3 6 10 m s 6 b For a proton p mv 9 v 1 67 10 kg The de Broglie wavelength is 0 20 10 m 0 20 10 m 9 9 v 2 0 10 m s 3 34 6 63 10 9 11 10 31 Js kg v 34 6 63 10 1 67 10 27 Js kg v h p h p 25 16 Model The momentum of a wave like particle has discrete values given by 1 2 3 Solve Because we want the smallest box and the momentum of the electron can not exceed a given value n must be minimum Thus where n np 2 n h L p mv 1 L h 2 L h mv 2 34 6 63 10 31 Js kg 10 m s 2 9 11 10 0 036 mm 25 17 Model A confined particle can have only discrete values of energy Solve From Equation 24 14 the energy of a confined electron is E n 2 h 2 8 mL 2 n n 1 2 3 4 The minimum energy is E 1 2 h 2 mL 8 L h mE 1 8 6 63 10 34 Js 8 9 11 10 31 kg 1 5 10 18 J 2 0 10 10 m 0 20 …
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