91.18 nm nm. λ = λ = nm. where n = 3, 4, 5, 6, and where we have used n = 6. Likewise for n = 8 and n = 10, λ = nm and.

Transcription

1 4.1. Model: Balmer s formula predicts a series of spectral lines in the hydrogen spectrum. Solve: Substituting into the formula for the Balmer series, nm λ = 1 1 n nm λ = = nm where n = 3, 4, 5, 6, and where we have used n = 6. Likewise for n = 8 and n = 10, λ = nm and λ = nm.

2 4.. Model: Balmer s formula predicts a series of spectral lines in the hydrogen spectrum. Solve: Balmer s formula is nm λ = 1 1 n As n, 1 n 0. Thus, λ n = 4( nm)= nm. n = 3, 4, 5, 6,

3 4.3. Model: Balmer s formula predicts a series of spectral lines in the hydrogen spectrum. Solve: Using Balmer s formula, λ = nm = nm 1 1 = n = n n

4 4.4. Model: The angles of incidence for which diffraction from parallel planes occurs satisfy the Bragg condition. Solve: The Bragg condition is dcosθ = mλ, where m = 1,, 3, For first and second order diffraction, Dividing these two equations, cosθ cosθ 1 m d cosθ = () 1 λ d cosθ λ 1 = ( ) 1 1 = θ = cos ( cosθ )= cos ( cos 68 )=

5 4.5. Model: The angles of incidence for which diffraction from parallel planes occurs satisfy the Bragg condition. Solve: The Bragg condition is dcosθ = mλ. For m = 1 and for two different wavelengths, Dividing these two equations, m d cosθ = () 1 λ d cos θ = () 1 λ cosθ1 λ cos = 1 θ 1 cos.. cos cos = 0.15 nm = 1 θ θ λ ( )= nm 1 1

6 4.6. Model: The angles corresponding to the various orders of diffraction satisfy the Bragg condition. Solve: The Bragg condition for m = 1 and m = gives Dividing these two equations, dcosθ 1 λ dcosθ λ = () = ( ) 1 cosθ cos45 1 cos45 cosθ1 = = θ1 = cos =

7 4.7. Model: The angles corresponding to the various diffraction orders satisfy the Bragg condition. Solve: The Bragg condition is dcosθm = mλ, where m = 1,, 3, The maximum possible value of m is the number of possible diffraction orders. The maximum value of cosθ m is 1. Thus, d d = m λ m = = ( nm ) 4. λ nm We can observe up to the fourth diffraction order. ( ) =

8 4.8. Model: Use the photon model of light. Solve: The energy of the photon is ( ) E = hf = h c = λ 34 photon Js m/s = m Assess: The energy of a single photon in the visible light region is extremely small. 19 J

9 4.9. Model: Use the photon model of light. Solve: The energy of the x-ray photon is 34 ( Js) E = hf = h c = 663. λ m/s = m Assess: This is a very small amount of energy, but it is larger than the energy of a photon in the visible wavelength region. 16 J

10 4.10. Model: Use the photon model of light. Solve: The energy of a photon with wavelength λ 1 is E1 = hf1 = hc λ 1. Similarly, E = hc λ. Since E be equal to E 1, hc hc λ1 = λ λ λ = = 600 nm = 300 nm 1 Assess: A photon with λ = 300 nm has twice the energy of a photon with λ = 600 nm. This is an expected result, because energy is inversely proportional to the wavelength.

11 4.11. Model: Use the photon model of light. Solve: The energy of the single photon is ( )( ) E hf h c Js m / s photon = = 34 8 = λ m = J Assess: Emol = NAEphoton = ( )( J )= J Although the energy of a single photon is very small, a mole of photons has a significant amount of energy.

12 4.1. 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 34 h Js λ = = 9 10 p 70 kg m / s 36 m

13 ( ) Solve: (a) For an electron, the momentum p = mv = kg v. The de Broglie wavelength is h λ = = m m = p ( ) Js v = kg v ( ) (b) For a proton, p = mv = kg v. The de Broglie wavelength is h 9 9 λ = = m m = p ( kg) 6 m/s Js 3 v = m/s v

14 4.14. Solve: The kinetic energy of a particle of mass m is related to its momentum by E = p m. Using λ = hp, for an electron we have ( ) ( )( ) 34 h Js E = = mλ kg m = J

15 4.15. Solve: (a) The baseball s momentum is p = mv = (0.00 kg)(30 m/s) = 6.0 kg m/s. The baseball s de Broglie wavelength is 34 h Js 34 λ = = = m p 6.0 kg m / s (b) Using λ = hp= hmv, we have h v = = mλ Js kg m ( ) ( ) = m/s

16 4.16. Model: The momentum of a wave-like particle has discrete values given by p n h L 1,, 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, 34 h h Js p1 = mv = L = = 31 L mv kg 10 m / s mm ( )( ) = n = ( )

17 4.17. Model: A confined particle can have only discrete values of energy. Solve: From Equation 4.14, the energy of a confined electron is The minimum energy is E 1 E h h = L = = 8mL 8mE 1 n = h ml n 8 n = 1,, 3, 4, Js kg J ( )( ) = 10 m = 0.0 nm

18 4.18. Model: Model the 5.0-fm-diameter nucleus as a box of length L = 5.0 fm. Solve: The proton s energy is restricted to the discrete values ( ) 1 = ( J) n ( )( ) = (. ) =. 34 h E ml n Js n n = = kg m where n = 1,, 3, For n = 1, E 1 = J, for n =, E J J, and for n = 11 3, E = 9E = J. 3 1

19 4.19. Model: The generalized formula of Balmer predicts a series of spectral lines in the hydrogen spectrum. Solve: (a) The generalized formula of Balmer m λ = 1 1 m n with m = 1 and n > 1 accounts for a series of spectral lines. This series is called the Lyman series and the first two members are λ1 =. m = λ = nm 11.6 nm = 10.6 nm For n = 4 and n = 5, λ3 = 97.3 nm and λ4 = nm. (b) The Lyman series converges when n. This means 1 n 0 and λ nm. (c) For a diffraction grating, the condition for bright (constructive interference) fringes is dsinθ p = pλ, where p = 1,, 3, For first-order diffraction, this equation simplifies to d sinθ = λ. For the first and second members of the Lyman series, the above condition is dsinθ1 = λ1 = 11.6 nm and dsinθ = λ = 10.6 nm. Dividing these two equations yields sinθ = 10.6 nm sin θ1 = ( ) sinθ nm The distance from the center to the first maximum is y = Ltanθ. Thus, y m tan θ1 = = θ1 = sin θ = ( ) sin ( ) θ = L 1.5 m Applying the position formula once again, y = Ltanθ = ( 1.5 m) tan ( )= m = 31.4 cm

20 4.0. Model: The generalized formula of Balmer predicts a series of spectral lines in the hydrogen spectrum. Solve: (a) The generalized formula of Balmer m λ = 1 1 m n with m = 3, and n > 3 accounts for a series of spectral lines. This series is called the Paschen series and the wavelengths are λ = nm = 80.6 n 1 1 n 9 3 n The first four members are λ1 = 1876 nm, λ = 18 nm, λ3 = 1094 nm, and λ4 = 1005 nm. (b) The Paschen series converges when n. This means 1 0 λ n ( ) nm = 81 nm (c) For a diffraction grating, the condition for bright (constructive interference) fringes is dsinθ p = pλ, where p = 1,, 3, For first-order diffraction, this equation simplifies to d sinθ = λ. For the first and second members of the Paschen series, the condition is d sinθ = λ and d sinθ = λ. Dividing these two equations yields 1 1 sinθ sinθ λ = 1 sin θ sinθ1 λ1 = 18 nm 1876 nm = ( ). The distance from the center to the first maximum is y = Ltanθ. Thus, y m tan θ 1 = = = L 1.5 m Applying the position formula once again, θ 1 = 03. sin θ sin. 03 θ = ( ) = y = Ltanθ = ( 1.5 m) tan = m = 39.8 cm

21 4.1. Model: Use the photon model of light. Solve: (a) The wavelength is calculated as follows: ( )( ) E hf h c Js m / s gamma = = 34 8 = λ 13 λ J (b) The energy of a visible-light photon of wavelength 500 nm is E visible The number of photons n such that E ( )( ) h c Js m / s = 34 8 = λ m gamma = ne is visible Egamma n = = E visible J = J = = J m

22 4.. Model: Use the photon model. Solve: The energy of a 1000 khz photon is Ephoton = hf = ( Js )( Hz )= J The energy transmitted each second is J. The number of photons transmitted each second is J/ J =

23 4.3. Model: Use the photon model for the laser light. Solve: (a) The energy is E hf h c m/s photon = = 8 Js = ( ) = λ m (b) The energy emitted each second is J = J. The number of photons emitted each second is J J/

24 4.4. Model: Use the photon model for the incandescent light. Solve: The photons travel in all directions. At a distance of r from the light bulb, the photons spread over a sphere of surface area 4πr. The number of photons per second per unit area at the location of your retina is s =. s m 4π m ( ) The number of photons that enter your pupil per second is 4 1 ( ) = s 1 m π m s

25 4.5. Model: Use the photon model of light and the Bragg condition for diffraction. Solve: The Bragg condition for the reflection of x-rays from a crystal is dcosθm = mλ. To determine the angles of incidence θ m, we need to first calculate λ. The wavelength is related to the photon s energy as E = hc λ. Thus, 34 8 hc ( Js )( m / s) 10 λ = = = m 15 E J From the Bragg condition, ( ) 10 mλ m m θ m = cos = cos = cos ( m) θ = cos (. )= d ( m) Likewise, θ 1 = cos ( )= and θ 3 = Note that θ 1 4 = cos ( ) is not allowed because the cosθ cannot be larger than 1. Thus, the x-rays will be diffracted at angles of incidence equal to 18.7, 50.8, and 71.6.

26 4.6. Model: The angles for which diffraction from parallel planes occurs satisfy the Bragg condition. Solve: We cannot assume that these are the first and second order diffractions. The Bragg condition is dcosθ = mλ. We have m ( ) dcos 456. = mλ dcos 10. = m+ 1λ Notice that θ m decreases as m increases, so 1.6 corresponds to the larger value of m. Dividing these two equations, cos cos 1. 0 = m + 1 = m m = Thus these are the third and fourth order diffractions. Substituting into the Bragg condition, d = m 10 = m = nm cos 45. 6

27 4.7. Model: The x-ray diffraction angles satisfy the Bragg condition. Solve: (a) The Bragg condition ( dcosθm = mλ) for normal incidence, θ m = 0, simplifies to d = mλ. For a thin film of a material on a substrate where n air < n material < n substrate, constructive interference between the two reflected waves occurs when d = mλ, where λ is the wavelength inside the material. (b) From a thin film with a period of 1. nm, that is, with d = 1. nm, the two longest x-ray wavelengths that will reflect at normal incidence are λ 1 = d λ = d 1 This means that λ 1 = ( 1. nm)=.4 nm and λ = 1. nm.

28 4.8. Solve: A small fraction of the light wave of an appropriate wavelength is reflected from each little bump in the refractive index. These little bumps act like the atomic planes in a crystal. The light will be strongly reflected (and hence blocked in transmission) if it satisfies the Bragg condition at normal incidence (θ = 0). 6 mλ dn m glass d = mλglass = λ = = n m 1 glass ( )( ) = 135. µm

29 4.9. Model: Particles have a de Broglie wavelength given by λ = hp. The wave nature of the particles causes an interference pattern in a double-slit apparatus. Solve: (a) Since the speed of the neutron and electron are the same, the neutron s momentum is mn p m v m mv mn m mv mn m p n = n n = e n = e e = e e e e where m n and m e are the neutron s and electron s masses. The de Broglie wavelength for the neutron is h h pe me λn = = = λe pn pe pn mn From Section. on double-slit interference, the fringe spacing is y = λ L/ d. Thus, the fringe spacing for the electron and neutron are related by 31 λn me y y m y kg 3 7 n = e = e = ( m )= m = µ m 7 λe n kg (b) If the fringe spacing has to be the same for the neutrons and the electrons, h h y y v v m 31 e kg 3 e = n λe = λn = n = e = ( m/s) = m/s 7 mv mv m kg e e n n n

30 4.30. Model: Electrons have a de Broglie wavelength given by λ = hp. The wave nature of the electrons causes a diffraction pattern. Solve: The width of the central maximum of a single-slit diffraction pattern is given by Equation.: ( )( ) ( )( )( ) = m = 34 L Lh Lh w = λ a = ap = amv = 1.0 m Js m kg m / s mm

31 4.31. Model: Neutrons have a de Broglie wavelength given by λ = hp. The wave nature of the neutrons causes a double-slit interference pattern. Solve: Measurements show that the spacing between the m = 1 and m = 1 peaks is 1.4 times as long as the length of the reference bar, which gives the real fringe separation y = 70 µm. Similarly, the spacing between the m = and m = is.8 times as long as the length of the reference bar and yields y = 70 µm. The fringe separation in a double-slit experiment is y = λ L d. Hence, 34 yd h h yd hl ( Js)( 3.0 m) λ = = = v = = L p mv L ymd m kg m 170 m / s ( )( )( ) =

32 4.3. Model: Electrons have a de Broglie wavelength given by λ = hp. Visualize: Please refer to Figure Notice that a scattering angle φ = 60 corresponds to an angle of incidence θ = 30. Solve: Equation 4.6 describes the Davisson-Germer experiment: Dsin( θm )= mλ. Assuming m = 1, this equation simplifies to Dsinθ = λ. Using λ = hmv, we have h D = = mvsinθ Js kg m / s sin 60 ( )( ) ( ) = = m nm

33 4.33. Model: A confined particle can have only discrete values of energy. Solve: (a) Equation 4.14 simplifies to 34 h ( E ml n Js ) n = = = ( ) ( kg )( m) ( J)( 1 )= J, E = ( )( )= Thus, E 1 = J n J J, and E 3 = J (b) The energy is E E1 = J J = J. 19 (c) Because energy is conserved, the photon will carry an energy of E E1 = J. That is, 34 8 hc hc ( Js )( m / s) E E1 = Ephoton = hf = λ = = = 539 nm 19 λ E E J 1

34 4.34. Model: A particle confined in a one-dimensional box has discrete energy levels. Solve: (a) Equation 4.14 for the n = 1 state is ( ) 34 h Js En = = 8mL kg 0.10 m ( )( ) 3 64 The minimum energy of the Ping-Pong ball is E 1 = J. (b) The speed is calculated as follows: E = J = mv = kg v v 1 ( ) = = J 64 (. J) kg = m/s

35 4.35. Model: A particle confined in a one-dimensional box has discrete energy levels. Solve: Using Equation 4.14 for n = 1 and, E h E = 1 8mL 1 19 ( ) J = L = Jm J ( Js) kg L ( ) ()= 3 9 = m = 1.35 nm L Jm

36 4.36. Model: A particle confined in a one-dimensional box has discrete energy levels. Solve: From Equation 4.14, Taking the ratio, Thus E h ml n 8 n = n n En E + 1 = n h E = ml n + n n + n = = = ( ) The two allowed energies have n = 3 and n = 4. Using n = 3 in the first energy equation, 34 ( Js 19 ) En = J = () 3 L = 1.3 nm kg L ( )

37 4.37. Model: A particle confined in a one-dimensional box of length L has the discrete energy levels given by Equation Solve: (a) Since the energy is entirely kinetic energy, (b) The first allowed velocity is E h ml n h 1 = = = mv v = pm ml n n = 1,, 3,K 8 n n n Js kg m 34 v 1 = 31 9 ( )( ) = For n = and n = 3, v = m/s and v 3 = m/s m / s

38 4.38. Model: The allowed energies of a particle of mass m in a two-dimensional square box of side L are h E = ml n + m nm ( ) 8 Solve: (a) The minimum energy for a particle is for n = m = 1: h h E = min E = ( ml )= 8 4mL (b) The five lowest allowed energies are E min, 5 E min (for n = 1, m = and n =, m = 1), 4E min (for n =, m = ), 5E min (for n = 1, m = 3 and n =3, m = 1), and 13 E min (for n =, m = 3 and n = 3, m = ).

39 4.39. Model: Sets of parallel planes in a crystal diffract x-rays. Visualize: Please refer to Figure 4.7. Solve: The Bragg diffraction condition is dcosθm = mλ. The plane spacing is d A = 0.0 nm and the x-ray wavelength is λ = 01. nm. Thus 9 mλ m( m) 1 cosθ m = = 03. m θ cos A1 d m A ( ) = ( ) = ( )= Likewise for m = and m = 3, θ A = cos 1 ( 0. 6 )= and θ = cos 1( 0. 9 )= These three angles for the A3 x-ray diffraction peaks match the peaks shown in Figure 4.7c. (b) The new interplaner spacing is db = da = nm (see Figure 4.7b). The Bragg condition for the tilted atomic planes becomes mλ cos θ m = = m d B For m = 1, θ B1 = cos 1 ( )= For m =, θ = cos 1( )= B (c) The crystal is already tipped by 45 to get the tilted planes (see Figure 4.7b). So, for m = 1, θ 1 = = θ = = also, but we can t see beyond 90. For m =, θ = = These two angles match the angles in the diffraction peaks of the tilted planes.

40 4.40. Model: Sets of parallel planes in a crystal diffract x-rays. Visualize: Please refer to Figure CP4.40. Solve: The Bragg diffraction condition is dcosθm = mλ, where d is the interplanar separation. Because smaller m values correspond to higher angles of incidence, the diffraction angle of 71.3 in the x-ray intensity plot must correspond to m = 1. This means m dcos = 1( m) d = = cos ( ) The cosines of the three angles in the X-ray intensity plot are cos =0.31, cos =0.64, and cos = These are in the ratio 1::3, which tells us that these are the m = 1,, and 3 diffraction peaks from a single set of planes with d = nm. We can see from the figure that the atomic spacing D of this crystal is related to the interplanar separation d by d D = = nm = nm sin60 sin60 10 m

41 4.41. Model: Electrons have a de Broglie wavelength given by λ = hp. Trapped electrons in the confinement layer behave like a de Broglie wave in a closed-closed tube or like a string fixed at both ends. Solve: (a) The four longest standing-wave wavelengths in the layer are λ = L, L, 3 L, and 1 L. This follows from the general relation for closed-closed tubes: λ = L/ n. Thus, λ = 10.0 nm, 5.0 nm, 3.33 nm, and.50 nm. (b) We have 34 h h Js p = mv = v = = = 31 λ mλ kg λ λ ( ) Using the above four longest values of λ we get the four smallest values of v. Thus, m /s 4 v 1 = = m / s m v = m / s, v 3 = m / s, and v 4 = m / s. 3 m /s

42 4.4. Model: As light is diffracted by matter, matter can also be diffracted by light. Visualize: Please refer to Figure CP4.4. Solve: The de Broglie wavelength of the sodium atoms is h h λ = = = p mv Js kg 50 m / s ( )( ) = The slit spacing of the diffraction grating is d = λ laser = 600 nm = 300 nm. Using the diffraction grating equation with m = 1, we have λ d sinθ = () 1 λ sin θ = = rad d In the small-angle approximation, sinθ tanθ = y L. We get y = Lsin θ = ( 1.0 m) ( )= 1.15 mm 10 m