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Doppler Shifts of Interstellar NaD lines

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1 Doppler Shifts of Interstellar NaD lines
589.00 589.59 (nm) / = / = v/c Harry Kroto 2004

2 The Discovery of the Interstellar Medium (ISM)
Harry Kroto 2004

3 Harry Kroto 2004

4 Hartman 1904 K Ti Fe Ca Na T (100K) Binary Diffuse cloud
Harry Kroto 2004

5 Blue Shifted Red Shifted Harry Kroto 2004

6 Blue Shifted Red Shifted Harry Kroto 2004

7 Several clouds in the line of sight
Harry Kroto 2004

8 Doppler Shifts in Interstellar lines
/ = / = v/c Harry Kroto 2004

9 Absorption of Interstellar CN radicals in a Stellar Spectrum
Harry Kroto 2004

10 Interstellar CN radicals
H-C≡N → H. + .C≡N hν .CN Binary Diffuse cloud Harry Kroto 2004

11 ΔE Harry Kroto 2004

12 ΔE n0 population of ground state Harry Kroto 2004

13 ΔE n1 population of excited state n0 population of ground state
Harry Kroto 2004

14 ΔE Boltzmann Equation n1 population of excited state
n0 population of ground state Harry Kroto 2004

15 ΔE n1 = noe –ΔE/kT Boltzmann Equation n1 population of excited state
n0 population of ground state n1 = noe –ΔE/kT Harry Kroto 2004

16 This is one of the most important equations of all
Boltzmann Equation This is one of the most important equations of all n1 = noe –ΔE/kT Harry Kroto 2004

17 This is one of the most important equations of all
Boltzmann Equation This is one of the most important equations of all so remember it n1 = noe –ΔE/kT Harry Kroto 2004

18 Only three lines observed R(0) R(1) P(1)
2 J’ Only three lines observed R(0) R(1) P(1) 1 R(0) 2 J” 1 Harry Kroto 2004

19 Only three lines observed R(0) R(1) P(1)
2 J’ Only three lines observed R(0) R(1) P(1) The detection of R(1) and P(1) indicates T > 0 K 1 R(1) R(0) 2 J” 1 Harry Kroto 2004

20 2 J’ 1 P(1) R(1) R(0) 2 J” 1 Harry Kroto 2004

21 I ∞ n The intensity I of an absorption line is proportional to n - the number of molecules in the lower state Harry Kroto 2004

22 Harry Kroto 2004

23 R(1)/R(0) = SR(1)N1/SR(0)N0
Harry Kroto 2004

24 R(1)/R(0) = SR(1)N1/SR(0)N0
SR(J) = (2J+2)/(2J+1) Harry Kroto 2004

25 R(1)/R(0) = SR(1)N1/SR(0)N0 SR(J) = (2J+2)/(2J+1)
so for the R(1) and R(0) lines SR(1) = 4/ SR(0) = 2/1 Harry Kroto 2004

26 R(1)/R(0) = SR(1)N1/SR(0)N0 SR(J) = (2J+2)/(2J+1)
so for the R(1) and R(0) lines SR(1) = 4/ SR(0) = 2/1 SR(1)/SR(0) = ⅔ Harry Kroto 2004

27 R(1)/R(0) = SR(1)N1/SR(0)N0 SR(J) = (2J+2)/(2J+1)
so for the R(1) and R(0) lines SR(1) = 4/ SR(0) = 2/1 SR(1)/SR(0) = ⅔ R(1)/R(0) = (⅔) N1/N0 Harry Kroto 2004

28 R(1)/R(0) = SR(1)N1/SR(0)N0 SR(J) = (2J+2)/(2J+1)
so for the R(1) and R(0) lines SR(1) = 4/ SR(0) = 2/1 SR(1)/SR(0) = ⅔ R(1)/R(0) = (⅔) N1/N0 = IR(1)/IR(1) Harry Kroto 2004

29 (⅔)(N1/N0) = I1/I0 Harry Kroto 2004

30 (⅔)(N1/N0) = I1/I0 N1/N0 = (3/2)(I1/I0) Harry Kroto 2004

31 (⅔)(N1/N0) = I1/I0 N1/N0 = (3/2)(I1/I0) NJ = N0(2J+1)e – BJ(J+1)/kT
Harry Kroto 2004

32 (⅔)(N1/N0) = I1/I0 N1/N0 = (3/2)(I1/I0) NJ = N0(2J+1)e – BJ(J+1)/kT
N1/N0 = 3 e –2B/kT Harry Kroto 2004

33 (⅔)(N1/N0) = I1/I0 N1/N0 = (3/2)(I1/I0) NJ = N0(2J+1)e – BJ(J+1)/kT
N1/N0 = 3 e –2B/kT 2B = GHz Harry Kroto 2004

34 (⅔)(N1/N0) = I1/I0 N1/N0 = (3/2)(I1/I0) NJ = N0(2J+1)e – BJ(J+1)/kT
N1/N0 = 3 e –2B/kT 2B = GHz (3/2)(29/86) = 3e-113.5/20.85T Harry Kroto 2004

35 (⅔)(N1/N0) = I1/I0 N1/N0 = (3/2)(I1/I0) NJ = N0(2J+1)e – BJ(J+1)/kT
N1/N0 = 3 e –2B/kT 2B = GHz (3/2)(29/86) = 3e-113.5/20.85T I1/I0 = 2 e-113.5/20.85T Harry Kroto 2004

36 Problem Determine the temperature of interstellar CN radicals by measuring the ratio of the heights of the R(1) and R(0) lines = I1/I0 I1/I0 = 2e – [F(1) - F(0)]/kT k = if F(J) in GHz For the CO molecule B = GHz Thus determine T (K) R(1) R(0) J =1 J = 0 F(J) = BJ(J+1) Harry Kroto 2004

37 Harry Kroto 2004

38

39 Problem Determine the temperature of interstellar CN radicals by measuring the ratio of the heights of the R(1) and R(0) lines = I1/I0 I1/I0 = 2e – [F(1) - F(0)]/kT k = if F(J) in GHz For the CO molecule B = GHz Thus determine T (K) R(1) R(0) J =1 J = 0 F(J) = BJ(J+1) Harry Kroto 2004

40

41

42 35/105 = Harry Kroto 2004

43 29/176 = = e-113.5/20.85T ln = /20.85T = /20.85T T = 113.5/(1.802 x 20.85) = 3.02 K error ca 3-4% k = cm-1 or GHz cm-1 = 30 GHz Harry Kroto 2004

44 29/176 = = e-113.5/20.85T ln = /20.85T = /20.85T T = 113.5/(1.802 x 20.85) = 3.02 K error ca 3-4% k = cm-1 or GHz cm-1 = 30 GHz Harry Kroto 2004

45

46 (⅔)(N1/N0) = 29/86 N1/N0 = (3/2)(29/86) NJ = N0(2J+1)e – BJ(J+1)/kT
N1/N0 = 3 e –2B/kT 2B = GHz (3/2)(29/86) = 3e-113.5/20.85T 29/176 = e-113.5/20.85T = 0.165 Harry Kroto 2004

47

48

49

50 Only three lines observed R(0) R(1) P(1)
2 J’ Only three lines observed R(0) R(1) P(1) The detection of R(1) and P(1) indicates T > 0 K 1 P(1) R(1) R(0) 2 J” 1 Harry Kroto 2004

51 Absorption of Interstellar CN radicals in a Stellar Spectrum
Harry Kroto 2004

52 Harry Kroto 2004

53 Only three lines observed R(0) R(1) P(1)
2 J’ Only three lines observed R(0) R(1) P(1) The detection of R(1) and P(1) indicates T > 0 K 1 P(1) R(1) R(0) 2 J” 1 Harry Kroto 2004

54 Harry Kroto 2004

55 Harry Kroto 2004

56 Harry Kroto 2004

57 Harry Kroto 2004

58 Harry Kroto 2004

59 Harry Kroto 2004

60 Harry Kroto 2004

61 Only three lines observed R(0) R(1) P(1)
2 J’ Only three lines observed R(0) R(1) P(1) The detection of R(1) and P(1) indicates T > 0 K 1 P(1) R(1) R(0) 2 J” 1 Harry Kroto 2004

62 Only three lines observed R(0) R(1) P(1)
2 J’ Only three lines observed R(0) R(1) P(1) The detection of R(1) and P(1) indicates T > 0 K 1 P(1) R(1) R(0) 2 J” 1 Harry Kroto 2004

63 Only three lines observed R(0) R(1) P(1)
2 J’ Only three lines observed R(0) R(1) P(1) The detection of R(1) and P(1) indicates T > 0 K 1 P(1) R(1) R(0) 2 J” 1 Harry Kroto 2004

64 Only three lines observed R(0) R(1) P(1)
2 J’ Only three lines observed R(0) R(1) P(1) The detection of R(1) and P(1) indicates T > 0 K 1 P(1) R(1) R(0) 2 J” 1 Harry Kroto 2004

65 Only three lines observed R(0) R(1) P(1)
2 J’ Only three lines observed R(0) R(1) P(1) The detection of R(1) and P(1) indicates T > 0 K 1 P(1) R(1) R(0) 2 J” 1 Harry Kroto 2004

66 IR(1) /IR(0) ~ 29/86 = 0.337 [Measured in mm]
Harry Kroto 2004

67 Only three lines observed R(0) R(1) P(1)
The detection of R(1) and P(1) indicates T> 0K Harry Kroto 2004

68 Only three lines observed R(0) R(1) P(1)
The detection of R(1) and P(1) indicates T> 0K  l  Io I I I = Ioe- l I = Io (1 - l + …) Io - I = I ~ l Harry Kroto 2004

69 IR(1) /IR(0) ~ R(1) /R(0)
Only three lines observed R(0) R(1) P(1) The detection of R(1) and P(1) indicates T> 0K  l  Io I I I = Ioe- l I = Io (1 - l + …) (Io – I)/ Io = I/ Io ~ l IR(1) /IR(0) ~ R(1) /R(0) Harry Kroto 2004

70 IR(1) /IR(0) ~ R(1) /R(0)
Only three lines observed R(0) R(1) P(1) The detection of R(1) and P(1) indicates T> 0K  l  Io I I I = Ioe- l I = Io (1 - l + …) (Io – I)/ Io = I/ Io ~ l IR(1) /IR(0) ~ R(1) /R(0) Harry Kroto 2004

71 IR(1) /IR(0) ~ R(1) /R(0)
Only three lines observed R(0) R(1) P(1) The detection of R(1) and P(1) indicates T> 0K  l  Io I I I = Ioe- l I = Io (1 - l + …) (Io – I)/ Io = I/ Io ~ l IR(1) /IR(0) ~ R(1) /R(0) Harry Kroto 2004

72 IR(1) /IR(0) ~ R(1) /R(0)
Only three lines observed R(0) R(1) P(1) The detection of R(1) and P(1) indicates T> 0K  l  Io I I I = Ioe- l I = Io (1 - l + …) (Io – I)/ Io = I/ Io ~ l IR(1) /IR(0) ~ R(1) /R(0) Harry Kroto 2004

73 IR(1) /IR(0) ~ R(1) /R(0) 29/86 = 0.337 [Measured in mm]
Harry Kroto 2004

74  = (4/3ħc) n em2  (Nm-Nn) (o-) Nn = 0 R(1) ~ R(0)
the Hönl London formulae J+1 eJ2  SR(J) J eJ+12  SP(J) R(1)  SR(1)N R(0)  SR(0)N0 SR(J) = (2J+2)/(2J+1) Harry Kroto 2004

75  = (4/3ħc) n em2  (Nm-Nn) (o-) Nn = 0 R(1) ~ R(0)
the Hönl London formulae J+1 eJ2  SR(J) J eJ+12  SP(J) R(1)  SR(1)N R(0)  SR(0)N0 SR(J) = (2J+2)/(2J+1) Harry Kroto 2004

76 Two serendipitous radio discoveries
Harry Kroto 2004

77

78 Harry Kroto 2004

79 Electronic Emission Spectrum of CN Radical in a Bunsen Burner Flame
v’= 0 C + N v”=3 2 1 r  Harry Kroto 2004

80 Harry Kroto 2004

81 Rotational Structure J’= 5 v’ = 0 v” = 0 J”= 5 R Branch P Branch
v” = 0 J”= 5 R Branch P Branch Harry Kroto 2004

82 J’+ 1 B’ (J+ 1)(J+2) R(J) P(1) B” J (J+ 1) J” R Branch If B” ~ B’
o + 2B J Harry Kroto 2004

83 J’= 5 v’ = 0 v” = 0 J”= 5 R Branch P Branch Harry Kroto 2004

84 Rotational Structure J’= 5 v’ = 0 v” = 0 J”= 5 R Branch P Branch R(0)
R(0) v” = 0 J”= 5 R Branch P Branch Harry Kroto 2004

85 Rotational Structure J’= 5 v’ = 0 v” = 0 J”= 5 R Branch P Branch R(1)
R(1) R(0) v” = 0 J”= 5 R Branch P Branch Harry Kroto 2004

86 Rotational Structure J’= 5 v’ = 0 v” = 0 J”= 5 R Branch P Branch R(2)
R(2) R(1) R(0) v” = 0 J”= 5 R Branch P Branch Harry Kroto 2004

87 Rotational Structure J’= 5 v’ = 0 v” = 0 J”= 5 R Branch P Branch R(3)
R(3) R(2) R(1) R(0) v” = 0 J”= 5 R Branch P Branch Harry Kroto 2004

88 Rotational Structure J’= 5 v’ = 0 v” = 0 J”= 5 R Branch P Branch R(3)
R(3) R(2) R(1) P(1) R(0) v” = 0 J”= 5 R Branch P Branch Harry Kroto 2004

89 Rotational Structure J’= 5 v’ = 0 v” = 0 J”= 5 R Branch P Branch R(3)
R(3) R(2) P(2) R(1) P(1) R(0) v” = 0 J”= 5 R Branch P Branch Harry Kroto 2004

90 Rotational Structure J’= 5 v’ = 0 v” = 0 J”= 5 R Branch P Branch R(3)
R(3) P(3) R(2) P(2) R(1) P(1) R(0) v” = 0 J”= 5 R Branch P Branch Harry Kroto 2004

91 Rotational Structure J’= 5 v’ = 0 v” = 0 J”= 5 R Branch P Branch P(4)
P(4) R(3) P(3) R(2) P(2) R(1) P(1) R(0) v” = 0 J”= 5 R Branch P Branch Harry Kroto 2004

92 Rotational Structure J’= 5 v’ = 0 v” = 0 J”= 5 Central section of the
P(4) R(3) P(3) R(2) P(2) R(1) P(1) R(0) v” = 0 J”= 5 Central section of the CN Band R Branch P Branch Harry Kroto 2004

93 Harry Kroto 2004

94 Electronic Emission Spectrum
3883 Å Å Å CN Violet Electronic Emission Spectrum 0-0 R(0) P(1) R Branch P Branch cm-1 25750 cm-1 Rotational Structure of the 0-0 band of CN at 3883Å observed from Comet Bennett (1970 II) Harry Kroto 2004

95 Harry Kroto 2004

96 Harry Kroto 2004

97  = (4/3ħc) n em2  (Nm-Nn) (o-) Nn = 0 R(1) ~ R(0)
the Hönl London formulae J+1 eJ2  SR(J) J eJ+12  SP(J) R(1)  SR(1)N R(0)  SR(0)N0 SR(J) = (2J+2)/(2J+1) Harry Kroto 2004

98  = (4/3ħc) n em2  (Nm-Nn) (o-) Nn = 0 R(1) ~ R(0)
the Hönl London formulae J+1 eJ2  SR(J) J eJ+12  SP(J) R(1)  SR(1)N R(0)  SR(0)N0 SR(J) = (2J+2)/(2J+1) Harry Kroto 2004

99  = (4/3ħc) n em2  (Nm-Nn) (o-) Nn = 0 R(1) ~ R(0)
the Hönl London formulae J+1 eJ2  SR(J) J eJ+12  SP(J) R(1)  SR(1)N R(0)  SR(0)N0 SR(J) = (2J+2)/(2J+1) Harry Kroto 2004

100  = (4/3ħc) n em2  (Nm-Nn) (o-) Nn = 0 R(1) ~ R(0)
the Hönl London formulae J+1 eJ2  SR(J) J eJ+12  SP(J) R(1)  SR(1)N R(0)  SR(0)N0 R(1)/R(0) = SR(1)N1/SR(0)N0 SR(J) = (2J+2)/(2J+1) Harry Kroto 2004

101  = (4/3ħc) n em2  (Nm-Nn) (o-) Nn = 0 R(1) ~ R(0)
the Hönl London formulae J+1 eJ2  SR(J) J eJ+12  SP(J) R(1)  SR(1)N R(0)  SR(0)N0 R(1)/R(0) = SR(1)N1/SR(0)N0 SR(J) = (2J+2)/(2J+1) Harry Kroto 2004

102 R(1)/R(0) = SR(1)N1/SR(0)N0 SR(J) = (2J+2)/(2J+1)
so for the R(1) and R(0) lines SR(1) = 4/ SR(0) = 2/1 SR(1)/SR(0) = ⅔ R(1)/R(0) = (⅔) N1/N0 = IR(1)/IR(1) = 29/86 Harry Kroto 2004

103 R(1)/R(0) = SR(1)N1/SR(0)N0 SR(J) = (2J+2)/(2J+1)
so for the R(1) and R(0) lines SR(1) = 4/ SR(0) = 2/1 SR(1)/SR(0) = ⅔ R(1)/R(0) = (⅔) N1/N0 = IR(1)/IR(1) = 29/86 Harry Kroto 2004

104 R(1)/R(0) = SR(1)N1/SR(0)N0 SR(J) = (2J+2)/(2J+1)
so for the R(1) and R(0) lines SR(1) = 4/ SR(0) = 2/1 SR(1)/SR(0) = ⅔ R(1)/R(0) = (⅔) N1/N0 = IR(1)/IR(1) = 29/86 Harry Kroto 2004

105 R(1)/R(0) = SR(1)N1/SR(0)N0 SR(J) = (2J+2)/(2J+1)
so for the R(1) and R(0) lines SR(1) = 4/ SR(0) = 2/1 SR(1)/SR(0) = ⅔ R(1)/R(0) = (⅔) N1/N0 = IR(1)/IR(1) = 29/86 Harry Kroto 2004

106 R(1)/R(0) = SR(1)N1/SR(0)N0 SR(J) = (2J+2)/(2J+1)
so for the R(1) and R(0) lines SR(1) = 4/ SR(0) = 2/1 SR(1)/SR(0) = ⅔ R(1)/R(0) = (⅔) N1/N0 = IR(1)/IR(1) = 29/86 Harry Kroto 2004

107 R(1)/R(0) = SR(1)N1/SR(0)N0 SR(J) = (2J+2)/(2J+1)
so for the R(1) and R(0) lines SR(1) = 4/ SR(0) = 2/1 SR(1)/SR(0) = ⅔ R(1)/R(0) = (⅔) N1/N0 = IR(1)/IR(1) = 29/86 Harry Kroto 2004

108 R(1)/R(0) = SR(1)N1/SR(0)N0 SR(J) = (2J+2)/(2J+1)
so for the R(1) and R(0) lines SR(1) = 4/ SR(0) = 2/1 SR(1)/SR(0) = ⅔ R(1)/R(0) = (⅔) N1/N0 = IR(1)/IR(1) = I(1)/I(0) Harry Kroto 2004

109 (⅔)(N1/N0) = 29/86 N1/N0 = (3/2)(29/86) NJ = N0(2J+1)e – BJ(J+1)/kT
N1/N0 = 3 e –2B/kT 2B = GHz (3/2)(29/86) = 3e-113.5/20.85T 29/176 = e-113.5/20.85T = 0.165 Harry Kroto 2004

110 (⅔)(N1/N0) = 29/86 N1/N0 = (3/2)(29/86) NJ = N0(2J+1)e – BJ(J+1)/kT
N1/N0 = 3 e –2B/kT 2B = GHz (3/2)(29/86) = 3e-113.5/20.85T 29/176 = e-113.5/20.85T = 0.165 Harry Kroto 2004

111 (⅔)(N1/N0) = 29/86 N1/N0 = (3/2)(29/86) NJ = N0(2J+1)e – BJ(J+1)/kT
N1/N0 = 3 e –2B/kT 2B = GHz (3/2)(29/86) = 3e-113.5/20.85T 29/176 = e-113.5/20.85T = 0.165 Harry Kroto 2004

112 29/176 = = e-113.5/20.85T ln = /20.85T = /20.85T T = 113.5/(1.802 x 20.85) = 3.02 K k = cm-1 or GHz cm-1 = 30 GHz Harry Kroto 2004

113 29/176 = = e-113.5/20.85T ln = /20.85T = /20.85T T = 113.5/(1.802 x 20.85) = 3.02 K k = cm-1 or GHz cm-1 = 30 GHz Harry Kroto 2004

114 29/176 = = e-113.5/20.85T ln = /20.85T = /20.85T T = 113.5/(1.802 x 20.85) = 3.02 K k = cm-1 or GHz cm-1 = 30 GHz Harry Kroto 2004

115 29/176 = = e-113.5/20.85T ln = /20.85T = /20.85T T = 113.5/(1.802 x 20.85) = 3.02 K k = cm-1 or GHz cm-1 = 30 GHz Harry Kroto 2004

116 29/176 = = e-113.5/20.85T ln = /20.85T = /20.85T T = 113.5/(1.802 x 20.85) = 3.02 K k = cm-1 or GHz cm-1 = 30 GHz Harry Kroto 2004

117 29/176 = = e-113.5/20.85T ln = /20.85T = /20.85T T = 113.5/(1.802 x 20.85) = 3.02 K error ca 3-4% k = cm-1 or GHz cm-1 = 30 GHz Harry Kroto 2004

118 “…restricted meaning” !
Harry Kroto 2004

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