Doppler Shifts of Interstellar NaD lines

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Presentation transcript:

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

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

Harry Kroto 2004

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

Blue Shifted Red Shifted Harry Kroto 2004

Blue Shifted Red Shifted Harry Kroto 2004

Several clouds in the line of sight Harry Kroto 2004

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

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

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

ΔE Harry Kroto 2004

ΔE n0 population of ground state Harry Kroto 2004

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

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

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

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

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

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

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

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

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

Harry Kroto 2004

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

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

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) = 4/3 SR(0) = 2/1 Harry Kroto 2004

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) = 4/3 SR(0) = 2/1 SR(1)/SR(0) = ⅔ Harry Kroto 2004

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) = 4/3 SR(0) = 2/1 SR(1)/SR(0) = ⅔ R(1)/R(0) = (⅔) N1/N0 Harry Kroto 2004

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) = 4/3 SR(0) = 2/1 SR(1)/SR(0) = ⅔ R(1)/R(0) = (⅔) N1/N0 = IR(1)/IR(1) Harry Kroto 2004

(⅔)(N1/N0) = I1/I0 Harry Kroto 2004

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

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

(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

(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 = 113.5 GHz Harry Kroto 2004

(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 = 113.5 GHz (3/2)(29/86) = 3e-113.5/20.85T Harry Kroto 2004

(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 = 113.5 GHz (3/2)(29/86) = 3e-113.5/20.85T I1/I0 = 2 e-113.5/20.85T Harry Kroto 2004

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 = 20.85 if F(J) in GHz For the CO molecule B = 56.75 GHz Thus determine T (K) R(1) R(0) J =1 J = 0 F(J) = BJ(J+1) Harry Kroto 2004

Harry Kroto 2004

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 = 20.85 if F(J) in GHz For the CO molecule B = 56.75 GHz Thus determine T (K) R(1) R(0) J =1 J = 0 F(J) = BJ(J+1) Harry Kroto 2004

35/105 = 0.336538461538461538461538461538 Harry Kroto 2004

29/176 = 0.165 = e-113.5/20.85T ln 0.165 = -113.5/20.85T -1.802 = -113.5/20.85T T = 113.5/(1.802 x 20.85) = 3.02 K error ca 3-4% k = 0.695 cm-1 or 20.85 GHz 1 cm-1 = 30 GHz Harry Kroto 2004

29/176 = 0.165 = e-113.5/20.85T ln 0.165 = -113.5/20.85T -1.802 = -113.5/20.85T T = 113.5/(1.802 x 20.85) = 3.02 K error ca 3-4% k = 0.695 cm-1 or 20.85 GHz 1 cm-1 = 30 GHz Harry Kroto 2004

(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 = 113.5 GHz (3/2)(29/86) = 3e-113.5/20.85T 29/176 = e-113.5/20.85T = 0.165 Harry Kroto 2004

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

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

Harry Kroto 2004

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

Harry Kroto 2004

Harry Kroto 2004

Harry Kroto 2004

Harry Kroto 2004

Harry Kroto 2004

Harry Kroto 2004

Harry Kroto 2004

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Two serendipitous radio discoveries Harry Kroto 2004

Harry Kroto 2004

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

Harry Kroto 2004

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

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

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

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

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

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

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

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

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

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

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

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 0-0 Band R Branch P Branch Harry Kroto 2004

Harry Kroto 2004

Electronic Emission Spectrum 3883 Å 4216 Å 4606 Å CN Violet Electronic Emission Spectrum 0-1 0-2 0-0 R(0) P(1) R Branch P Branch 25850 25750 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

Harry Kroto 2004

Harry Kroto 2004

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

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

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

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

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

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) = 4/3 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

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) = 4/3 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

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) = 4/3 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

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) = 4/3 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

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) = 4/3 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

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) = 4/3 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

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) = 4/3 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

(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 = 113.5 GHz (3/2)(29/86) = 3e-113.5/20.85T 29/176 = e-113.5/20.85T = 0.165 Harry Kroto 2004

(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 = 113.5 GHz (3/2)(29/86) = 3e-113.5/20.85T 29/176 = e-113.5/20.85T = 0.165 Harry Kroto 2004

(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 = 113.5 GHz (3/2)(29/86) = 3e-113.5/20.85T 29/176 = e-113.5/20.85T = 0.165 Harry Kroto 2004

29/176 = 0.165 = e-113.5/20.85T ln 0.165 = -113.5/20.85T -1.802 = -113.5/20.85T T = 113.5/(1.802 x 20.85) = 3.02 K k = 0.695 cm-1 or 20.85 GHz 1 cm-1 = 30 GHz Harry Kroto 2004

29/176 = 0.165 = e-113.5/20.85T ln 0.165 = -113.5/20.85T -1.802 = -113.5/20.85T T = 113.5/(1.802 x 20.85) = 3.02 K k = 0.695 cm-1 or 20.85 GHz 1 cm-1 = 30 GHz Harry Kroto 2004

29/176 = 0.165 = e-113.5/20.85T ln 0.165 = -113.5/20.85T -1.802 = -113.5/20.85T T = 113.5/(1.802 x 20.85) = 3.02 K k = 0.695 cm-1 or 20.85 GHz 1 cm-1 = 30 GHz Harry Kroto 2004

29/176 = 0.165 = e-113.5/20.85T ln 0.165 = -113.5/20.85T -1.802 = -113.5/20.85T T = 113.5/(1.802 x 20.85) = 3.02 K k = 0.695 cm-1 or 20.85 GHz 1 cm-1 = 30 GHz Harry Kroto 2004

29/176 = 0.165 = e-113.5/20.85T ln 0.165 = -113.5/20.85T -1.802 = -113.5/20.85T T = 113.5/(1.802 x 20.85) = 3.02 K k = 0.695 cm-1 or 20.85 GHz 1 cm-1 = 30 GHz Harry Kroto 2004

29/176 = 0.165 = e-113.5/20.85T ln 0.165 = -113.5/20.85T -1.802 = -113.5/20.85T T = 113.5/(1.802 x 20.85) = 3.02 K error ca 3-4% k = 0.695 cm-1 or 20.85 GHz 1 cm-1 = 30 GHz Harry Kroto 2004

“…restricted meaning” ! Harry Kroto 2004