In Voltaire’s “Elémens de la Theorie de Newton“ (1738)

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Presentation transcript:

In Voltaire’s “Elémens de la Theorie de Newton“ (1738) Harry Kroto 2004

Fraunhofer Absorption Lines in the Sun’s Spectrum Na D lines Orange street lamps contain sodium Harry Kroto 2004

Harry Kroto 2004

Emission Spectra from Nebulae Harry Kroto 2004

Harry Kroto 2004

Photon from local star Harry Kroto 2004

photoionisation Harry Kroto 2004

recombination Harry Kroto 2004

Harry Kroto 2004

Harry Kroto 2004

Harry Kroto 2004

H Atom ElectronicTransitions in the Radio Range Harry Kroto 2004

Hydrogen atom E = 2R/n3 Harry Kroto 2004

Bohr radius an = aon2 ao = 0.5 Å (1Å = 10-8cm) a300 = 0.5x10-3 cm = 0.005 mm Harry Kroto 2004

Energy Levels E(n) = - R/n2 Harry Kroto 2004

Energy Levels E(n) = - R/n2 F(n) = - R/n2 (in cm-1) Harry Kroto 2004

Energy Levels E(n) = - R/n2 F(n) = - R/n2 (in cm-1) R = 109678 cm-1 Harry Kroto 2004

For n= -1  emission lines Energy Levels E(n) = - R/n2 F(n) = - R/n2 (in cm-1) R = 109678 cm-1 For n= -1  emission lines Harry Kroto 2004

For n= -1  emission lines n2 = n+1 Energy Levels E(n) = - R/n2 F(n) = - R/n2 (in cm-1) R = 109678 cm-1 For n= -1  emission lines n2 = n+1 n1 = n Harry Kroto 2004

For n= -1  emission lines n2 = n+1 Energy Levels E(n) = - R/n2 F(n) = - R/n2 (in cm-1) R = 109678 cm-1 For n= -1  emission lines n2 = n+1 n1 = n Transition Frequencies F(n1) = - R[ 1/n22 – 1/n12] Harry Kroto 2004

For n= -1  emission lines n2 = n+1 101 Energy Levels E(n) = - R/n2 F(n) = - R/n2 (in cm-1) R = 109678 cm-1 For n= -1  emission lines n2 = n+1 101 100 n1 = n 100 Transition Frequencies F(n1) = - R[ 1/n22 – 1/n12] Harry Kroto 2004

Transition Frequencies F(n1) = - R[1/n22 – 1/n12] Harry Kroto 2004

Transition Frequencies F(n1) = - R[1/n22 – 1/n12] Harry Kroto 2004

Transition Frequencies F(n1) = - R[1/n22 – 1/n12] F(n) = - R[n2 – (n+1)2]/[(n+1)2n2] Harry Kroto 2004

Transition Frequencies F(n1) = - R[1/n22 – 1/n12] F(n) = - R[n2 – (n+1)2]/[(n+1)2n2] F(n) = R[(n+1)2 – n2]/[(n+1)2n2] Harry Kroto 2004

Transition Frequencies F(n1) = - R[1/n22 – 1/n12] F(n) = - R[n2 – (n+1)2]/[(n+1)2n2] F(n) = R[(n+1)2 – n2]/[(n+1)2n2] F(n) = R[(n2 + 2n +1 - n2)/[(n+1)2n2] Harry Kroto 2004

Transition Frequencies F(n1) = - R[1/n22 – 1/n12] F(n) = - R[n2 – (n+1)2]/[(n+1)2n2] F(n) = R[(n+1)2 – n2]/[(n+1)2n2] F(n) = R[(n2 + 2n +1 - n2)/[(n+1)2n2] F(n) = R[(2n+1)/(n+1)2n2] Harry Kroto 2004

Transition Frequencies F(n1) = - R[1/n22 – 1/n12] F(n) = - R[n2 – (n+1)2]/[(n+1)2n2] F(n) = R[(n+1)2 – n2]/[(n+1)2n2] F(n) = R[(n2 + 2n +1 - n2)/[(n+1)2n2] F(n) = R[(2n+1)/(n+1)2n2] F(n) ~ R2n/n4 (for large n, n~n+1) Harry Kroto 2004

Transition Frequencies F(n1) = - R[1/n22 – 1/n12] F(n) = - R[n2 – (n+1)2]/[(n+1)2n2] F(n) = R[(n+1)2 – n2]/[(n+1)2n2] F(n) = R[(n2 + 2n +1 - n2)/[(n+1)2n2] F(n) = R[(2n+1)/(n+1)2n2] F(n) ~ R2n/n4 (for large n, n~n+1) F(n) ~ 2R/n3 Harry Kroto 2004

F(n) ~ 2x109678/n3 (cm-1) Harry Kroto 2004

F(n) ~ 2x109678/n3 (cm-1) For n = 100 F(100) ~ 0.2194cm-1 ~ 6581MHz Harry Kroto 2004

F(n) ~ 2x109678/n3 (cm-1) For n = 100 F(100) ~ 0.2194cm-1 ~ 6581MHz Accurate calculation 1/1012 -1/1002 Harry Kroto 2004

F(n) ~ 2x109678/n3 (cm-1) For n = 100 F(100) ~ 0.2194cm-1 ~ 6581MHz Accurate calculation 1/1012 -1/1002 = (0.980296 – 1)10-4 = -0.019704x10-4 Harry Kroto 2004

F(n) ~ 2x109678/n3 (cm-1) For n = 100 F(100) ~ 0.2194cm-1 ~ 6581MHz Accurate calculation 1/1012 -1/1002 = (0.980296 – 1)10-4 = -0.019704x10-4 0.216109 cm-1 = 6483.27 MHz Harry Kroto 2004

Hydrogen atom E = 2R/n3 Harry Kroto 2004