Modern Physics Unit 6: Hydrogen tom - Radiation Lecture 6.5: Optical bsorption Ron Reifenberger Professor of Physics Purdue University 1
We now have a simple quantum model for how light is emitted. How is light absorbed? EM radiation at wavelength λ I o (ugust eer; 185-1863) Classically we know from measurements that: I(x) I(x+Δx) Δx material medium 1 J / s Io( ω) cεoeo ( ) ( ) ( ) ( ) ( ) I x I x+ x I x I x α I x x m 1 di( λ) αλ ( ) I( λ) dx Transmission I I o e L α x x L I x
Typical results http://www.pveducation.org/pvcdrom/pn-junction/absorption-coefficient 3
What is absorbing the light? The absorption problem including atomic quantum states EM radiation at frequency ω I(x) I(x+Δx) I I o Δx I( x) I( x+ x) I( x) I( x) α I( x) x 1 di( ω) I( ω) dx material medium αω ( ) x No. atoms in given state Degeneracy of given state What are atomic contributions to? Collection of atoms: Δx α Model as a two-state system: E N, g N 1, g 1 4
Einstein: Three fundamental TOMIC Optical Processes E N, g N 1, g 1 Spontaneous Emission Rate probability per unit time to go from 1 E N, g N 1, g 1 bsorption Rate probability per unit time to go from 1 E N, g Stimulated Emission Rate* probability per unit time to go from 1 in the presence of another photon N 1, g 1 * new process invented by Einstein in 1917. Required to derive Planck s blackbody law. The new process needed to: i) be proportional to the intensity of the incident radiation field 5 and ii) cause emission of a photon.
What to calculate? Focus on relative populations 1. How many transitions per second to go from 1 by spontaneous emission? proportional to number of atoms N in state E. How many transitions per second to go from 1? proportional to the energy density of the radiation field u(f) how many photons are there? proportional to number of atoms N 1 in state N u(f) N 1 6
Defining the Transition Probabilities (assume g 1 g ) I. Spontaneous Emission E N f E E h 1 N 1 Rate 1 spontaneous Number of photons of energy E by spontaneous emission N - 1 E emitted per second Change in N over a time interval dt due to spontaneous emission dn N dt : 7
II. bsorption u(f) E N N 1 f E 1 h E Rate 1 1 absorption 1 1 1 1 u( f ) u( f ) is the energy density of the radiation field at frequency f Number of photons with energy E N u ( f) - 1 E absorbed per second Change in N over a time interval dt due to absorption dn 1 u( f ) N 1 dt : Units: [ u(f) ] J s/m 3 [ u tot u(f)df ]J/m 3 8
III. Stimulated (Induced) Emission E N u(f) N 1 Rate 1 stimulated emission u( f ) u( f ) is the energy density of the radiation field at frequency f Number of photons of energy E by stimulated emission - 1 N u( f ) E emitted per second Change in N over a time interval dt due to stimulated emission dn u( f ) N dt : 9
How does the number of electrons in the two levels change with time? E N N 1 in equilibrium ( describes lackbody radiation) 1 1 dn dn dn + dt dt dt 1 absorption spontaneous emission N u( f) N N u( f) 1 1 stimulated emission u( f) solve for u( f ) N 1 1 N 1 1 10
The oltzmann factor For a system with a large number of atoms in thermal equilibrium at some temperature T, how are the atoms distributed as a function of energy? T? E E o N n Ne o E n kt k 1.38x10-3 J/K E E o T N N 1 N o For a system with two energy states: E N? N 1? 0 u( f) E1 E ( ) kt ( kt) N e N e N N 1 1 ( E E ) ( hf ) 1 kt e e e 1 hf kt 1 kt 1 11
ut from first week of course: Planck lackbody Radiation 8π u( f ) df df ; units : u( f ) 3 / c u( f ) e f hf Js hf kt 1 ( hf k 1) T e 1 previous slide 1 m 3 evidently, by direct comparison ( Einstein, 1917) : 8π hf 1 3 c 3 Find one, the other two are determined 1
E dn dn dn + dt dt dt 1 N u( f) N N u( f) 1 1 u( f) 0 dn N How to measure? consider only two states in H atom If N is the number of atoms in the state E, how does N change with time in the absence of radiation? dt absorption N ( t) N ( t 0) e spontaneous t stimulated emission N (t) N (t) N 1 (t) f Change in N with t is fundamental quantity of interest; typically ranges between 10-3 s -1 (microwaves) to 10 15 s -1 (X-rays). time E E 1 h 13
Let s say is measured to be x10 8 s -1 (τ5 ns) for the 3p s transition in atomic H. What is? 3p s N (t) N 1 (t) f 13.6eV 13.6eV E E ( 3) ( ) 1.51eV + 3.4eV h h h 19 1.89eV 1.60 10 J 14 4.6 10 Hz 34 6.66 10 Js 1eV 1 8π hf c 3 3 3 8 1 c 1 3.0 10 m/ s 3 10 s 14 1 8π h f 8π 6.66 10 4.6 10 s 3.3 10 J ms 1 3 1 ( 34 Js) Can now calculate rate of stimulated emission u(f) 3 8 1 14
u(f) depends on conditions. If H gas atoms are in thermal equilibrium at some temperature T, then u( f) 8π f c ( hf k 1) T e 3 / Suppose the H is in a gas cloud at say T750 K. Then Rate for stimulated emission ( )( K ) hf 3 c 8π f hf ( ) 3 3 / / u f kt hf k T hf kt h f kt c ( e ) ( e ) ( )( ) 8π hf 1 1 8 1 34 14 1 19 x10 s hf 6.66 10 Js 4.6 10 s 3.05 10 J 1.38x10 J / K 750 1.04 10 19 3.05 10 J 0 1.04 10 J 3 0 e 9.5 9.5 8 1 x10 s u( f) 3. 10 s 1 π 10 5 1 J π 10 1 CONCLUSION: Rate of spontaneous emission is 10 8 /s Rate of stimulated emission at 750 K is 3 10-5 /s 15
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