4E : The Quantum Universe Lecture 19, May 3 Vivek Sharma modphys@hepmail.ucsd.edu
Ceparated in Coppertino Oxide layer Wire #1 Wire # Q: Cu wires are seperated by insulating Oxide layer. Modeling the Oxide layer as a square barrier of height U1.eV, estimate the transmission coeff for an incident beam of electrons of E7. ev when the layer thickness is (a) 5. nm (b) 1.nm Q: If a 1. ma current in one of the intwined wires is incident on Oxide layer, how much of this current passes thru the Oxide layer on to the adjacent wire if the layer thickness is 1.nm? What becomes of the remaining current? 1 U 4 EU ( E) T(E) 1+ sinh ( L) 1 m(u-e),k me 5
1 U 4 EU ( E) Use 1.973 kev.a/c, m T(E) 1+ sinh ( L) e 3 me ( U E) 511 kev / c (3. 1 kev ) 1.973 kev.a/c Substitute in expression for TT(E) 1 511 kev/c 1 1-1 38 T 1+ sinh (.8875 A )(5A).963 1 ( small)!! 4 7(1 7) -7 However, for L1A; T.657 1 1.8875A Reducing barrier width by 5 leads to Trans. Coeff enhancement by 31 orders of magnitude!!! QNq N electrons t N # of electrons that escape to the adjacent wire (past oxide layer) e 15 1 ma current I 6.5 1 T 15 N T N. T (6.5 1 electrons) For L1 A, T; -7 T.657 1 NT 4.11 1 IT 65.7 pa!! Current Measured on the first wire is sum of incident+reflected currents and current measured on "adjacent" wire is the I T Oxide layer Wire #1 Wire # Oxide thickness makes all the difference! That s why from time-to-time one needs to Scrape off the green stuff off the naked wires 53-1
A Complicated Potential Barrier Can Be Broken Down Multiplicative Transmission prob, ever decreasing but not U(x) Integration x Can be broken down into a sum of successive Rectangular potential barriers for which we learnt to find the Transmission probability T i The Transmitted beam intensity thru one small barrier becomes incident beam intensity for the following one So on & so forth till the particle escapes with final Transmission prob T T T dx e i m U ( x ) Edx 54
Radioactivity: The -particle & Steve McQueen Compared T T dx e i m ke Z E dx r In a Nucleus such as Ra, Uranium etc particle rattles around parent nucleus, hitting the nuclear walls with a very high frequency (probing the fence ), if the Transmission prob T>, then eventually particle escapes Within nucleus, particle is virtually free but is trapped by the Strong nuclear force Once outside nucleus, the particle sees only the repulsive (+) columbic force (nuclear force too faint outside) which keeps it within nucleus Nuclear radius R 1-14 m, E 9MeV Coulomb barrier U(r) kq 1 q /r At rr, U(R) 3 MeV barrier -particle, due to QM, tunnels thru It s the sensitivity of T on E that accounts for the wide range in half-lives of radioactive nuclei 55
Radioactivity Explained Roughly (..is enough!) Protons and neutrons rattling freely inside radioactive nucleus (R 1-15 m) Constantly morphing into clusters of protons and neutrons Proto-alpha particle (p+n) of 9 MeV prevented from getting out by the imposing Coulombic repulsion of remaining charge ( 3MeV) Escapes by tunneling thru Coloumb potential but some puzzling features: particles emitted from all types of radioactive nuclei have roughly same KE 4-9 MeV In contrast, the half live T(N e -1 N) differ by more than orders of magnitude! T T dx e i m ke Z E dx r Element KE of emitted Half Life 1 Po 8.95 MeV 3x1-7 s 4 Cm 6.4 MeV 7 days 6 Ra 4.9 MeV 1.6x1 3 Yr 3 Th 4.5 MeV 1.41x1 1 yr 56
Radioactivity Explained Crudely m U ( x) E dx ez T( E) e, U( x) 4πε r b ez ln T m E dr, 4 πε r limits of integration correspond to values of r when EU ez ez E b 4πε b 4πε E r r Define ξ ; lnt b ez/4πε E ( m E ) ( ) m E b Substitute sin in integration, change limits ln T 4 ξ θ b π / cos 1 1 1 dξ ξ π / cos θ θ π / 4 & θdθ ; use d E mv ln T π Ze m 4π Ze 1 4πε E 4πε V T e 4π Ze 1 4πε V 1 V Z T e and T e...sharp DEPENDENCE!! 57
Radioactivity A more eloberate calculation (Bohm) yields T( E) e where r and E mke E 4π Z + 8 E 8fm is the "Bohr Radius" of alpha particles ke.993mev Nuclear "Rydberg" r To obtain decay rates, need to multiply T(E) by the number of collisions particle makes with the "walls" of the nuclear barrier. This collission frequency V f transit time for particle crossing the nuclear barrier (rattle time) R 1 Typically f 1 collissions/second ZR r Decay rate (prob. of emission per unit time) λ f T(E) λ 1 1 e E ZR 4π Z + 8 E r Definition : Half life t 1/ ln λ 58
Half Lives Compared: Sharp dependence on E particles emerge with (a) E4.5 MeV in Thorium (b) E8.95 MeV in Polonium. The Nuclear size R 9 fm in both cases. Which one will outlive you? Thorium ( Z9) decays into Radium (Z88) { } T(E) exp -4 π ( 88 ) (.993/4.5) + 8 88 (9. / 7.5) 1.3 1-39 1 18.693 1 Taking f1 Hz λ 1.3 1 emission t1/ 1.7 1 yr!!! 18 1. 3 1 Polonium (Z84) decays into Lead (Z8) { } T(E) exp -4 π ( 8 ) (.993/8.95) + 8 8 (9./7.5) 8. 1-13 1 8.693 1 Taking f1 Hz λ 8. 1 emission t1/ 8.4 1 s!!! 8 8. 1 59
Potential Barrier : An Unintuitive Result When E>U Region I Incident Beam Reflected Beam A B Description Of WaveFunctions in Various II L U F Transmitted Beam Region III In Region I : Ψ I ( xt, ) Ae + Be ; In Region III: Ψ ( xt, ) Fe In Region II of Potential U: d ψ ( x) d ψ ( x) m TISE: - + Uψ ( x) Eψ ( x) ( ) ( ) ( ) m U E ψ x ψ x dx dx m(u-e) with ; U<E < m( E U ) Define i k ; ( k ) ; k i( k x ωt) i( k x ωt) Ψ II Ce + De Oscillatory Wavefunction Apply continuity condition at x & xl i ( kx ωt) i( kx ω t ) i( kx ω t ) III x regions: Simple Ones first A+BC+D ; ka - kb k D - k C ; Ce + De Fe ; k De k Ce kfe ikl ikl ikl ikl ikl ikl Eliminate B, C,D and write every thing in terms of A and F 1 k k k k A Fe + e + + + e 4 k k k k ikl ik L ik L 6
Potential Barrier : An Unintuitive Result When E>U L * 1 AA 1 k k 1 U coskl i sinkl 1 sin kl * + + T F F 4 k k 4 E( E U) Only when Since k Region I Incident Beam Reflected Beam sin kl, T 1; t kl n En U + n his happens when me ( U) me ( U) nπ π is the condition ml for particle to be completely transmitted For all other energies, T<1 and R>!!! A B This is Quantum Mechanics in your face! II U F Transmitted Beam π Region III x > 1 General Solutions for R & T 61
Special Case: A Potential Step U Ae ikx-iωt U for x < U U for x Be ikx-iωt X X XL ikx iωt In region I (X<) : Ψ ( x, t) Ae + Be In region II (X ) : Ψ ( x, t) I II x iωt Ce + De ikx iωt x iωt dψ Applying Continuity conditions of Ψ and at x dx A + B C & ika - ikb -C; Eliminating C ika - ikb - ( A + B) 1 Defining Penetration Depth δ, mu ( E) rewrite as ikδa - ikδb - (A+B) A(1+ik δ) -B(1-ik δ) * B (1+ik δ ) B B Reflection Coeff R * A (1-ik δ ) AA 1 ; as expected 6
Transmission Probability in A Potential Step U Ae ikx-iωt U for x < U U for x Be ikx-iωt Ψ II Ce -x-iωt X XL X Since Ψ I( x, t) Ae + Be ; Ψ ( x,) t Ce dψ Applying Continuity conditions of Ψ and at x : dx C B (1+ik δ ) A + B C 1 + 1 - A A (1-ik δ ) ikx iωt ikx iωt x iωt II C ikδ C C T >!!! A 1 ikδ A A The particle burrows into the skin of the step barrier. If one has a barrier of width L δ, particle escapes thru the barrier. penetration distance x distance for which prob. drops by 1/e. ψ (x x) * - x -1 1 C e C e ; happens when x1 or x m(u-e) 63