Quantum Theory of the Atomic Nucleus

Transcription

1 G. Gamow, ZP, 51, Quantum Teory of te tomic Nucleus G. Gamow (Received 1928) It as often been suggested tat non Coulomb attractive forces play a very important role inside atomic nuclei. We can mae many ypoteses concerning te nature of tese forces. Tey can be te attractions between te magnetic moments of te individual constituents of te nucleus or te forces engendered by electric and magnetic polarization. In any case, tese forces diminis very rapidly wit increasing distance from te nucleus, and only in te immediate vicinity of te nucleus do tey outweig te Coulomb force. From te scattering of α particles we may conclude tat for eavy elements te forces of attraction are still not measurable down to a distance of cm. We may terefore tae te potential energy as being correctly represented by te curve in [Fig. 1]. Here r gives distance down to wic it as been sown experimentally tat te Coulomb repulsion alone exists. From r down te deviation (r is unnown and peraps muc smaller tan r ) from te Coulomb force is pronounced and te U curve as a maximum at r 0. For r < r 0 te attractive forces dominate; tis region, te particle circles around te rest of te nucleus lie a satellite. Tis motion, owever, is not stable since te particle s energy is positive, and, after some time, te α-particle will fly out (α -emission). Here, owever, we meet a fundamental difficulty. 1

2 Figure 1: To fly off, te α particle must overcome a potential barrier of eigt U 0 [Fig. 1]; its energy may not be less tan u 0. But te energy of te emitted α particle, as verified experimentally, is muc less. For example, we find on analyzing te scattering of Ra c α particles by uranium tat for te uranium nucleus te Coulomb law is valid down to distances of cm. on te oter and, te α-particles emitted by uranium itself ave an energy wic represents a distance of cm ( r 2 in [Fig. 1] ) on te repulsive curve. If an α-particle, coming from te interior of te nucleus, is to fly away, it must pass troug te region r 1 and r 2 were its inetic energy would be negative, wic, naturally, is impossible classically. To overcome tis difficulty, Ruterford assumed tat te α-particles in te nucleus are neutral, since tey are assumed to ave two electrons tere. Only at a certain distance from te nucleus, on te oter side of te potential barrier s maximum, do tey, according to Ruterford. lose teir two electrons. wic fall bac into te nucleus wile te α-particles fly on impelled by te Coulomb repulsion. But tis assumption seems very unnatural, and can ardly be a true picture. If we consider te problem from te wave mecanical point of view, te above difficulties disappear by temselves. In wave mecanics a particle always as a finite probability, different from zero, of going from one region to anoter region of te same energy, even troug te two regions are separated by an arbitrarily large but finite potential barrier. s we sall see furter, te probability for suc a transition, all tings considered, is very small and, in fact, is smaller te iger te potential barrier is. To clarify tis point, we sall analyze a simple case [see Fig. 2]. We ave a rectangular potential barrier and we wis to find te solution 2

3 Figure 2: of Scroedinger s equation wic represents te penetration of te particle from rigt to left. For te energy E we write te wave function ψ in te following form: ψ = Ψ(q)e (2πi/)Et, were Ψ(q) satisfies te amplitude equation For te region I we ave te solution 2 Ψ q 2 + 8π2 m 2 (E U) Ψ = 0 (1) Ψ 1 = cos (q + α) were and α are two arbitrary constants and In te region II te solution reads = 2π 2m E (2a) were Ψ II = B 1 e q + B 2 e q = 2π 2m U 0 E (2b) t te boundary q = 0 te following conditions apply: Ψ I (0) = Ψ II (0) and [ ] ΨI = q q=0 [ ] ΨII q q=0, 3

4 from wic we easily obtain were B 1 = 2 sin θ sin(α + θ); B 2 = sin(α θ), 2 sin θ 1 sin θ = 1 + ( ) 2 Te solution in region II terefore reads Ψ II = In III we again ave 2 sin θ [sin(α + θ) e q sin(α θ)e q ]. Ψ III = C cos(q + β) t te boundary q = l we ave from te boundary conditions and Hence 2 sin θ [sin(α + θ)e l ] sin(α θ)e +l ] = C cos(l + β) 2 sin θ [sin(α + θ)e l ] sin(α θ)e +l ] = C sin(l + β). {[ ( C 2 = 2 ] 4 sin 2 θ 1 + sin 2 (α θ) e 2l [ ( ] 1 2 sin(α θ) sin(α + θ) [ ( ] } sin 2 (α + θ) e 2l. (3) Te calculation of β is of no interest to us. We are interested only in te case in wic l is very large so tat we need consider only te firs term in (3). We tus ave te following solution: left: rigt: cos(q + α)... sin(α θ) 2 sin θ [ ( ] 1/2 1 + e +l cos(q + β). 4

5 If we now write α π 2 instead of α multiply te obtained solution by i, and ten add te two solutions, we obtain on te left Ψ = e i(q+α), (4a) on te rigt, owever, Ψ = [ ( ] 1/2 2 sin θ 1 + e +l {sin(α θ) cos(q + β) i cos(α + θ) cos(q + β ) }, (4b) were β is te new pase. If we multiply tis solution by e 2πi E t, we obtain for Ψ on te left te (from rigt to left) advancing wave; on te rigt, owever, te complex oscillatory penomenon wit a very large amplitude (e l ) tat departs only sligtly from a standing wave. Tis means noting oter tan tat te wave coming from te rigt is partly reflected and partly transmitted. We tus see tat te amplitude of te transmitted wave is smaller te smaller is te total energy E, and in fact te factor 2m 2π e l = e U0 E.l plays an important role in tis connection. We can now solve te problem for two symmetrical potential barriers [Fig. 3]. We sall see two solution. Figure 3: One solution is to be valid for positive q, and for q > q 0 + l is to give te wave: e i(2πe t q+α) 5

6 Te oter solution is valid for negative q, and for q < (q 0 + l) gives te wave ( ) 2πEl e i +q α. We cannot attac te two solutions eac oter continuously at q = 0 since we ave ere two boundary conditions to fulfill and only one arbitrary constant α to adjust. Te pysical reason for tis impossibility is tat te Ψ function constructed from tese two solutions does not satisfy te conservation law t +(q 0 +l) (q 0 +l) ψ ψdq = 2 [ ] ψ grad ψ ψ grad ψ 4π i m I To overcome tese difficulties we must assume tat te vibrations are damped and mae E complex E = E 0 + i λ 4π were E 0 is te usual energy and λ is te damping decrement (decay constant). We ten see from te relations (2a) and (2b) tat and are complex, tat is, tat te amplitude of our wave also depends exponentially on te coordinate q. For example, for te running wave te amplitude in te direction of te diverging wave will increase. Tis means noting more tan tat if te vibrations are damped at te source of te wave, te amplitude of te wave segment tat left earlier must be larger. We can now determine α so tat te boundary conditions are fulfilled. But te exact solution does not interest us. If λ is small compared to E (forra c E = 10 5 sec 10 1 = sec 1 and λ = 10 5 sec 1 ) te cange 27 in Ψ(q) is very small and we can simply multiply te old solution wit e α 2 t. Te conservation principle ten reads t e λt +(q 0 +l) (q 0 +l) from wic we obtain Ψ (q) II,III Ψ(q) II, III dq = 2 2 4π i m 2i e λt, λ = 4 sin 2 θ [ ( ) ] πm e 4πl 2m U0 E, (5) 2(l + q 0 0 ) were is a number of order of magnitude one. 6

7 Tis formula gives te dependence of te decay constant on te decay energy for our simple nuclear model. Now we can go over to te case of te actual nucleus. We cannot solve te corresponding wave equation because we do not now te exact formula for te potential in te neigborood of te nucleus. But even witout an exact nowledge of te potential we can carry over to te actual nucleus results obtained from our simple model. s usual, in te case of a central force, we see te solution in polar coordinates and, in fact, in te form Ψ = u(θ, φ)χ(r). For u we obtain te sperical armonics, and χ must satisfy te differential equation d 2 χ dr [ ] dχ r dr + 8π2 m 2 E U 2 n(n + 1) 8π 2 m r 2 χ = 0, were n is te order of te sperical armonic. We can place n = 0, since if n > 0, tis would really be just as troug te potential energy were enlarged, and because of tis damping for tese oscillations is muc smaller. Te particle must first pass over to te state n = 0 and can only ten fly away. It is quite possible tat suc transitions are te cause of te γ rays wic always accompany α emission. Te probable sape of U is sown in [Fig. 4]. For large values of r, we sall tae for χ te solution χi = r ei(eπe t r) Even troug we cannot obtain te exact solution of te problem in tis case, we can still say tat on te average in te regions I and I, χ does not decrease very rapidly (in te tree dimensional case lie l/r). In te region III, owever, χ decreases exponentially, and in analogy wit our simple case we may state tat te relation between te amplitude decrease and E is given by te factor e 2π 2m r 2 r 1 U E dr 7

8 Figure 4: If we use te conservation principle we can again write down te formula λ = D. e 2π 2m r 2 r 1 U Edr (6) were D depends on te particular properties of te nuclear model. We can neglect te dependence of D on E compared to its exponential dependence. We may also replace te integral r 2 U E dr by te approximate integral r 1 2Ze 2 E 0 2Ze 2 r E dr Te relative error we introduce in tis way is of te order of r 1 r 2. Since r 1 /r 2 is small, tis error is not very large. Since E does not differ muc for different radioactive elements, we write as an approximation log λ = const E + B E. E, 8

9 ... were B E = π2 2m 2Ze 2 E 3/2. (7) We wis to compare tis formula wit te experimental results. It is nown tat if we plot te logaritm of te decay constant against te energy of te emitted α-particle, all te points for a definite radioactive family fall on a straigt line. For different families we obtain different parallel lines. Te empirical formula reads log λ = const + be were b is a constant tat is common to all radioactive families. Te experimental value of b is b exp = (calculated from Ra and Ra). If we put te energy value for Ra into our formula, we get b teoretical = Tis order of magnitude agreement sows tat te basic assumptions of our teory must be correct... 9