FIG.II. 2.1 MATERIAL Location o f tnmirrtom in
t o t a l obsorV*tio»i
TVMckrv*ss F ”, in g /c m ' ioir V z.7 (A uction in in te n jity
Alummom ~ 3 5 M i V - 4 7 8/cn.» o t -aSM^v
Colo(3er ~ I O ^ 3 1 II " ~10m»v
Lead ~ 3 '/t « .2 5 » ·. -SJiMev
Problem: Suppose a cosmic ray electron of energy 1000 mc^
passes tiirouGb a lead plate 1 mm. thick. What Is the probability that the electron leaves the plate accompanied by a pair whose total enercy (electron plus positron) Is larger than 500 mc^?
Assume the particles change direction neclicibly.
Ui® 1 1 .5 3 , writing it in terms o f ^ : ^__
^ ' jpR Kt
ro Get the cross-section for all the nuclei In thickness dx, multiply by N dx, ^ettlnc ^ - ¡Lx A v '
^ = — — · Jk V'·
* From H eitler, p. 2 1 6 .
On. II
A\g|»enclix U .l51
)K>TIAL D\RE'CTI0N TO WHICH IS MEASURED
FIG. 11.22
/-Shorter derivaxlon of 11.22:
\fe make use of the fact -that m a ll angles are vectors.
P
Then ~ - i
0;
In averaging o v e / many traversals, ®t is positive as often as It is negative,
thus __ ^
a " = in J
^ Cli· XX
52 Appendix II
Atioendix I I . 2 . Derivins 11.25, © * = 2 '7 f N D j ^
In detail, P, the average mmber of collisions in one t2?aversal is given by
•p_ j’'^*'*2Tr’b O K ¿)o
I f / is a function of" x and ia the normalized probability density f o r X , then the mean of -f is
^ * J/ibfJ ¿k I I . 69
applied to calculate 9^- J [©OOl'^C^Ddb.
The probability density for ^0°) is proportional to cross- sectional ?.rea at the distance b, 2"irb db, to the nuclear, density li, and to the path length D. Therefore the normalized
s _ _ 3 r b _ D N _ _ _ _ a i r b OH
^ ■ P I I .
Therefore
F I I . 71
Finally
5 ,« * = fmrn H .20
Appendix I I . 3 . Equation 11.26 can be derived by conslderlxig the projections of the angles ^ on some chosen plane and deriving a value for , the mean square of the resultant projected angle.
Then the unproJected, true scattering angle can be found from the
re stilt that = ( ^ « 1 ) I I •72
The fac to r/^ comes from averaging cos^jz! , the angle beins random.
A similar relation holds between the actual linear deviation of
Ch. II Appendix II.3 53
the point of exit of the p artic le measured from the geometrical shadow of the point of entrance, and the pixijectlon o f that d i s placement on some chosen plane; i . e . ^ __ ___
emor^ence
I (plane o f paper normal to incident -- 1-- d ir e c t io n . Looking toward source
of p a r tic le s )
^ 5WqcIow cf" entmnce Jaoint
In a cloud chamber photograph it is the projected angle of scattering that is shown.
This approach (use of projected angles) I s used by J^n o ss y . Problem; Electrons of energy
ICP e .v . (from cosmic rays) pass through a lead plate 1 cm.
thick. Calculate the mean square of the linear deviation of the emergent points from the geo
metrical shadovi point on the· last surface of the lead p la t e . Assume the path is almost straight.
io®€y. peviATioN
Let the collisions take place at d istances from the^sur-
last
face and result In d e fle ctio n s 0 j « l . The azimuth angle i s ^ . The projected angle of d e fle c t io n at /i| = O is % cosg^ . The projection of linear displacement at i^ = 0 i s _ ^ 4 · ©-cce
______ _____ ___ I -i »
f o r c o l l i -
SION ;
POINT or I
KMtWFNCE · II
FAce view, LpokiNC
iOWARO SOURCC
OP
FVSRTICLBS.
e^ & jC o s^ . cos«^· , bi± COS c o s ^ · - ^ 4 j
5m « y^· a n d a c e i n c l e l » € w d e n t .
=riV;^ = -r
/ . p p a s a v e r o3e rKJ. o -f c o l l i s i o n s
V y l l .7 1
The unnro.lected lin e a r (d isp la ce m e nt)^, , w i l l be twice th is .
a X H .3 I)
Appendix I I . 4 . Proof that momentum cannot be conserved In crea
tion of an electron-positron p a ir by a single Iso lated photon.
By conservation of energy;
niovneMtum o’f/3'*’ 1 1 .7 4
.1
By conservation of momentum;
(Hvf ^
(The equality holds I f g·*· and p" go o ff In the same d ir e c t io n ) Prom 1 1 .7 5 0^"^)^=
54 Appendix II.4 Ch. II
From 11.74. M ' =
The radical term Is larger thc.n 'S! , so both cannot be s a t is fie d .
References; Books and articles referred to more than once i n th is ciiapter are liste d .
Abraham-Becker, Theorie der E le c tr lzlt ilt. V o l. I I (Tsubner, EdTOrds) H e i t l e r , Quantum Theory of Radlg-tlon. 2nd ed. (Oxford)
Janossy, Cosmic Rays (Oxford)
L iv in g sto n and Bethe, Rev. Mod. P h y s . . £ , 245 (1937) (experim ental jmclear p hy sics)
R u th erford, Chadwick and E l l i s , Radiations from Radioactive Substances" ( Cambridge!
Ro h sI and G-relsen, Rev. Mod. Phva. 1 3 . 249 (l 9 4 l)
ALPI'A PARTICLE EMISSION
A. Penetration of ]Reotang;ul·aΓ Barriers
A quaiituiii mécnanTcal explanation was ap p lied to~the problem of alpha emission independently by Gamow and by Gurney and Con
den in 1928. This explanation makes use of the a b i l it y of a quantum mechanical particle to penetrate a p otential b a r r i e r · Fir s t the problem of the one dimensional UC^;>
rectangular barrier w ill be considered·
C lassically an incident particle of energy E (see F ig · III^ l)^ coming from the l e f t , could not penetrate into region I I , since Uo> E · It would be completely reflected. However,
quantijm mechanically, the p a r t ic le , which is an incident wave, is partially reflected and p a r t ia lly transm itted·
Let r be the incident p article tr a v e llin g to th©
^ ^ right
then ^ = ^ 0 is the reflected wave, / y»
and ^ - 0 ^ 'nr is the transmitted wave.
The probability of penetration or the ^ treinstriitted inte nsity _ i£_
transparency = in c id e n t 'in t e n s it y “ 1°^
Considering the time Independent part of ,
Chapter III
¿ A - X
Those are solutions of the tiwe independent wave equation:
dy.^ ^
In region I I the wave equation becomes
dx-L e t b =u. -e:
Then
Now the constants (3, ^ , K, and L w i l l be determined by mak
ing use of the fact that the wave fu n ctio n must be w e ll behaved physically.
Thus and I I are continuous across tho boundaries x = 0 and X = bf
55
56
Thas oí· +/3 = K +L
re
Rectangalar barrier
since J^(o)=^íoJ
Ch. III
lÁír
í'(e )
í i í ^ - é » · )
Eliminating fi between the first two of these equations gives
Solving the last two of these equations for K and L gives
K ' f e‘*Y'*f;e·^
Thus ,
Since
^ for 1 ^ » ,
J l f
Ul^
III- l
Notice that as h -► 0, this approaches the classical limit of zero transparency· Also the trans
parency is greater for particles of smaller mass. This explains why electrons have lit t le trouble in going from one atom to another In diatomic molecules and metals.
See P ia I I I ·? .·
PIG III^2 Diatomic molecule with one electron B· Barrier of Arbitrary Shape
The order of magnitude of the transparency of an arbitrarily shaped barrier can be obtained by finding the av©rage"‘height’* and treating as a rectangular barrier·
Ch. I l l
III»l· then gives
Rectangular b a r r ie r 57
I I I . 2
Problem! A very slowly moving oar of 1 ton (kinetic energy con
sidered almost zero) encounters a sin uso id al xjcx) bump in the road which is 1 f t . high and
100 f t . long. C l a s s i c a l l y , the car c a n ’ t get past. However I I I . 2 shows that there actually is a f i n i t e p r o b ab ility (w ) that the car can overcome the bump.
rh- .
W ^ ^
U (x) = Mg X 12 X 2 .5 4 s in ai:(x-a) 3050 XxT dx 5 . 3 X 10^
- / t J »10^
w ■10
39
In the follo w in g flR u r e , I I I . 2 is used to calculate the
transparency for protons penetrating the coulomb barrier of nuclei of A =· 1, 8 , 9 0 , and 2 3 8 . These are plotted as a function of proton enerp,y div ided by b a r r ie r energy.
CUKVE /4 2
a ./ 1 O.^Aiev
Jj- 4
c <io ■40
d i a n z
7s
Transparency to Protons of Nuclear Coulomb Barriers
C. Seml-classlcal A p p lic a tio n to Alpha Decay