Alpha Particle Emission 

by Reinaldo Baretti Machín

 

http://www1.uprh.edu/rbaretti

http://www1.uprh.edu/rbaretti/methodsoftheoreticalphysics.htm

http://www1.uprh.edu/rbaretti/methodsoftheoreticalphysicsPart2.htm

http://www1.uprh.edu/rbaretti/methodsoftheoreticalphysicsPart3.htm

http://www1.uprh.edu/rbaretti/methodsoftheoreticalphysicsPart4.htm

reibaretti2004@yahoo.com

 

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References:

1.  R W Gurney and E U Condon, "Quantum Mechanics and Radioactive Disintegration" Nature 122, 439 (1928); Phys. Rev 33, 127 (1929)

2.  http://en.wikipedia.org/wiki/WKB_method

3. Schaum's Outline of Quantum Mechanics, 2d edition page 247 ,Yoav Peleg (Author), Reuven Pnini (Author), Elyahu Zaarur (Author), Eugene Hecht (Author) , chapter 11 .

4. Introduction to Quantum Mechanics with Applications to Chemistry by Linus Pauling and E. Bright Wilson Jr.

5. Fundamentals of Modern Physics by Robert Martin Eisberg , page 238-239, also pages 612 -13.

6. http://www1.uprh.edu/rbaretti/WKB7dic2009.htm

 

An alpha particle ( mass= 4mproton ) with energy, E, inside a nucleus may be picture with a wave function ( see Fig 1)

                               Ψinside (x) ~ exp( i k r )         r < R (nuclear size)             (1)

k = (2m(E - (-Vn) )1/2 /h' .  It is bouncing from one side to the other but there is a  possibility of tunneling through the barrier.

Immediately outside outside ( i.e. r>R)  ,   E < Vc and  k →  i (2m(Vc -E  )1/2 /h' , where Vc = (Z-2)(2)kcoulombe2/r

Under the barrier the wave function is

                                      Ψ 2 (x) ~ exp( -  [(2m/h'2 ) (Vc -E  ) ]1/2  r )         .    (2)

 

 

                               

Fig 1. Alpha particle emission is a problem of transmission through a potential barrier, also called tunnelling.

 

The transmission factor is defined by

                                     T  = exp( - 2 r0 R [(2m/h'2 ) (Vc -E  ) ]1/2  dr                    (3)

where r0 is the distance where the particle comes out of the barrier. It is defined by  E = (Z-2)(2) kcoulomb e2 /r0 .

The decay rate λ ~ probabilty/time is given by the expression

                                     λ  = (v/(2R)) T                                                               (4)

where v is the speed of the alpha particle v = ( 2E/m)1/2 .

The integral

                        I = ∫ ( (Z-2)(2)kce2 /r -E  ) ]1/2  dr               R ≤  r ≤ r0  ,

                          = ∫ ( c/r -E  ) ]1/2  dr  

                          = ( c /E1/2 )arctan {(c/R-E)1/2 /E1/2 }  - R (c/R - E)1/2           ,   (5)

where   c = (Z-2) 2 e2 kcoulomb  and  kcoulomb = 9 E9 Nm2 /C2 .

Taking the natural logarithm of (4) gives

    

            ln(λ) = ln(v/2R) -(8m/h'2 )1/2 {  ( c /E1/2 )arctan [ (c/R-E)1/2 /E1/2 ]  - R (c/R - E)1/2 } .  (6)

 

   We have plotted in Fig 1,   ln(λ) vs  (Z-2)/E1/2 ( E~ Mev) .

Fig 1. Alpha decay rates.

 

The lineae regression yields                             y  = -2.7919x + 78.828

                          ln (λ) = -2.7919(Z-2)/E1/2(Mev) + 78.828

 

 

FROM WIKIPEDIA    ********

Alpha particles have a typical kinetic energy of 5 MeV (that is, ≈ 0.13% of their total energy, i.e. 110 TJ/kg) and a speed of 15,000 km/s. This corresponds to a speed of around 0.05 c. There is surprisingly small variation around this energy, due to the heavy dependence of the half-life of this process on the energy produced (see equations in the Geiger–Nuttall law).

Formulated in 1911 by Hans Geiger and John Mitchell Nuttall, in its modern form the Geiger-Nuttall law is

λ = ln(2) / τ     ,  τ ~ half- life 

                                                                \ln\lambda=-a_1\frac{Z}{\sqrt{E}}+a_2

where λ is the decay constant (λ = ln2/half-life), Z the atomic number, E the total kinetic energy (of the alpha particle and the daughter nucleus), and a1 and a2 are constants.

********

 

FORTRAN code for natural log (lambda)

 

c alpha decay formula Gamow, Condon , Gurney
real m ,kcoul
data kcoul ,q /9.e9,1.602E-19/
c(z)=(z-2.)*2.*q**2*kcoul
phase(z)=((c(z)/R -en) /en)**.5
Fln(z)=log(v/(2.*R))-A*( (c(z)/En**(1./2.) )*atan(phase(z))
$ - R*(c(z)/R - En)**(1./2.) )
emev=5.
en=5.E6*1.6E-19
m=4.*1.67E-27
hbar=1.05e-34
v=sqrt(2.*en/m)
R=8.E-15
A=(8.*m/hbar**2)**.5
do 10 iz=50,90
z=float(iz)
x=(z-2.)/sqrt(emev)
print 100,x,fln(z)
10 continue
100 format('(z-2)/E^(1/2),ln(lambda)=',2(3x,e10.3))
stop
end