(4) Tema 3-1 Desintegracion Alfa

7/28/2019 (4) Tema 3-1 Desintegracion Alfa

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Para esta lectura:

Summary from last time:

- Coocer ley de desintegracin radiactiva, vida media y periodo de semidesintegrac.

-Ser capaz de calcular los factores de Q de la desintegracin alfa y comprender-las implicaciones para la estabilidad nuclear

Una mirada de la mecnica cuntica en la desintegracin alfa


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By 1928, George Gamouw (brilliant Russian born theorist)

developed a quasi-classical model for alpha decay

Good friend ofLev Landau and another

Russian physicist Dmitri Ivanenkothe Three Musketeers

Worked in several different fields

Early 1900s: radioactive decay known to have characteristic rates and energies.

However, no clear explanation as to why......

Attempted twice to defect in 1932, trying to kayak ~250 km over the Arctic Sea toNorway. Both attempts failed.......

In 1933, tried a less dramatic approach: disappeared when attending a conference in

Brussels. Turned up In 1934 in the United States.

One of the most significant physicists never to have won NP.

contributed to big bang theory and theoretical biology


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Cmo algunas partculas se forman en el ncleo?

Consider 2 protons + 2 neutrons in heavy nucleus:Liquid drop modelBE~ 8 MeV/nucleon for all nucleons

However, shell model BE~ 6 MeV/nucleon forouter nucleons

Total BE of 4 outer nucleons ~ 4 x 6 MeV = 24 MeV

Imagine 4 nucleones amalgamate to form -particleDetails not really clear.... (QM plays a large role)

As que por qu no acaba de salir??

Why dont heavy nuclei release -particles and decompose instantaneously?

BE of an a particle ~ 28.3 MeV

PotentialEnergy

4 nucleons in nucleus

a particle outside nucleus

BEis depth of the potential energy well in which the nucleons sit,i.e. Piense of as a negative potential energy:

Distance

Barrier at edge of nucleus due to coulomb potential

i.e. -particle has (+) kinetic energy and en principio podria dejer el nucleo


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Pictorial representation of potentials:

Potential energy of nucleons:

Difference in potential

energy of 4 nucleons and a:


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Entonces, cmo la partcula alfa nunca abandona ?

QM tunel a traves de la barrera:

Probability of tunnelling out =

Square of ratio of wavefunctions (Y)

inside and outside barrier

Tunnelling through 1/r Coulombpotential difficult to analyze

The solution: approximate to

sequence of square barriers:

By classical physics, there is no possibility for alpha particle a subir la barrier


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Reminder of square barrier tunnelling:

Transmission (T) probability given by:

U0 = barrier height

t = barrier thicknessm = particle mass

E = particle energy

By multiplying transmission probabilities:

1 2 32 ( )2 2 2 ..........

t dtt t tT e e e e

(where integral is over thickness for which U(t) > E)

Through sequence of barriers:

Simply rewriting this byreplacing t with radial coordinate r:

GT ewith Gamow factor:

R is radius of strong potential (i.e. nucleus)

b is distance at which U(r) = E

Note: sqrt (barrier height)


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Tunnelling through a Coulomb barrier:

For r > R, Coulomb barrier potential is: (z = 2 foraparticle)

Substitute into Gamow factor:

Integrate:

( do this with change of variable k using r = b cos2(k) )

Typically, coulomb barrier potential is much larger than particle kinetic energy

Then b >> R, and one can make the approximations: 1)

2)

1

cos 2R Rb b

2

2

R R Rb bb

Gamow factor becomes:

Z ofdaughternucleus


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gives the transmission probability ofa penetrating theCoulomb barrier when it approaches it

However, to know decay rate we must also know how often particle approaches

barrier:

Recall that b is the distance r at which U(r) = E, i.e.

Also, since particles are non-relativistic:

Substituting these obtains

the simplest form of G:

(fine structure constant)

T = Exp(-G)

Assume -particle has velocity V0wi th in nucleus, radius R

Will then make V0/2R approaches to barrier per second

Thus probability of leaving nucleus per unit time:

Have their normal meaning

=


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Kinetic energy ofa particle

Shorter t1/2 for lower Z and higher Q

1) Z

2) Q

Lifetime behaviour within alpha decay groups:

1) Z dependence:

=

Potenti

al

Energy

r

Lower potential barrier for lower Z nuclei

- G factor lower

- Higher probability of escape: lower t1/2In nucleus

Outside nucleus

2) Q dependence Higher Q higher V0 inside nucleus

More attempts to escape, lower t1/2


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Functional form of tunnelling probability function:

Where the variables f and g are given by:

Approximately constant: ln(R) and RZ vary only slightlybetween common -emitters.V0 assumed to be constant.

Constant

Recall the empirical Geiger-Nuttall relation:

Take the logarithm of both sides:

functionally identical to tunnelling equation above (assuming KE of

liberated -particle takes lions share of Q released in decay)

Log dependence ofl and E get enormous range in decay constant for relatively

small changes in KE of -particle due to 1/r dependence of coulomb barrier

=

b1

=


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Only one unknown left: V0.

For simplicity, assume V0 = V:

If the mass of daughter nucleus is >> a particle, then E ~ Q

Finalizing the tunnelling rate equation:

Get finalized equation:

=

fits well to experimental decay rates

best fits yield empirical relation for radius:

R = 1.53 x A1/3 fm.

Compare this to empirical relation found from

scattering measurements: R = 1.2 x A1/3 fm.

Empirical radius from decay fits are aconvolution of nuclear and -particle radii

ln(l)Z (m/Q)1/2


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Worked example:

2Gv e

Rl

24 8Zc mc RZGv c

a a

Calculate the half life for the emission, , which has Q = 6 MeV235 231 493 91 2Np Pa He

(Neptunium protactinium)

Aparent = 235

Zdaughter= 91

= 0.8 x 10-5 s-1

Rparent = 1.53 x A1/3 fm = 9.44 fm

Determines height of coulomb barrier that

alpha particle has to tunnel through

a) Calculate energy of emitted alpha particle:Q KE of alpha = m

av2

ma= 6.64 10-27 kg

v = 1.7 x 107 m/s

b) Calculate emission probability from nucleus:

Fine structure

constant = 1/137

= 147.3 87.3 = 60

Careful to convert

R from fm to m

T1/2 = ln(2) / l = 86,600 s = 24 hours


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works well in broad terms:

Explains Geiger-Nuttall relation and other caracteristicas claves

predicts most emission lifetimes fairly accurately

find that Q


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For next lecture:

A closer look at beta decay

Summary:

- explain alpha emission probability in terms of barrier tunnelling

- be able to apply tunnelling probability function to determine decay rates


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2. Potencial pozo infinito

3. Potencial pozo finito cuadrado

4. Potencial pozo redondeado (Wood-Saxon)

5.Potencial exponencial

6. Potencial de Yukawa

1. Potencial tipo oscilador armnico:

7. Potencial con centro repulsivo impenetrable.


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(4) Tema 3-1 Desintegracion Alfa

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