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# For the alpha particle D m= 0.0304 u which gives 28.3 MeV binding energy! - PowerPoint PPT Presentation

For the alpha particle D m= 0.0304 u which gives 28.3 MeV binding energy!. 236 94. Is Pu unstable to -decay?. 236 94. 232 92. 4 2. Pu  U + . + Q. Q = ( M Pu – M U - M  ) c 2. = ( 236.046071 u – 232.037168 u – 4.002603 u ) 931.5MeV/ u. = 5.87 MeV > 0.

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For the alpha particle Dm= 0.0304 u which gives

28.3 MeV binding energy!

94

Is Puunstable to -decay?

236

94

232

92

4

2

Pu U + 

+ Q

Q = (MPu – MU - M)c2

= (236.046071u – 232.037168u – 4.002603u)931.5MeV/u

= 5.87 MeV > 0

Examine the stability against beta decay by plotting the rest mass

energy M of nuclear isobars (same value of A) along a third axis

perpendicular to the N/Z plane.

42Mo

A = 104 isobars



43Tc

103.912

A

Z

A

Z+1

?

?

-decay: X  Y + -

N

N-1

?

A

Z

A

Z-1

e-capture: X + e  Y

48Cd

103.910

N

N+1

Odd Z

47Ag

Mass, u

103.908

45Rh

Even Z

103.906

44Ru



46Pd

103.904

42

44

46

48

Atomic Number, Z

Generalization of ordinary “Fourier expansion” or “Fourier series”

Note how this pairs canonically conjugate variables  and t.

1.0

MAX

0.5

= FWHM

E

Eo

R=roA1/3

ro1.2 fm

Incident mono-energetic beam

v Dt

A

dW

N = number density in beam

(particles per unit volume)

Solid angle dWrepresents

detector counting the dN

particles per unit time that

scatter through qinto dW

Nnumber of scattering

centers in target

intercepted by beamspot

FLUX = # of particles crossing through unit cross section per sec

= NvDt A / Dt A = Nv

Notice: qNv we call current, I, measured in Coulombs.

dN NF dW dN = s(q)NF dW dN =NFds

-

incident particle velocity, v

Nscattered= NFdsTOTAL

The scattering rate

per unit time

Particles IN (per unit time) = FArea(ofbeamspot)

Particles scattered OUT (per unit time) = F NsTOTAL

D. R. Nygren, J. N. Marx, Physics Today 31 (1978) 46

a

p

p

d

m

dE/dx(keV/cm)

e

Momentum [GeV/c]

and the transitionrate

vz

E = ½ mv2 = ½ m(vx2 + vy2 + vz2 )

vy

Notice for any fixed E, m this defines

a sphere of velocity points all which

give the same kinetic energy.

vx

The number of “states” accessible by that energy are within the

infinitesimal volume (a shell a thickness dv on that sphere).

dV = 4v2dv

E = ½ mv2 = ½ m(vx2 + vy2 + vz2 )

We just argued the number of accessible states (the “density of states”)

is proportional to

4v2dv

dN

dE

 E1/2

already there when the rock was formed?

Which we can rewrite as:

y = x  m + b

Allows for

the presence

of initial 87Sr

of the kinetic

energy of an

alpha particle

emitted by the

nucleus 238U.

The model for

this calculation

is illustrated

on the

potential

energy

diagram

at right.

V

E

x = r1

x = r2

III

II

I

probability of tunneling to here

E

r2

R

So let’s just write

as

When the result is substituted into the exponential the expression for the transmission becomes

(2 expression for the transmission becomes l + 1)( l- m)!

4p( l + m)!

ml(q,f ) = Pml (cosq)eimf

Pml (cosq) = (-1)m(1-cos2q)m[()mPl (cosq)]

d

d (cosq)

1

2l l!

d

d (cosq)

Pl(cosq) = [()l(-sin2q)l ]

So under the parity transformation:

P:ml(q,f ) =ml(p-q,p+f)=(-1)l(-1)m(-1)m ml(q,f )

= (-1)l(-1)2m ml(q,f ) )=(-1)l ml(q,f )

An atomic state’s parity is determined by its angular momentum

l=0 (s-state)  constant parity = +1

l=1 (p-state)  cos parity = -1

l=2 (d-state)  (3cos2-1) parity = +1

Spherical harmonics have (-1)l parity.

spin: Iinitial = sX

and: Ifinal = sX'+ sa + ℓa

1p1/2

1p3/2

1s1/2

4He

So |sX'– sX| < ℓa< sX'+ sX

Sa = 0

Since the emitted is just itsa is described by a wavefunction:

the parity of the emitted a particle is(-1)ℓ

Which defines a selection rule:

restricting us to conservation of angular momentum and parity.

If P X' = P X then ℓ= even

If P X' = -P X then ℓ= odd

This does is just itsnot take into account the effect of the nucleus’ electric charge

which accelerates the positrons and decelerates the electrons.

Adding the Fermi function F(Z,pe) ,

a special factor (generally in powers of Z and pe),

is introduced to account for this.

This phase space factor determines the decay electron momentum spectrum.

(shown below with the kinetic energy spectrum for the nuclide).

the shortest half-lifes ( momentum spectrum. most common) b-decays “super-allowed”

0+ 0+

10C 10B*

14O 14N*

The space parts of the initial and final wavefunctions are idenitical!

What differs?

The iso-spin space part (Chapter 11 and 18)

|MN|2 =

Note: momentum spectrum.

the nuclear matrix element depends on how alikeA,ZandA,Z±1are.

When A,ZA,Z±1|MN|2~ 1

otherwise |MN|2 < 1.

If the wavefunctions correspond to states of

different J or different parities

then |MN|2= 0.

Thus the Fermi selection rules for beta decay

DJ = 0 and

'the nuclear parity must not change'.

Total momentum spectrum. S = 0 (anti-parallel spins)

Total S = 1 parallel spins)

Fermi Decays

Gamow-Teller Decays

Nuclear I = 0

Ii = If + 1

I = 0 or1

With Pe, = (-1)ℓ = +1

PA,Z = PA,Z1

I = 0,1 with no P change

10 momentum spectrum. C10B*

14O14N*

0+ 0+

Fermi Decays

0+ 1+

6He6Li

13B13C

Gamow-Teller Decays

3/2- 1/2-

e, pair account for I = 1 change

carried off by their parallel spins

n  p

3H3He

13N13C

1/2+ 1/2+

1/2+ 1/2+

1/2- 1/2-

Forbidden Decays momentum spectrum.

ℓ=1 “first forbidden”

With either Fermi decays s = 0

Gamow-Teller decays s = 1

with Parity change!

Forbidden Decays momentum spectrum.

even rarer!

ℓ=2 “second forbidden”

With either Fermi decays s = 0

Gamow-Teller decays s = 1

With no Parity change!

Fermi and Gamow-Teller already allow (account for)

I= 0, 1 with no parity change

M momentum spectrum. össbauer Effect

If this change is large enough, the  will

not be absorbed by an identical nucleus.

In fact, for absorption, actually need to exceed the step between energy

levels by enough to provide the nucleus with the needed recoil:

pN2

2mN

p2

2mN

TN =

=

p=Eg/c

The photon energy is mismatched by

As an example consider the distinctive 14.4 keV momentum spectrum. gfrom 57Fe.

~90% of the 57Fe* decays are through this intermediate level produce 14.4 keV s.

t=270d

7/2

57Co

The recoil energy of the iron-57 nucleus is

EC

136keV

5/2

t=10-7s

3/2

14.4keV

1/2

57Fe

With  = 10-7 s,  =10-8eV

this is 5 orders of magnitude greater than the natural linewidth

of the iron transition which produced the photon!

The Compound Nucleus momentum spectrum.

[ Ne]*

20

10

The Optical Model momentum spectrum.

To quantum mechanically describe a particle being absorbed,

we resort to the use of a complex potential

in what is called the optical model.

Consider a traveling wave moving in a potential V then

this plane wavefunction is written

where

If the potential V is replaced by V + iW then k also becomes complex

and the wavefunction can be written

and now here

A possible ( momentum spectrum. and observed) spontaneous fission reaction

8.5 MeV/A

7.5 MeV/A

Gains ~1 MeVper nucleon!

2119 MeV = 238 MeV

released by splitting

238U

119Pd

Z momentum spectrum. 2/A=36

such unstable states

decay in characteristic

nuclear times ~10-22sec

Z2/A=49

Tunneling does allow spontaneous fission, but it must compete with other decay mechanisms (-decay)

The potential energy V(r) = constant-B

as a function of the separation, r, between fragments.

At smaller values of momentum spectrum. x, fission by barrier penetration can occur,

However recall that the transmission factor (e.g., for -decay) is

where

m

while for  particles (m~4u)

this gave reasonable, observable

probabilities for tunneling/decay

for the masses of the nuclear fragments we’re talking about,

 can become huge and X negligible.

only momentum spectrum.

the

Natural uranium (0.7% 235U, 99.3% 238U)

undergoes thermal fission

Fission produces mostly fast neutrons

Mev

but is most efficiently

induced by slow neutrons

E (eV)

The proton-proton cycle momentum spectrum.

The sun 1st makes deuterium through the weak (slow) process:

Q=0.42 MeV

then

Q=5.49 MeV

2 passes through both of the above steps then can allow

Q=12.86 MeV

This last step won’t happen until the first two steps have built up

sufficient quantities of tritium that the last step even becomes possible.

2(Q1+Q2)+Q3=24.68 MeV

plus two positrons whose

annihilation brings an extra

4mec2 = 40.511 MeV

The CNO cycle momentum spectrum.

Q=1.95 MeV

Q=1.20 MeV

Q=7.55 MeV

Q=7.34 MeV

Q=1.68 MeV

Q=4.96 MeV

carbon, nitrogen and oxygen are only catalysts

The 1 momentum spectrum. st generation of stars (following the big bang) have no C or N.

The only route for hydrogen burning was through the p-p chain.

In later generations

the relative

importance

of the two processes

depends upon

temperature.

Rate of energy production

Shown are curves

for solar densities

105 kg m-3 for protons

and 103kg m-3 for 12C.