# Spin Parity Assignment Of Mortgage

## Nuclear Physics *PHY303*

## Solutions 2

### Constants

Charge of the electrone = 1.6 10^{-19}C

Mass of the electronm_{e}= 9.11 10^{-31}kg = 511 keV/c^{2}

Mass of the protonm_{p}= 1.673 10^{-27}kg = 938.272 MeV/c^{2}

Mass of the neutronm_{n}= 1.675 10^{-27}kg = 939.566 MeV/c^{2}

Planck constanth = 6.626 10^{-34}J s = 4.136 10^{-15}eV s

Boltzmann constantk = 1.38 10^{-23}J K^{-1}= 8.617 10^{-5}eV K^{-1}

Speed of light in free spacec = 3.00 10^{8}m s^{-1}

Permittivity of free space_{0}= 8.85 10^{-12}F m^{-1}

Permeability of free spaceµ_{0}= 4 10^{-7}H m^{-1}

Avogadro constantN_{A}= 6.02 10^{26}kg-mol^{-1}

Rydberg constantR = 1.10 10^{7}m^{-1}

Bohr magnetonµ_{B}= 9.27 10^{-24}J T^{-1}

Nuclear magnetonµ_{N}= 5.0508 10^{-27}J T^{-1}= 3.1525 10^{-14}MeV T^{-1}

Fine structure constant= 1/137

### Useful data

Unified atomic mass unit1u = 1.66 10^{-27}kg = 931.502 MeV/c^{2}

Energy conversion1 eV = 1.6 10^{-19}J

Year in seconds1 yr = 3.16 10^{7}s

Atmospheric pressure1 atmosphere = 1.01 10^{5}N m^{-2}

Acceleration due to gravity on Earth's surfaceg = 9.81 m s^{-2}1 gram moleculeatSTPoccupies22.4 litres

**Solutions 3**

Go back to the Problems or to the constants and useful data.

Go back to the Problems or to the constants and useful data.

- This problem can be handled by focusing on the right hand side of the energy level figure given in the notes and repeated here

Note this latter asignment follows the ordering given in the question which slightly differs at this point from that in the figure. This just underlines the fact that the ordering of the levels does exhibit some local variations and so the figure should be considered as no more than a reasonable approximation.

Across these three isotopes the number of protons remains the same of course but the number of neutrons rises from

**7**to

**9**. This means that

**has one neutron missing in the**

^{15}O**1p**level;

_{1/2}**has a complete**

^{16}O**1p**neutron level;

_{1/2}**has a single neutron in the**

^{17}O**1d**level. From this and the given data we can deduce:

_{5/2}The binding energy of a neutron in 1p level is _{1/2}127.62 - 111.96 = 15.66 MeV |

The binding energy of a neutron in 1d level is _{5/2}131.76 - 127.62 = 4.14 MeV |

Thus the (1p difference is _{1/2} - 1d_{5/2})15.66 - 4.14 = 11.52 MeV |

The numbers of nucleons associated with each of the listed levels are:

1s_{1/2} | 1p_{3/2} | 1p_{1/2} | 1d_{5/2} | 2s_{1/2} | 1d_{3/2} | 1f_{7/2} | 2p_{3/2} |

2 | 4 | 2 | 6 | 2 | 4 | 8 | 4 |

2 | 6 | 8 | 14 | 16 | 20 | 28 | 32 |

From this it can be seen that the

**3rd proton**is in the

**1p**level; the

_{3/2}**11th proton**is in the

**1d**level; the

_{5/2}**17th neutron**is in the

**1d**level; the

_{3/2}**21st proton**is in the

**1f**level. Hence the ground and excited state spin and parity assignments are:

_{7/2}^{7}Li_{3} | ^{23}Na_{11} | ^{33}S_{16} | ^{41}Sc_{21} |

(3/2)^{-} | (5/2)^{+} | (3/2)^{+} | (7/2)^{-} |

(1/2)^{-} | (1/2)^{+} | (7/2)^{-} | (3/2)^{-} |

(1/2)^{+} | (1/2)^{-} | (1/2)^{+} | (3/2)^{+} |

The nucleus

**has an odd number of protons and an even (**

^{209}Bi_{83}**magic**) number of neutrons. Looking at the energy level diagram given above it would be expected that the odd proton would be in the

**1h**level indicating a

_{9/2}**(9/2)**ground state, as is observed to be the case.

^{-}In this simple model the magnetic moment comes from the single proton, being a combination of its orbital and intrinsic moments. For the ground state

**= 5**and

**j = 9/2**thus we need the expression for the

**j = - 1/2**Schmidt limit. This is:

µ = [g_{} j(j + 3/2)/(j + 1) - g_{s}j/(2(j + 1))] µ_{N} |

For the proton

**g**and

_{}= 1**g**. Inserting these and the value of

_{s}= 5.5857**j**gives

**µ = 2.62 µ**which is quite a bit smaller than the observed value of

_{N}**4.1 µ**. This may well reflect a mixing of states - the nearest other state is

_{N}**(7/2)**which has the same parity and so is a likely candidate. This is also a

^{-}**j = + 1/2**level and so a proton in it would produce a relatively large magnetic moment (

**5.79 µ**). It should also be noted that the value of

_{N}**g**which has been used is that for a free proton. The appropriate value for a proton inside a nucleus is most likely to be different and an examination of the distribution of the measured moments relative to the Schmidt limits suggests a value of something like

_{s}**0.6 g**. Both of these factors would enlarge the magnetic moment of

_{s}**beyond the simple Shell Model prediction.**

^{209}Bi_{83}The Electric Quadrupole moment can be simply thought of, in these units, as (minus) the area enclosed by the orbit of the proton. Estimating the radius of the orbit as approximately that of the nucleus

**r**would give a quadrupole moment

_{o}A^{1/3}**-0.5 x 10**. This is pretty close to the observed value considering the crudeness of the model. A quantum mechanical treatment reveals the dependence on

^{-28}m^{2}**j**and gives

**-0.4 x 10**for this case.

^{-28}m^{2}As indicated above this nucleus has a magic number (

**126**) of neutrons and so the small neutron capture cross-section is not surprising.

The nuclide

**has an odd number of both protons and neutrons. From the energy level diagram above the 21st nucleon should be in the**

^{42}Sc_{21}**1f**level and the combination of two

_{7/2}**(7/2)**states will give

^{-}**(7)**if all the angular momenta are "parallel".

^{+}In terms of magnetic moments the expression for the

**j = + 1/2**Schmidt limit is required. This is:

µ = [g_{} (j - 1/2) + g_{s}/2] µ_{N} |

For the proton

**g**and

_{}= 1**g**and for the neutron

_{s}= 5.5857**g**and

_{}= 0**g**. Thus the two contributions to the magnetic moment are

_{s}= -3.8261**5.79 µ**from the proton and

_{N}**-1.91 µ**from the neutron - a total of

_{N}**3.88 µ**.

_{N}This is a straightforward if somewhat tedious classical slog. For those who like that sort of thing the quadrupole moment is given by an integral of the form:

Then making use of the axial symmetry:

We can equate

**dV**to the elemental ring volume illustrated.

The elipsoidal shape is represented by

where

**b**is the deformation parameter and the spherical harmonic

Writing

**p**for

**cos()**and integrating over

**r**from

**0**to

**R**reduces the integral to

Inserting the expression for

**R**and substituting

**Ze 4(R**

_{av}

**)**

^{3}

**/3**for the density produces the integral

Assuming that

**b**is small, the expansion is only taken up to

**b**

^{2}. Also note

which just leaves two terms in the integral

Evaluation of this yields

**Solutions 4**

Go back to the Problems or to the constants and useful data.

Go back to the Problems or to the constants and useful data.

- The decay scheme is

^{14}O_{8} (^{14}N_{7})* + e^{+} + _{e} + e^{-} |

where the latter is the excess atomic electron and the excited nitrogen isotope decays via gamma emission. The rest mass energies before and after are related by

M(14,8)c^{2} = M(14,7)c^{2} + 2m_{e} + Q |

The

**Q = 1.835 + 2.313 = 4.148 MeV**and converting this into unified mass units gives the mass of the original atom as

**14.008625 u**.

The missing piece of information is the binding energy of

**. Using the given coefficients the following values are obtained for the terms**

^{231}Th_{90}3187.8 | -489.42 | -757.16 | -225.19 | 0 | 1716.0 MeV |

Using this with the binding energies given for the other nuclei gives

**4.6 MeV**for the alpha particle energy and

**0.3 MeV**for the maximum beta energy.

Allowing a possible error of about

**1 keV**and ensuring that each state can reach the ground state a possible scheme is

By looking at the curve for the binding energy per nucleon

**B/A**as a function of

**A**

it can be deduced that fission is generally not favoured in terms of energy release unless

**A**is greater that approximately

**60**. In order to pin down at which value of

**A**alpha particle decay becomes energetically favoured it is necessary to be a little more specific. Consider just those nuclides which are stable against beta decay. To find the value of

**Z**at the bottom of the valley for a given

**A**essentially requires the differentiation of the expression for the binding energy with respect to

**Z**at fixed

**A**.

a_{v}A | - a_{s}A^{2/3} | - a_{c}Z^{2}/(A)^{1/3} | - a_{a}A(1 - 2Z/A)^{2} |

0 | 0 | - 2Za_{c}/(A)^{1/3} | + 4a_{a}(1 - 2Z/A) |

Equating the differential to zero produces a relationship between

**Z**and

**A**for the nuclei which lie on the line of beta stability. That relationship can be expressed as

The term in the brackets shows how the ratio

**Z/A**increasingly differs from

**0.5**as

**A**gets larger. This expression means that

**B**can be written as a fuction of

**A**alone. My prefered method at this point is to use a spreadsheet program such as Excel to calculate binding energies for a range of

**A**values, and hence determine where the increase in

**B**for an decrease in

**A**of 4 becomes less than the binding energy of the deuteron. This occurs for

An alternative, approximate method can be used. From the above sketch for

**B/A**it appears that for

**A**greater than about

**60**the curve can be approximated by a straight line. An example of how this could be done using the values of

**a**etc given in

_{v}**question 2**and making the appropriate substitution for

**Z/A**is illustrated below.

a_{v} | -a_{s}A^{-1/3} | -a_{c}(Z/A)^{2}A^{2/3} | -a_{a}(1 - 2Z/A)^{2} | |

13.8 | -2.80 | -2.34 | -0.37 | 8.29 MeV |

13.8 | -2.22 | -3.18 | -0.79 | 7.61 MeV |

The straight line passing through these two points has the equation

Finally, this approximation together with the given binding energy of

**28.3 MeV**for the alpha particle, enables the simple calculation of the energy released in the alpha decay of a nucleus with mass number

**A**. Setting this energy to zero yields about

**141**for the limiting value of

**A**.

The actual lightest alpha emitter of this type is

**.**

^{146}Sm_{62}The deduction for the normal density of states factor is given in the notes and from consideration of coordinate and momentum space for both the electron and the neutrino an expression

dn = (V^{2}16^{2}/h^{6}). (p_{e})^{2}dp_{e}(p_{})^{2}dp_{} |

is obtained, which through the substitutions

p_{} = (T_{0} - T_{e})/c : dp_{}/dE = 1/c |

yields

dn/dE = (V^{2}16^{2}/(h^{6}c^{3})). (T_{0} - T_{e})^{2}(p_{e})^{2}dp_{e} |

In order to find the effect of a neutrino mass it is necessary to re-evaluate the above substitutions which assumed zero mass. The expression

**E**must now be used for the neutrino where

^{2}= (pc)^{2}+ (mc^{2})^{2}**E**is

**(T**. The substitutions become

_{0}- T_{e})p_{} = [(T_{0} - T_{e})/c].[1 - m^{2}c^{4}/(T_{0} - T_{e})^{2}]^{1/2} : dp_{}/dE = [1/c].[1 - m^{2}c^{4}/(T_{0} - T_{e})^{2}]^{-1/2} |

giving the required overall multiplicative factor.

The slope of the normal spectrum goes to zero as

**T**approaches its limiting value

_{e}**T**. This is due to the presence of the factor

_{0}**(T**throughout. In the case where the neutrino has mass the limit of

_{0}- T_{e})**T**is

_{0}- T_{e}**m**rather than zero. There is now a term in the slope of the spectrum which is proportional to the reciprocal of the multiplicative factor and since this latter goes to zero at the limit, the slope there is infinite.

_{}c^{2}The maximum energy available is

**1.29 MeV**. Neglecting the recoil of the proton and assigning this energy to the electron gives a momentum of

**[(1.29)**. The maximum momentum that the anti-neutrino can carry is less because the electron still has its rest mass energy. Thus for the anti-neutrino the momentum is

^{2}- (0.51)^{2}]^{1/2}= 1.185 MeV/c**1.29 - 0.51 = 0.78 MeV/c**.

Back to previous page

In the case of difficulties with this course contact Prof Neil Spooner.If you find this material at all helpful please let me know!

© 1999 - FH Combley, 2000 - CN Booth, 2009 - NJC Spooner

Сьюзан услышала глухой хлопок, когда уже спустилась на несколько пролетов. Звук показался очень далеким, едва различимым в шуме генераторов. Она никогда раньше не слышала выстрелов, разве что по телевизору, но не сомневалась в том, что это был за звук. Сьюзан словно пронзило током. В панике она сразу же представила себе самое худшее.

## Comments