The Deuteron

05/01/13 5:21 PMThe Deuteron - ICT-Wiki

Page 1 of 13http://ictwiki.iitk.ernet.in/wiki/index.php/The_Deuteron

The DeuteronDeuteron is the simplest nuclear system where nuclearforces come into play. It is the Hydrogen Atom of nu-clear physics. It may also be viewed as the simplest odd-odd nucleus, having one proton and one neutron. It ex-hibits features which require the use of a number of newconcepts such as exchange force, tensor force etc. Of threepossible states of two nucleon system, di-neutron, di-pro-ton, and deuteron only deuteron is known to be stable.Since there are only two particles in the nucleus, thedeuteron problem can be reduced to a one body problemin the center of mass frame and easily solved. We first dis-cuss the various properties of deuteron.

Properties of Deuteron

It exists in its ground state only and has no excited states.Thus when energy is supplied to it, it readily disintegratesinstead of going into higher energy states.

Binding Energy: The binding energy of the deuteron ismeasured to be 2.22MeV. The measurements are carriedout by using the atomic-mass spectroscopy and nuclear re-actions. Two direct methods involve the slow neutron cap-ture by proton or, the inverse process i.e., the photo disin-



tegration of deuteron.

As the binding energy is very small, the deuteron is aloosely bound system. The depth of the potential well be-ing 36MeV and the binding energy being 2.22MeV , it canbe termed as a loosely bound system. It is almost sitting al-most at the mouth of the potential and hence there is no ex-cited state.

Angular momentum and parity: The angular momentum Jhas been measured by a number of optical, radio frequency

and microwave methods. The deuteron has angularmomentum. Since the parity of the nuclear state cannot bemeasured directly, one uses the parity selection rules ofnuclear disintegrations and reactions to determine the pari-ty. An even parity thus emerges as the most plausible as-signment for the ground state of deuteron.

Radius of deuteron. High energy electron scattering givesa root-mean square (rms) charge radius of deuteron to beabout 2.13 Fm. This is pretty large for a two particle sys-tem. The size of deuteron is thus more than 4 Fm.

Magnetic and Quadrupole Moments:

The magnetic dipole and the electric quadrupole moments



of deuteron are most revealing properties as far as the na-ture of nuclear force is concerned.

It possesses a finite magnetic dipole moment of "d =0.8574376(4)"N. This is almost the sum of neutron and pro-ton magnetic moments. The sum , a value quite close to the

measured value but not ex-actly the same. It tells us

two things. One, the contribution of orbital motion to mag-netic moment is zero implying a L = 0 state. Two, the val-ues are not exactly same and, therefore, there is an admix-ture of non-zero L state.

It also possesses a small but finite quadrupole moment ofQ = 0.00282(1) barns. A non-zero quadrupole momentagain implies the presence of a non-zero L state.

Solution of the Deuteron problem

The two body problem of a neutron and a proton interact-ing with a force represented by a potential can

be transformed into the problem of a single particle in apotential by the well known classical mechanical

transformation to the centre of mass frame of reference.

As the binding energy is very small compared to the depth



of the potential, we may neglect it and this approach istermed as the zero potential approximation. For simplicitywe use a square well potential, where the potential insidethe well is constant and negative and the potential outsideis zero. The square well potential is completely defined byits depth V0 and its range r0 . We have,

V = V0 for r < r0 --------------------(1)

= 0 for r > r0 ---------------------(2)

A graphical represen-tation of the potential

We intend to find outthe values of V0 and r0that will reproduce thebinding energy of thedeuteron to be

2.22MeV .

It is natural to assume that the ground state of thedeuteron is primarily (n = 1,l = 0) state. To obtain J = 1, wemust necessarily have S = 1 with L = 0. This assignment



also gives us the positive parity for the ground state ofdeuteron. The Schrdinger equation in the centre of massframe is given by

,----------------(3)

where the reduced mass m is given by,

------------------(4)

Here mp and mn are the masses of proton and neutron re-spectively. The average of the two masses is considered tobe M and r is the relative distance from the centre of mass.For a central potential, the wave function can be separatedinto its radial and angular parts,

-----------------------(5)

The radial wave function ul is the solution of

------------------(6)

This is the radial wave equation. The centrifugal potentialpulls the particles apart and to achieve binding we need anattractive potential. The l = 0 state proves to be the easiest



way to obtain such a binding wavefunction

-------------------(7)

Using E = W = 2.22MeV, the binding energy of thedeuteron, we obtain

for r < r0---------------(8)

for r > r0------------(9)

where,

and .

The general solution is given by,

u(r) = Asinkr + Bcoskr -----------------------(10)

The wave function is finite at r = 0 and zero at ,which leads to u = r# vanishing at the origin and making itsquare integrable. Another consequence of the aboveboundary condition is that B turns out to be zero. Hencethe general solution becomes



u(r) = Asinkr ---------------------(11)

The general solution of the wave function outside thesquare well potential is,

u(r) = Ce r + Der -------------------(12)

Boundary con-dition at infin-ity demandsD = 0 so thatu(r) remainsfinite. Thus

u(r) = Ce r --------------(13)

At r = r0 both the wave function and its derivative shouldbe finite and continuous (and match with each other), sothat

--------------------(14)



--------------------(15)

Taking the ratio of the two equations we obtain,

kcotkr0 = --------------------(16)

Zero binding approximation:

Since ------------(17)

This leads to ---------------------(18)

And --------------------(19)

We thus obtain,

.---------------(20)

Thus we get an approximate value of . From neutronproton scattering, we know that r0 is approximately 2 Fer-mi. Putting the value into the equation (20), we obtain thevalue of V0 = 36MeV . Also since the interior wave functionmust match the external wave function, the value of kr0

must be slightly greater than (116 deg). Also for r0 =2fermi , , which is an estimate of the size of the

deuteron, and is called the relaxation length.



Graphical Solution:

We may write

, where x = kr0

Taking , and . Plot cotx and .The intersecting points give

the solutions for x . We can see that Now, let

Substituting



And

i.e the value of .

Normalization of the Deuteron wave function

The constants A and C are left undetermined since the rela-tionship between well depth and width is independent ofthem. To determine these constants, we impose the nor-malization condition

---------------------(21)

-----------------(22)

----------------------(23)

Combining these equations with the previously obtainedvalues from the differentiability and continuity conditions,we obtain the values of the constants to be



.

To calculate the rms radius, we calculate the average of thesquare of the proton to center of mass distance. This dis-tance is half of the separation r between the proton and theneutron, so that

.-------------------(24)

Both the integrals are solved by partial integration methodand inserting the value of the normalization constants andreplacing the trigonometric functions by simple functionsof $ and k by using kcotkr0 = $ equation, we get,

--------------------(25)

We thus get the rms radius as a function of r0 .

In this derivation we have assumed that the proton is apoint charge, which is not true. It can be shown that if thecharge density of proton is spherically symmetric, the ef-fect of this distributed charge on the rms radius of thedeuteron can be accounted for by adding the square of therms proton radius. Equation (26) therefore becomes



----------------------(26)

Choosing the value of r0 = 2fm , , and , we obtain. So, the size of the deuteron is

of the order of 4 fermi.

Tensor Forces and the deuteron problem

The ground state of deuteron has been observed to be a 3S1state implying a spherically symmetric charge distribution.This in turn implies that the electric quadrupole momentshould be zero. However deuteron is known to possess a

small positive quadrupole moment of.This implies that the deuteron wave function is not spheri-cally symmetric necessitating the introduction of an inter-action potential which is not purely central. The magnetic

moment data also supports this conjecture since.

The total angular momentum of the ground state of thedeuteron is J = 1 and must have a definite parity. The vari-ous possibilities which may lead to J = 1 are listed below:

l = 0,S = 1(3S1)

l = 1,S = 0(1P1)



l = 1,S = 1(3P1)

l = 2,S = 1(3D1)

Configurations of different parity cannot mix. Thus theonly admixtures possible are 3S1 with 3D1 or, 3P1 with 1P1.It is the first admixture of states, which is able to reproducethe observed values of "d and Q. This problem is beyondthe scope of these lectures but may be found in the refer-ence given below.

Reference

1. R.R. Roy and B.P. Nigam, Nuclear Physics (Wiley East-ern, 1979).

Documents

The Deuteron