Nuclear Physics and A. K. Jain Assignment

Nuclear Physics and A. K. Jain Assignment Words: 1600

The Deuteron (Lecture time: 2 Hours) Deuteron is the simplest nuclear system where nuclear forces come into play. It is the “Hydrogen Atom” of nuclear physics. It may also be viewed as the simplest odd-odd nucleus, having one proton and one neutron. It exhibits features which require the use of a number of new concepts such as exchange force, tensor force etc. Of three possible states of two nucleon system, did-neutron, did-proton, and deuteron only deuteron is known to be stable.

Since there are only two particles in he nucleus, the deuteron problem can be reduced to a one body problem in the center of mass frame and easily solved. We first discuss 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 disintegrates instead of going into higher energy states. Binding Energy: The binding energy of the deuteron is measured to be 2. Move . The measurements are carried out by using the atomic- mass spectroscopy and nuclear reactions.

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Two direct methods involve the slow neutron capture by proton 0 n p d 0 0 or, the inverse process I. . , the photo disintegration of deuteron. As the binding energy is very small, the deuteron is a loosely bound system. The depth of the potential well being move and the binding energy being 2. Move , it can be termed as a loosely bound system. It is almost sitting almost at the mouth of the potential and hence there is no excited state. Angular momentum and parity: The angular momentum J has been measured by a number of optical, radio frequency and microwave methods.

The deuteron has J 0 1 angular momentum. Since the parity of the nuclear state cannot be measured erectly, one uses the parity selection rules of nuclear disintegrations and reactions to determine the parity. An even parity thus emerges as the most 1 Lecture notes by Proof. A. K. GAIN, IT ROOKIE plausible assignment for the ground state of deuteron. Radius of deuteron. High energy electron scattering gives a root- mean square (arms) charge radius of deuteron to be about 2. 13 Fm. This is pretty large for a two particle system. The size of deuteron is thus more than 4 Fm.

Magnetic and Quadruple Moments: The magnetic dipole and the electric quadruple moments of deuteron are most revealing properties as far as the nature f nuclear force is concerned. It possesses a finite magnetic dipole moment of Odd 0 0. 8574376(4) ON . This is almost the sum of neutron and proton magnetic moments. The sum 1 2 p 0 On N (g SP 0 g sin ) 0 0. 8798040 N , a value quite close to the measured value but not exactly the same. It tells us two things. One, the contribution of orbital motion to magnetic moment is zero implying a state. Two, the values are not exactly same and, therefore, there is an admixture of non-zero L state.

It also possesses a small but finite quadruple moment of Q 0 0. 00282(1) barns. A non-zero quadruple moment again implies the presence of a non-zero L state. Solution of the Deuteron problem The two body problem of a neutron and a proton interacting with a force single particle in a potential V (r ) 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 is termed as the zero potential approximation.

For simplicity we use a square well potential, where the potential inside the well is constant and negative and the potential outside is zero. The square well potential is completely defined by its depth VOW and its range or . We have, 2 Lecture notes by proof. A. K. GAIN, IT ROOKIE v o vivo ROR for r or A graphical representation of the potential We intend to find out the values of VOW and or that will reproduce the binding energy of the deuteron to be 2. Move . It is natural to assume that the ground state of the deuteron is primarily Is (n 0 1, 1 0 0) state. To obtain J=l, we must necessarily have S=l with L=O.

This assignment also gives us the positive parity for the ground state of deuteron. The Scarödinner equation in the centre of mass frame is given by 2 m 0 20 ABOVE, where the reduced mass m is given by, Here m p and MN are the masses of proton and neutron respectively. The average of the two masses is considered to be M and r is the relative distance from the centre of mass. For a central potential, the wave function can be separated into its radial and angular parts, 3 Lecture notes by Proof. A. K. GAIN, o lour IT ROOKIE The radial wave function LU is the solution of d o I (1 DRP runner This is the radial wave equation.

The centrifugal potential pulls the particles apart and to achieve binding we need an attractive potential. The 1 0 0 state proves to be he easiest way to obtain such a binding habituation d oho (r ) O O (r ) o EYE (r ) M DRP 2 (7) Using E 0 OWE 0 02. 22 Move, the binding energy of the deuteron, we obtain d oho (r oho (r ) 002 DRP where, for r 0 or and M(VIVO) M (VOW) MM The general solution is given by, B coos Or The wave function is finite at (10) and zero at r 00, vanishing at the origin and making it square integrally. Another consequence of the above boundary condition is that B turns out to be zero.

Hence the general solution becomes Or u(r) CEO r o Doe r The general solution of the wave function outside the square well potential is, (12) 4 Boundary condition at infinity demands D 0 0 so that u(r) remains finite. Thus At r 0 or both the wave function and its derivative should be finite and continuous (and match with each other), so that (13) A SST or AAA coos CEO or Taking the ratio of the two equations we obtain, (14) (15) Zero binding approximation: Since VOW 00 W , cot 0 or 0 This leads to And We thus obtain, (16) o w vivo 030 or-ћ. …. Etc 22 5 Provo 02 MM (20) Thus we get an approximate value of VOW or 2 .

From neutron proton scattering, we know that or is approximately 2 Fermi. Putting the value into the equation (20), we obtain the value of VOW 0 move . Also since the interior wave function must match the external wave function, the value of 0 or must be slightly greater than 0 1 (116 deg). Also for or 2 Fermi , 0 4. 3 Fermi , which is an estimate of the size 2 of the deuteron, and Graphical Solution: We may write 1 is called the relaxation length. 0 cot cot x 0 Taking Rorer , where and sorrow 0. 7, 20 or intersecting points give the solutions for x .

We can see that Now, let o ROR 6 Substituting And ROR 20 or I. E the value of 0 0. 8 . Normalization of the Deuteron wave function The constants A and C are left undetermined since the relationship between well depth and width is independent of them. To determine these constants, we impose the normalization condition 7 Rocco]2 sin rodeo 02 or 020 r droll (21) [1 coco 20 r r droll 20 (22) 1 c 2 roar 20 or]IEEE 20 2 (23) Combining these equations with the previously obtained values from the differentiability and continuity conditions, we obtain the values of the constants to be 20 sin 20 iron center of mass distance.

This distance is half of the separation r between the proton and the neutron, so that 0 r r r DRP . Arrear (24) Both the integrals are solved by partial integration method and inserting the value of he normalization constants and replacing the trigonometric functions by simple functions of and 0 by using equation, we get, (25) 3rd r 01 280 80 2401 We thus get the arms radius as a function of or . In this derivation we have assumed that the proton is a point charge, which is not true.

It can be shown that if the charge density of proton is spherically symmetric, the effect of this distributed charge on the arms radius of the deuteron can be accounted for by adding the square of the arms proton radius. Equation (26) therefore becomes 8 r 2 order Choosing 2 obtain rd 1 2 (26) and the of 2 FM , VOW move , 0. 8 FM, 0 2. Fermi . 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 ASS state implying a spherically symmetric charge distribution.

This in turn implies that the electric quadruple moment should be zero. However deuteron is known to possess a small positive quadruple moment of Q 0 2. 88 010027 CM . This implies that the deuteron wave function is not spherically symmetric necessitating the introduction of an interaction potential which is not purely central. The magnetic moment data also supports this conjecture since 0 p 0 On 0 Odd . The total angular momentum of the ground state of the deuteron is and must have a definite parity.

The various possibilities which may lead tog=1 are listed below: 1 00, S O I(ASS 102, S O I(ADD )configurations of different parity cannot mix. Thus the only admixtures possible are ASS with 3 ODL or, 3 P with IP . It is the first admixture of states, which is able toll 1 reproduce the observed values of 0 d and Q. This problem is beyond the scope of these lectures but may be found in the reference given below. References 1. R. R. Roy and B. P. Enigma, Nuclear Physics (Wiley Eastern, 1979). 9

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