Looking for a Tutor Near You?

Post Learning Requirement » x
Ask a Question
x

Choose Country Code

x

Direction

x

Ask a Question

x

Hire a Tutor
Delhi

Nuclear Physics Notes

Home > Study Notes > Nuclear Physics Notes
Published in: Physics
18,330 Views

Notes On Nuclear Physics.

Akhilesh K / Lucknow

4 years of teaching experience

Qualification: M.Sc (NIT Rourkela - 2019)

Teaches: All Subjects, English, Mathematics, Science, Chemistry, Physics, Algebra, IIT JEE Mains, AIPMT, NEET

Contact this Tutor
  1. Nucleus The Deuteron 1. 2. 3. 1 eli A deuteron (2H nucleus) consists of a neutron and a proton. (A neutral atom of 21-1 is called deuterium.) It is only and the simplest bound state of nucleons and therefore gives us an ideal system for studying the nucleon-nucleon interaction. Exist only in the ground state which is stable. No excited states. An interesting feature of the deuteron is that it does not have excited states because it is a weakly bound system. Constituents Mass Binding energy Angular momentum Magnetic dipole moment Electric quadruple moment RMS separation Parity 1 proton 1 neutron 2.014732 u 2.226 ± 0.002 MeV 1 0.85735 ± 0.00003 +(2.82 ± 0.01)x 10-27 cm2 4.2 fm even
  2. The Deuteron — Measured Binding energy Mass doublet method m(2H) = 2.014101789 ± 0.000000021 u m(2H) = 2.014101771 ± 0.000000015 u B = [m(1H) + m(n) - = 2.22463 ± 0.00004 Mev Measure energy released (gamma) on formation from a neutron and proton Photodissociation 2.2245 ± 0.000002 Mev B = 2.224 ± 0.002 MeV (1) (2) (3) For experimental descriptions see Krane's textbook in chapters 3 and 4.
  3. Binding energy of the deuteron is 2.2 MeV. If the neutron in the deuteron were to decay to form a proton, electron and antineutrino, the combined mass energies of these particles would be 2(938.27 MeV) + 0.511 MeV = 1877.05 MeV But the mass of the deuteron is 1875.6 MeV !! As we have discussed previously, the average binding energy per nucleon is about 7 8 MeV for typical nuclei. The binding energy of the deuteron, B = 2.224 MeV, is away too small when compared with typical nuclei. This means that the deuteron is very weakly bound. The existence of quadruple moment indicates that the deuteron is not truly spherical. So the bound state cannot be said to be a pure S-state but should be considered to have a small admixture of D-state component The mixing of S and D states is possible only if the nucleon-nucleon potential has a tensor component. Since the deuteron has even parity, the mixing of odd-I states is not possible. The nuclear force being a strong interaction, conserves parity. Here we want to explore more about this result and study the properties of the deuteron.
  4. Deuteron Wave function 1 c Phenomenologied potential b Nuclear model potential for studying the deuteron To simplify the analysis of the deuteron, we assume that the nucleon- nucleon potential is a three-dimensional square well, as shown in the figure a:
  5. Quantum mechanical description of the weak binding for the deuteron —o Enew Binding enerw of deuteron well Distance for r < R for r > R (1) Here r represents the separation between the proton and the neutron, so R is in effect a measure of the diameter of the deuteron. The dynamical behavior of a nucleon must be described by the Schrödinger's equation: — V (r) + — E V (r) where m is the nucleon mass. (2) 2m If the potential is not orientationally dependent, a central potential, then the wave function solution can be separated into radial and angular parts: 9U(r) = p) (3)
  6. Substitute R(r) = u(r)/r in to the Schrödinger's equation the function u(r) satisfies the following equation ; h2 d2u I (l + 1)h2 u(r) = Eu(r) 2 2 2m dr 2mr (4) The solution u(r) is labeled by two quantum numbers n and I so that: u (r) unl (r) (5) The full solution W(r) then can be written as unl (r) (6) V (r) = vnlm(r, 0, p) = Rnl with (r) Three quantum numbers to define an eigenstate n: l: m: the principal quantum number which determines the energy of an eigenstate. the orbital angular momentum quantum number. the magnetic quantum number, —l m l.
  7. The angular part of the solution is called the "spherical harmonic" of order l, m and satisfies the following equations: and where C —h2 and L sin 0 DO sin2 0 Dp 2 sino DO (7) (8) For the case of a three dimensional square well potential with zero angular momentum (l = 0), which we use as the model potential for studying the ground state of the deuteron, the Schrödinger's equation can be simplified into: h2 d2u 2 2m dr h2 d2u 2 2m dr vou(r) Eu(r) Eu(r) , forrR
  8. I. When r < R The Schrödinger's equation is — + kl 0 with kl h2 d2u 2 2m dr This equation can be rearranged into: 2 And the solution is -Vou(r) = Eu(r) 2m(E+V0) u(r) (10) (11) (12) limv(r) = lim To keep the wave function finite for r 0 The coefficient B must be set to zero. Therefore the acceptable solution of physical meaning is (13)
  9. Il. When r > R The Schrödinger's equation is: The solution is with h2 d2u = Eu(r) 2 2m dr ( 14) (15) To keep the wave function finite for r 00 2mE limu(r) = 0 The coefficient D must be set to zero. Therefore the acceptable solution of physical meaning is (16)
  10. Applying the continuity conditions on u(r) and du/dr at r = R, we obtain (17) This transcendental equation gives a relationship between Vo and R. From electron scattering experiments, the rms charge radius of the deuteron is known to be about 2.1 fm. Taking R = 2.1 fm we may solve from equation (17) the value of the potential depth Vo. The result is Vo = 35 MeV. The bound state of the deuteron, at an energy of about -2 MeV, is very close to the top of the well. H.W: Expression for Vo ?. R = 2.1 fm Binding engo of deuteron 0-35 Mel/ v- Depth of potential well Oisunce
  11. o 1 1 2 4 6 8 Here we show the deuteron wave function for R = 2.1 fm. The exponential joins smoothly to the sine at r = R, so that both u(r) and du/dr are continuous. .10 If the nucleon-nucleon force were just a bit weaker the deuteron bound state would not exist at all. In this situation the whole universe would be all quite different from the one we are observing.
  12. Spin and parity of the deuteron e The measured spin of the deuteron is 1= 1. 0 By studying the reactions involving deuterons and the property of the photon emitted during the formation of deuterons, we know that its parity is even. The total angular momentum I of the deuteron should be like 1=Sn+S +1 (21) where sn and sp are individual spins of the neutron and proton. The orbital angular momentum of the nucleons as they move about their common center of mass is l.
  13. There are four ways to couple sn, sp, and I to get a total I of 1. a sn and sp parallel with 0 (b) sn and sp antipara//e/ with 1= 1 (c) sn and sp parallel with 1= 1 ) sn and sp parallel with 2 parallel antiparallel e Since we know that the parity of the deuteron is even and the parity associated with orbital motion is determined by (-1)/ we are able to rule out some options. O Orbital angular momentum I = 0 and I = 2 give the correct parity determined from experimental observations. e The observed even parity allows us to eliminate the combinations of spins that include I =1, leaving 0 and 2 as possibilities.
  14. The magnetic dipole moment of the deuteron If the I = 0 is perfectly correct description for the deuteron, there should be no orbital contribution to the magnetic moment. We can assume that the total magnetic moment is simply the combination of the neutron and proton magnetic moments: g spßN g snßN Sn -k h h (22) where g = -3.826084 and = 5.585691. If we take the observed magnetic moment to be the z component of when the spins have their maximum value (+—h) (23) = 0.879804/1N The observed value is 0.8574376 ± 0.0000004 PN, in good but not quite exact agreement with the calculated value.
  15. In the context of the present discussion we can ascribe the tiny discrepancy to the small mixture of d state ( I = 2) in the deuteron wave function: (24) IV = asl/J(l = O) -k a dl/J(l = 2) Calculating the magnetic moment from this wave function gives (25) = as "(l = 0) -Fad = 2) The observed value is consistent with This means that the deuteron is 96% I = 0 ( s orbit) and only 4% 2 (d orbit). a: = 0.96 and ad 2 = 0.04 (26)
  16. The electric quadrupole moment of the deuteron The bare neutron and proton have no electric quadrupole moment, and so any measured nonzero value for the quadrupole moment must be due to the orbital motion. — The pure I = 0 wave function would have a vanishing quadrupole moment. The observed quadrupole moment for the deuteron is When the mixed wave function [equation (24)] is used to calculate the quadrupole moment of the deuteron (Q) the calculation gives two contribution terms. One is proportional to (a and another proportional to the cross-term (asad). (28) r2Rs (r)Rd (r)r2dr where r o calculate Q we must know the deuteron d-state wave function and it is obtainable from the realistic phenomenological potentials. The d-state admixture is of several percent in this calculation and is consistent with the 4% value deduced from the magnetic moment.
  17. Some comments concerning the (I-state admixture obtained from the studies of magnetic moment and the quadrupole moment Q: 1. 2. 3. 4. 5. This good agreement between the d-state admixtures deduced from and Q should be regarded as a happy accident and not taken too seriously. In the case of the magnetic dipole moment, there is no reason to expect that it is correct to use the free-nucleon magnetic moments in nuclei. Spin-orbit interactions, relativistic effects, and meson exchanges may have greater effects on than the d-state admixture (but may cancel one another's effect). For the quadrupole moment, the poor knowledge of the d-state wave function makes the deduced d-state admixture uncertain. Other experiments, particularly scattering experiments using deuterons as targets, also give d-state admixtures in the range of 4%. Thus our conclusions from the magnetic dipole and electric quadrupole moments may be valid after all. It is important that we have an accurate knowledge of the d-state wave function because the mixing of I values in the deuteron is the best evidence for the noncentral (tensor) character of the nuclear force.
  18. Nucleon-Nucleon Scattering e The total amount of information about nucleon-nucleon interaction that we acquire from the study of the deuteron is very limited. As far as we know there is only one weakly bound state of a neutron and a proton. The configuration of the deuteron is I = 0, parallel spins, and 2 fm separation. e To study the nucleon-nucleon interaction in different configurations we need to perform nucleon-nucleon scattering experiments. There are two ways to perform nucleon-nucleon experiments. (a). An incident beam of nucleons is scattered from a target of nucleons. The observed scattering of a single nucleon will include the complicated effects of the multiple encounters and is very difficult to extract the properties of the interaction between individual nucleons. (b). An incident beam of nucleons is scattered from a target of hydrogen. Incident nucleons can be scattered by individual protons. Multiple encounters are greatly reduced by large spatial separations between nucleons. Characteristic properties of nucleon-nucleon interactions can therefore be deduced without complications.
  19. N/A
  20. As in the case of electron scattering the nuclear scattering problem is analogous to the diffraction problem in optics. There are three features worth mentioning: 1. The incident wave is represented Detector Outgoing scattered wave Tgeet pocroscopicaity beam Figure 19-1 Schematic layout for scattering experiment. The scattering angle is the laboratory angle. by a plane wave, while far from the target (obstacle) the scattered wave fronts are spherical. The total energy content of any expanding spherical wave front cannot vary; thus its intensity (per unit area) must decrease like r2 and its amplitude must decrease like rl. 2. Along the surface of any spherical scattered wave front, the diffraction is responsible for the variation in intensity of the radiation. The intensity thus depends on angular coordinates 0 and (b. 3. A radiation detector placed at any point far from the target would record both incident and scattered waves.
  21. To solve the nucleon-nucleon scattering problem using quantum mechanics we assume the nuclear interaction by a square-well potential, as we did for the deuteron. In fact, the only difference between this calculation and that of the deuteron is that here we concerned with free incident particles with E > 0. For low energy nucleon-nucleon scattering we may simplify the Schrödinger's equation by assuming I = 0. Consider an incident nucleon striking a target nucleon just on the surface so that the impact parameter is of the order of b = 1 fm. If the incident particle has velocity v, its angular momentum relative to the target is mvR. We have mvR=1h where I is the angular momentum quantum number If mvR
  22. The Schrödinger's equation of the two nucleons system is v2V(r) - EV(r) 2m (29) The mass appearing in the equation is the reduced mass and is about half of the nucleon mass. 1 2 (30) By defining the radial part wave function as u(r)/r, the Schrödinger equation is h2 d2u 2 2m dr h 2 d2u 2m dr2 vou(r) = Eu(r) , for r < R Eu(r) , for r > R The acceptable solution in the region r < R is u(r) = Asin/qr with (31) (32)
  23. The acceptable solution in the region r < R is u(r) = Asin/qr with For r > R, the wave function is u(r) = C' sink2r + D' cosk2r with k2 (32) (33) (35) v For further discussions it is convenient to rewrite Equation (33) as (34) where Ccosö and D'= Csinö The boundary condition on u and du/dr at r = R give and E Dividing then we have a transcendental equation to solve: (36) Given E ö is called the Vo, and R, we can in principle solve for 5. "phase shift"
  24. u(r) (free particle) (attrætive ptentjal) (repulsive Phöe shift (attractive potential) (repulsive potential) The effect of a scattering potential is to shift the phase of the scattered wave at points beyond the scattering regions, where the wave function is that of a free particle. Phag shift
  25. A more general scattering theory with zero angular momentum (l = 0). Incident particles are described quantum mechanically the incident plane wave. Mathematically the incident plane wave can be described with spherical waves eikr/r and e-ikr/r. By multiplying with the time-dependent factor e-lWt it is easily recognized that elkr gives an outgoing wave whereas e Ikr gives an incoming wave. For 0 we can take, V Incident 2ik r i kr (37) The minus sign between the two terms keeps the incident wave function finite for r 0, and using the coefficient A for both terms sets the amplitudes of the incoming and outgoing waves to be equal. We further assume that the scattering does not create or destroy particles, and thus the amplitudes of the elkr and e-ikrterms should be the same. All that can result from the scattering is a change in phase of the outgoing wave: v(r) 2ik where ß is the change in phase. (38)
  26. If we want to find the amplitude of the scattered wave we need to subtract the incident amplitude from v(r) V scattered = l/" — V Incident (39) In terms of V scattered , the current of scattered particles per unit area can thus be calculated by the following equation: DIV* h scattered 2im (40) The scattered current is uniformly distributed over a sphere of radius r. The probability do that an incident particle is scattered into dQ is the ratio of the scattered current to the incident current: Incident plane Figure 4.5 The basic geornetry of scattering, ( J scattered Jincident (41. a) The differential cross section do/ dQ, which is the probability per unit solid angle, can thus be written as: r (Jscattered (41. b) Jincident
  27. Taking the equation (34) we know that in the region r > R the wave function is of the form -1) u(r) 1/J(r) — Sin(kr -k ) This form can be manipulated in the following way: (42) i kr — Sin(kr + v(r) i(kr+öo) i(kr where ß=2öo and A—kCe i(kr+2ö0) iöo e (43) iöo when compared with the equation (38). By subtracting the incident part of the wave function from Eq. (43) we have the scattered wave. V scattered — V incident ikr 2iö0 2ik (44)
  28. Using Eq. (40) the current of scattered particles per unit area is: h J scattered 2mi 2 hIA sin 2 mkr 2 hkIA The incident current is Jincident m dc sin2 The differential cross section is (45) (46) (47) In general, do/ dQ varies with direction over the surface of the sphere; in the special case of I = 0 scattering, do/ dQ is constant and the total cross section o is Incident plane pcm Figure 4.5 The basic geometry of scattering, sin2 47T sin k (48) The I = 0 phase shift is directly related to the probability for scattering to occur. That is, we can evaluate 50 from our simple square-well model, Eq. (36), and compare with the experimental cross section.
  29. 10 103 IOS Neutron kinetic erWw (eV) In order to understand the data taken from the low-energy neutron-proton scattering we may return to the analysis of Equation (36) by putting in proper values for all related quantities. (36) Agur• 4.6 The neutron-proton cross section at low energy. Data taken from a review by R, K, -Adair, Rev. Phys. 22, 249 (1950). with additional recent results from T. L. Houk. Phys. Rev. C S. 1886 (1970). Assume that the incident energy is small, say E 10 kev and take Vo = 35 MeV from our analysis of the deuteron bound state. 2m(V0 + E)/h2 = 0.92fm -1 2mE/h2
  30. and so 2 (52) Using R 2 fm from the study of the 21-1 bound state gives a = 0.2 fm-I. Thus 1

Need a Tutor or Coaching Class?

Post an enquiry and get instant responses from qualified and experienced tutors.

Post Requirement

Related Study Notes

Upload Study Notes

If you have your own Study Notes which you think can benefit others, please upload on LearnPick. For each approved Study Note you will get 50 Credit Points and 50 Activity Score which will increase your profile visibility.

Upload Now