An Introduction to Nuclear Physics (1)

1.The Liquid Drop model

Nuclear models occur under a set of hypotheses, each of which explains some aspect of nuclear behavior. The most basic is perhaps the “liquid drop model” which is used to account for nuclear fission, radioactivity. This is illustrated in Fig, 1 below:

Fig. 1: The Liquid Drop model showing U235 Fission

As the name implies, the nucleus is depicted in terms of a “liquid drop”, which analogous to a liquid macroscopic drop, exhibits surface tension and an excitation energy when disturbed.

This model is premised on the exceedingly short range of nuclear forces which requires that nearest neighbor attractions predominate. This is somewhat similar to the type of attraction between the molecules of a liquid which leads to the property of its surface tension.  Thus, in a liquid drop nucleus each nucleon shares its total binding energy with every nucleon.


As the simplified model diagram (for U 235 fission) shows, we expect  a spherical configuration in the minimum energy or unexcited state.  The unstable (compound) nucleus then yields two “daughter nuclei” : La 139 and Mo 95, with two additional neutrons released

    The total binding energy initially holding all the nucleons together in the liquid drop nucleus model is comprised of a sum of separate energies, including:  surface or E(s), volume or E(v), and Coulombic repulsion or E(c) – the latter due to the electrostatic repulsion between protons. Thus, the binding energy:

 E(B) = E(s) + (E(v) + E(c)

     A reduction in any one or all of these constituent energies reduces the overall binding energy and hence the bonding energy between any pair of nucleons. To account for energy release in the fission of heavy nuclei we need only represent the nucleus as a liquid drop and its transition from the unexcited to the excited state as shown in Fig. 1.

 Thus, between the spherical and compound nucleus phase, the “drop” may be said to undergo oscillations which lead it to overshoot sphericity in two directions – vertically (converting to an elliptical configuration with semi-major axis in the vertical sense, and horizontally, with semi-major axis in the horizontal sense. It finally reaches a point – the compound nucleus phase – at which the surface tension restoring force is no longer able to contain the long –range Coulombic repulsion force and the inertia of nuclear matter.

    At this point, the drop reaches a stage of distortion in which the nucleus splits into two fission fragments, denoted by [A1, Z1] and [A2, Z2]. In order to further assess the nuclear changes, we use what is called a “semi-empirical mass formula” which sets out terms as follows:


(1) f_o(Z, A) = 1.008142Z + 1.008982 (A – Z)

     Where the Z coefficient is the mass of the hydrogen atom in atomic mass units and the (A – Z) coefficient is the mass of the neutron in amu. The remaining terms listed correct for several effects which contribute to the total nuclear binding energy:

(2) f₁ (Z, A) = -a₁ A

      This term accounts for the binding energy and is essentially proportional to the nuclear mass or the volume of the nucleus.

(3) f₂ (Z, A) = +a₂ A^2/3

     This term is a positive correction to the surface area of the nucleus, i.e.  the effect of surface tension energy.

(4) f₃ (Z, A) = + a₃ (Z²/ A^1/3 )

     This term accounts for the positive Coulomb energy of the charged nucleus, assumed to be a sphere of radius proportional to  A^1/3.


(5) f₄ (Z, A) = + a₄ (Z  -  A/2)²/ A

     This term and the last (6) introduce properties specific to the nucleus. This term is zero for the case of Z = (A – Z) or 2Z = A.

(6) f₅ (Z, A) = 0, - f(A), +f(A)

    Where  0 is the result when ‘Z even’ pairs with (A – Z) odd, or ‘Z odd’ pairs with (A – Z) even; and –f(A) is the result when Z even pairs with (A – Z) even, and f(A) is the results when Z odd is paired to (A – Z) odd.  The form of f(A) is determined by fitting the data. It is found that the best fit for a simple power law relation is:

f (A) = a₅ A ^{- 1/2}

Combining all the terms we get:

M(Z, A) = 1.008142Z + 1.008982 (A – Z) -a₁ A+ a₂ A^2/3

+ a₃ (Z²/ A^1/3 ) + a₄ (Z  -  A/2)²/ A +  (-1, 0 , 1) a₅ A ^{– 1/2}

    The parameter values designated are:  a₁ = 0.01692,  a₂ = 0.01912,  a₃ = 0.000763,  a₄  = 0.10178, and  a₅  = 0.012.

2. The  Nuclear Shell Model:

The shell model of the nucleus treats a different aspect, namely specific energy levels within the nucleus.  To achieve this the nucleons are treated as independent particles, each of which moves in its own spherically symmetrical potential well about 50 MeV deep with rounded edges. 

Fig. 2: The potential well for the Shell model


The Schrodinger Equation  is then set up for this potential and solved. The solution yields stationary quantum states, somewhat like the states which are associated with electrons in the outer shells of the atom. 

Like the electrons occupying atomic shells, the nucleons are fermions, i.e. particles each with spin ½  and therefore obeying the Pauli Exclusion Principle so that no more than two nucleons of the same type can occupy the same energy level (e.g. one with spin +½, the other with spin -½.). This leads to particles pairing up in the nucleus: spin up (+ ½) protons with spin down (-½) protons, and spin up neutrons with spin down neutrons. 

The form or expression for the energy of a nucleus is very closely approximated by  the energy associated with a “square well” potential. The Fermi energy in more explicit form (which we will not elaborate upon too much beyond this) is:

E_F = 3 ^2/3 p ^4/3 ħ² r ^2/3/ 2M

 

Where M is the nucleon mass, ħ = h/2p is the adjusted Planck number (h divided by 2p) and r is the nucleon density. Performing the operation with known or estimated values, the result is E_F =  30 MeV.   If we assume a spherical nucleus then the radius is just:

R = r _o A ^1/3

This is also the dimension on either side of the central symmetry line of Fig. 2. The evidence for the shell model includes the following:

i)Particularly stable nuclei are borne out by “closed shells”, based on our observations.

ii) The model predicts (correctly) that the even N (or A – Z) nuclei will be most stable and the odd Z, odd N nuclei the least stable. Experiments bear this out: there are 160 stable nuclides (with even Z, N) and only 4 with odd Z, N.

iii) The model predicts that for even N, Z nuclides the total angular momentum J = 0, and that for odd nuclides it is half-integral. (Borne out by measurements of nuclear magnetic moments.)

Problem: Find the diameter of the oxygen (O ¹⁶) nucleus.

Solution:

We apply:  R = r _o A ^1/3

Where:  r _o  =  1.2 x 10 ^-15 m  is the Bohr radius, and A = 16.  Then:

R = (1.2 x 10 ^-15 m)  (16)  ^1/3= 3.0238 x 10 ^-15 m

 

And the diameter D = 2R = 2(3.0238 x 10 ^-15 m)

D = 6.05 x 10 ^-15 m

The preceding result can also be generalized to a situation in which the ratio of radii (between two nuclei) is sought. In this case, we can write:

R1/ R2  =  [r _o A1 ^1/3] /  [r _o A2  ^1/3]

Or:

R1/ R2 = (A1/ A2) ^1/3

Example: How much larger is a copper nucleus than an oxygen nucleus?

 

Let R1 denote  the radius of the copper nucleus, and R2, oxygen’s. Then: A1 = 64, and A2 = 16. So:

R1/ R2 = (64/16) ^1/3 =   4/ 2.51 = 1.587

Thus, the copper nucleus is about 1.587 times the size of the oxygen nucleus.

Problems:

 1) Evaluate the terms of the semi-empirical mass formula for the U 238 nucleus,  i.e. if A = 238 and Z = 92. Use this information to find the total mass in atomic mass units (u) and compare it to the standard mass expression. Thence or otherwise, obtain the mass defect and the binding energy as well as the binding energy per nucleon.

2)The sketch graph shown below plots the mass number A vs. the binding energy per nucleon (BE/nucleon) on the vertical axis.

Using any models or other explanations, account for the region of greatest stability indicated. 

3) A quantum square well potential is  schematically defined according to: 

It is found that the Schrodinger equation to solve becomes:

-ħ²/ 2m ( dy²/dx² ) = Ey

Thence:

dy²/dx² + 2m/ ħ² (Ey) = 0

And the quantized energy is found to be:

E _n = (h²/ 8m L²) n²

This can be modified to yield a simplified Fermi energy in one dimension by using instead:

E _n = (h²  p²/ 8m L²) (N/2)²

 Where all the Fermi shell energy levels are presumed occupied up to N/2.  Use this to obtain a simplified estimate of the energy associated with the oxygen nucleus if we assume its shells filled up to N =4 and use an estimate for L as:

  R = r _o A ^1/3

(Take m = 1.7 x 10 ^-27 kg)

4)(a) Compare the value you obtain in (3) to the actual Fermi (shell) energy for oxygen, if the nuclear density r » 2.3 x 10¹⁷ kgm^-3.

(b) Write out at least one technique that might be used to find the density of an atomic nucleus. State clearly any assumptions made and how you quantify them.

5)(a) Find the ratio of the helium nucleus radius to that of the uranium 238 nucleus.

(b) Estimate, using any technique you can think of, the ratio of the nuclear densities for part (a).

6) An element has mass number A = 202 and atomic number Z = 80.

a) Find the diameter of the nucleus and how many times it is greater than that of hydrogen.

b) Find the mass defect D M for this nucleus

c) From the result in (b) obtain the binding energy and the binding energy per nucleon, E_B/ A.  Using the graph on the opposite page, mark the location of E_B/ A.

  

An Introduction to Quantum Mechanics (3)

(Continued from previous section)

4. The Wave-Particle Duality & Heisenberg Microscope

     We now look in somewhat more detail at wave-particle duality as it arises in quantum mechanics.  In the particle interpretation, electrons  fired from a device such as an electron gun would not all follow the same path since the trajectory of an electron – unlike a missile- can’t be predicted from its initial state. We consider here the case of electron diffraction, whereby (based on Fig. 7) electrons are emitted from an electron gun and pass through a slit toward a detector or photographic plate onto which a diffraction pattern appears. This pattern will also coincide with an intensity distribution such as shown in Fig. 10.

In effect, the intensity distribution basically describes the probability for an individual particle (electron) to strike each of several areas designated on the photographic film. This discloses a fundamental indeterminacy that has no counterpart in Newtonian mechanics. Now, consider an electron striking at some angle q, such as indicated:

                                                                             

Fig. 10: Showing electron diffraction and intensity pattern on screen.

We have, from the quantities shown:

p _y/ p _x = tan q  or   p _y =  p _x  q  (in limit of small q)

Therefore, the y-component of momentum can be as large as:

p _y =  p _x  (l/ a)

Where a denotes the slit width. The narrower the dimension of a the broader the diffraction pattern, and the greater D p. From Louis de Broglie’s matter wave hypothesis (already introduced into the Bohr atom, as we saw, cf. Fig. 7) :   l_D = h/ p _x

Therefore:

p _y =  p _x  (h/ p _x  a)  = h/ a  or: p _y a =  h

But ‘a’ represents uncertainty in electron position vertically (D y), i.e. as it passes through the slit. We can reduce D p_y  only by narrowing the slit width a and vice versa. Thus we get:

p _y a =   D p_y  D y »  h

Which is one form of the Heisenberg Uncertainty Principle which states that the momentum of a quantum particle and its positions cannot simultaneously be known to the same arbitrary precision. One corollary is that to detect a particle any given detector must interact with it thereby altering the motion of the particle.

     This view is no longer taken in any literal way because we understand that the quantum measurements are statistical in nature and hence a particular measurement is the result of a vast statistical assembly. Paul Dirac, in his book Quantum Mechanics, defined the “principle of superposition” thusly[1]:

“A state of a system may be defined as a state of undisturbed motion that is restricted by as many conditions or data as are theoretically possible without mutual interference or contradiction"

     But let’s examine this in more detail. By “undisturbed motion” Dirac meant the state is pure and hence no extraneous observations are being made such that the state experiences interference effects to displace or disturb it. In the Copenhagen Interpretation, “disturbance” of mutually defined variables, say x, p or position and momentum, occurs only if:

[x, p] = -i h =  -i h/ 2p

If it were the case that [x, p] = 0, one would say the variables “commute” and hence there’s no interference. If the condition doesn’t hold, then interference exists. Hence, Dirac’s setting of an upper limit in the last portion of his definition, specifying as many conditions as theoretically possible “without mutually interfering interference.” This state is undisturbed.  We have a statistical perspective!

Fig. 11: Sketch of Heisenberg Microscope and key parameters.

    It is important to see from the preceding, how the Heisenberg Uncertainty Principle arises not just from an ad hoc assumption, but from the limits (or “tolerance thresholds”) of explicit quantities (e.g. p, x), when considered in the quantum limit. Hence, the model of the Heisenberg “microscope” provides a useful (although not practical, since it can’t actually be constructed) means of deriving the statistical principle of superposition based on an observational ansatz.

    Consider a measurement made to determine the instantaneous position of an electron by means of a microscope. In such a measurement the electron must be illuminated, because it is actually the light quanta (photon) scattered by the electron that the observer sees. The resolving power of the microscope determines the ultimate accuracy with which the electron can be located.  This resolving power is known to be approximately:

 l/ 2 sin q

Where  l is the wavelength of the scattered light and q is the half-angle subtended by the objective lens of the microscope.  Then:

Δx   =  l/ 2 sin q

In order to be collected by the lens, the photon must be scattered through any range of angle from -q to q. In effect, the electron’s momentum values range from:

+ h sin q/ l   to   -   h sin q/ l  

Then the uncertainty in the momentum is given by:

D p_x  =  [ h sin q/ l  -  (-   h sin q/ l)]    =   2 h sin q/ l  

Then the Heisenberg Uncertainty Principle product is:

D p_x   Δx   =    (2 h sin q/ l )  l/ 2 sin q   = h

4. Probability density and Expectation Values

Earlier we saw:

P = ½y (1s) y (1s) *½

Which is the probability density and a quantity we can actually measure, e.g. for the 1s state of hydrogen. Then this needs to be generalized to apply to more than one case.

Since the electron locations can’t be computed from Newtonian mechanics but more plausibly based on an analogous probability density to what  we saw above, then we can generalize and write:

P _ab  =     ò ^b_a   ‖y(x) ‖²  dx

Where x is the state under consideration and this system is 1-dimensional with the probability assessed from a to b.  Note that we define the normalization condition as:

 ò ^b_a  ‖y‖²  dx   =  1

Normalization is simply a condition stating that the particle exists at some point at all times. Thus if we had:

ò ^b_a   ‖y‖²  dx   =  0

The probability would not exist. The probability condition then allows us to specify the probability of observing a particle even though we cannot specify the position. The normalization then gives the probability of finding the particle in the range a <  x  < b, say in one dimension.

The wave function, y(x) satisfies the Schrodinger equation. For the simple one dimensional case we can write:

d² y/dx² + F(x)   y = 0

Though the wave function y(x) it self is not a measurable quantity, other measurable quantities such as the energy E and momentum of the particle can be derived from it. Also, if the wave function is known it is possible to compute the average position of the particle, known as the expectation value:

  =       ò^¥_-¥   x ‖y(x) ‖²  dx

This expression implies the particle is in a definite state so that the probability density is time –independent.

Example Problem: Consider the 1D quantum system shown, and a particle confined therein:

With maximum dimension L in direction +x. Find: a) the probability P _ab  the particle is between x = 0 and x  = L, b) the expectation value and (c) show the energy for the particle can be quantized according to:

E _n = (h²/ 8m L²) n²

Let the wave function be:

y(x)  = Ö2/ ÖL    sin (kx)

Solution:

We rewrite the wave function as:

y(x)  =   Ö2/ ÖL   [sin (px/L)]   where k = p/L

Then:

P _ab  =     ò ^b_a   ‖y(x) ‖²  dx =   ò ^L ₀   (Ö2/ ÖL )²   sin² (px/L) dx

=    2/ L ò ^L ₀    ½ [1 - cos (px/L)] dx

(Let q = px/L  and use:  sin² q= ½ (1 – cos 2q))

P _ab  =     2/ L [½  ò ^L ₀    dx  - ò ^L ₀   cos (px/L)] dx

P _ab  =      1 -  1/p  sin (2px/L)] ^L ₀   = 1 -  1/p  sin (2p)

But : sin (2p) = 0 so P _ab  =      1  

The expectation value is:

  =       ò ^¥_-¥   x ‖y(x) ‖²  dx

=  2/ L [ ò  ^L ₀   x sin (px/L)²] dx

= L²/ 4  -  [x sin (2px/L)/ 4p/L - cos (2px/L)/8(p²/L²)]

= 2/L  (L² /4)  =  L/ 2

Finding the energy:  We have the Schrodinger equation:

dy²/dx²  + K² y = 0

where K =  Ö [2mE] / ħ

If we examine the sketch below:

We see plots of the wave function y(x)   vs. position x (far left),  and of the probability density (middle) and the energy levels. Since we have represented the wave function by a sinusoidal function then it follows that the allowed wavelengths are those for which the length L is equal to an integral number of half wavelengths, or:

L = nl/ 2

These allowed states are called stationary states and represent standing waves (analogous to the ones seen earlier for the Bohr atom). Thus, the wavelengths of the particle are restricted by the condition:

l  = 2L/ n

Then the magnitude of the momentum p is also restricted to specific values (e.g. using p = h/ l) [2] such that:

p = h/ l =  h/ 2L/ n = nh/ 2L

The energy associated with the particle is then:

E = 1/2 mv²  =  p²/ 2m  = (nh/ 2L)² / 2m

E= ( h²/ 8mL²) n²

(n= 1, 2, 3 etc.)

Thus the energy is quantized with the energy of the lowest energy state corresponding to n =1 so:

E1= ( h²/ 8mL²)

This least energy that the particle can have is called the “zero point energy” and means the particle can never be at rest.

Note that the above energy result can also obtained through the use of differential equations, e.g.

http://brane-space.blogspot.com/2013/05/applications-of-differential-equations.html

The probability density can be extended to 3 dimensions by writing:

P   =     ò ^¥_{- ¥}  ‖y(x) ‖²  dV

The quantized energy will  then be (for a 3D box, for which dV = dx dy dz):

E= ( h²/ 8mL²)[ n _x ²  +  n _y ²    + n _z ²  ]

Problems:

1)For a 1D box, let one electron inside have the wave function:

y(x)  =   Ö2/ ÖL   [sin (2px/L)]  

Find the probability of locating the electron between x = 0 and x = L/4.

2)Use the uncertainty principle to estimate the uncertainty in momentum for a particle in a 1D box. Estimate the ground state energy using this means and compare it to the actual ground state energy.

3) The wave function for a particle confined to moving in a 1D box is given by:

y(x)  =   A [sin (npx/L)]  

Use the normalization condition on y(x)   to show the constant A is given by:  A =  Ö2/ ÖL  

4) It is known from quantum mechanics that a particle in a one dimensional potential well (such as shown in the diagram) can exist in a number of energy states. Imagine an electron confined between the boundaries x and x +  Dx, where Dx is 0.5 Angstroms.

Approximately, what is the uncertainty in the x-component of  the  momentum of the electron?

5)(a) Consider a free particle confined between two impenetrable walls at x and x + L.  What is the probability according to classical physics that the particle will be found between x and x + L/3 if no other information is given?

b) What is the probability according to quantum mechanics that the particle in its lowest energy state will be found between x and x + L/3?

c) What is the probability according to quantum mechanics that the particle in the second lowest energy state will be found between x and x + L/3?

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[1] Dirac, P.A.M.: 1941, Quantum Mechanics, Oxford University Press, 11.

[2] Recall from Planck’s law: E = hc/ l and p = Ö [2mE].