IPhO 2010 · Theory problem 3

Simple model of an atomic nucleus

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Introduction

Although atomic nuclei are quantum objects, a number of phenomenological laws for their basic properties (like radius or binding energy) can be deduced from simple assumptions: (i) nuclei are built from nucleons (i.e. protons and neutrons); (ii) strong nuclear interaction holding these nucleons together has a very short range (it acts only between neighboring nucleons); (iii) the number of protons (\(Z\)) in a given nucleus is approximately equal to the number of neutrons (\(N\)), i.e. \(Z\approx N\approx A/2\), where \(A\) is the total number of nucleons (\(A\gg1\)). Important: Use these assumptions in Tasks 1-4 below.

Task 1 - Atomic nucleus as closely packed system of nucleons

In a simple model, an atomic nucleus can be thought of as a ball consisting of closely packed nucleons [see Fig. 1(a)], where the nucleons are hard balls of radius \(r_N=0.85\ \mathrm{fm}\) (\(1\ \mathrm{fm}=10^{-15}\ \mathrm{m}\)). The nuclear force is present only for two nucleons in contact. The volume of the nucleus \(V\) is larger than the volume of all nucleons \(AV_N\), where \(V_N=\frac43r_N^3\pi\). The ratio \(f=AV_N/V\) is called the packing factor and gives the percentage of space filled by the nuclear matter.

Figure 1

Fig. 1. (a) An atomic nucleus as a ball of closely packed nucleons. (b) The SC packing.

a) Calculate what would be the packing factor \(f\) if nucleons were arranged in a “simple cubic” (SC) crystal system, where each nucleon is centered on a lattice point of an infinite cubic lattice [see Fig. 1(b)]. (0.3 points)

Important: In all subsequent tasks, assume that the actual packing factor for nuclei is equal to the one from Task 1a. If you are not able to calculate it, in subsequent tasks use \(f=1/2\).

b) Estimate the average mass density \(\rho_m\), charge density \(\rho_c\), and the radius \(R\) for a nucleus having \(A\) nucleons. The average mass of a nucleon is \(1.67\cdot10^{-27}\ \mathrm{kg}\). (1.0 points)

Task 2 - Binding energy of atomic nuclei – volume and surface terms

Binding energy of a nucleus is the energy required to disassemble it into separate nucleons and it essentially comes from the attractive nuclear force of each nucleon with its neighbors. If a given nucleon is not on the surface of the nucleus, it contributes to the total binding energy with \(a_V=15.8\ \mathrm{MeV}\) (\(1\ \mathrm{MeV}=1.602\cdot10^{-13}\ \mathrm{J}\)). The contribution of one surface nucleon to the binding energy is approximately \(a_V/2\). Express the binding energy \(E_b\) of a nucleus with \(A\) nucleons in terms of \(A\), \(a_V\), and \(f\), and by including the surface correction. (1.9 points)

Task 3 - Electrostatic (Coulomb) effects on the binding energy

The electrostatic energy of a homogeneously charged ball (with radius \(R\) and total charge \(Q_0\)) is \[ U_c=\frac{3Q_0^2}{20\pi\varepsilon_0R}, \] where \(\varepsilon_0=8.85\cdot10^{-12}\ \mathrm{C^2N^{-1}m^{-2}}\).

a) Apply this formula to get the electrostatic energy of a nucleus. In a nucleus, each proton is not acting upon itself (by Coulomb force), but only upon the rest of the protons. One can take this into account by replacing \(Z^2\rightarrow Z(Z-1)\) in the obtained formula. Use this correction in subsequent tasks. (0.4 points)

b) Write down the complete formula for binding energy, including the main (volume) term, the surface correction term and the obtained electrostatic correction. (0.3 points)

Task 4 - Fission of heavy nuclei

Fission is a nuclear process in which a nucleus splits into smaller parts (lighter nuclei). Suppose that a nucleus with \(A\) nucleons splits into only two equal parts as depicted in Fig. 2.

a) Calculate the total kinetic energy of the fission products \(E_{\mathrm{kin}}\) when the center of two lighter nuclei are separated by the distance \(d\geq2R(A/2)\), where \(R(A/2)\) is their radius. The large nucleus was initially at rest. (1.3 points)

b) Assume that \(d=2R(A/2)\) and evaluate the expression for \(E_{\mathrm{kin}}\) obtained in part a) for \(A=100,\ 150,\ 200\) and \(250\) (express the results in units of MeV). Estimate the values of \(A\) for which fission is possible in the model described above? (1.0 points)

Figure 2

Fig. 2. A schematic description of nuclear fission in our model.

Task 5 – Transfer reactions

a) In modern physics, the energetics of nuclei and their reactions is described in terms of masses. For example, if a nucleus (with zero velocity) is in an excited state with energy \(E_{\mathrm{exc}}\) above the ground state, its mass is \(m=m_0+E_{\mathrm{exc}}/c^2\), where \(m_0\) is its mass in the ground state at rest. The nuclear reaction \({}^{16}\mathrm{O}+{}^{54}\mathrm{Fe}\rightarrow{}^{12}\mathrm{C}+{}^{58}\mathrm{Ni}\) is an example of the so-called “transfer reactions”, in which a part of one nucleus (“cluster”) is transferred to the other (see Fig. 3). In our example the transferred part is a \({}^{4}\mathrm{He}\)-cluster (\(\alpha\)-particle). The transfer reactions occur with maximum probability if the velocity of the projectile-like reaction product (in our case: \({}^{12}\mathrm{C}\)) is equal both in magnitude and direction to the velocity of projectile (in our case: \({}^{16}\mathrm{O}\)). The target \({}^{54}\mathrm{Fe}\) is initially at rest. In the reaction, \({}^{58}\mathrm{Ni}\) is excited into one of its higher-lying states. Find the excitation energy of that state (and express it units of MeV) if the kinetic energy of the projectile \({}^{16}\mathrm{O}\) is \(50\ \mathrm{MeV}\). The speed of light is \(c=3\cdot10^8\ \mathrm{m/s}\). (2.2 points)

1. \(M({}^{16}\mathrm{O})\) 15.99491 a.m.u.
2. \(M({}^{54}\mathrm{Fe})\) 53.93962 a.m.u.
3. \(M({}^{12}\mathrm{C})\) 12.00000 a.m.u.
4. \(M({}^{58}\mathrm{Ni})\) 57.93535 a.m.u.

Table 1. The rest masses of the reactants in their ground states. \(1\ \mathrm{a.m.u.}=1.6605\cdot10^{-27}\ \mathrm{kg}\).

b) The \({}^{58}\mathrm{Ni}\) nucleus produced in the excited state discussed in the part a), deexcites into its ground state by emitting a gamma-photon in the direction of its motion. Consider this decay in the frame of reference in which \({}^{58}\mathrm{Ni}\) is at rest to find the recoil energy of \({}^{58}\mathrm{Ni}\) (i.e. kinetic energy which \({}^{58}\mathrm{Ni}\) acquires after the emission of the photon). What is the photon energy in that system? What is the photon energy in the lab system of reference (i.e. what would be the energy of the photon measured in the detector which is positioned in the direction in which the \({}^{58}\mathrm{Ni}\) nucleus moves)? (1.6 points)

Figure 3

Fig. 3. The schematics of a transfer reaction.

Official English text of the problem.

Answers

For compactness, define \[ S(A)=A-\left(A^{1/3}-2f^{1/3}\right)^3,\qquad r_0=r_Nf^{-1/3}=1.0546\ \mathrm{fm}, \] and \[ a_C=\frac{3e^2}{20\pi\varepsilon_0r_0} =0.81955\ \mathrm{MeV}. \]

  • 1(a). \[ \boxed{f=\frac{\pi}{6}\approx0.5236=52.36\%.} \]

  • 1(b). \[ \boxed{\rho_m\approx3.40\times10^{17}\ \mathrm{kg\,m^{-3}},\qquad \rho_c\approx1.63\times10^{25}\ \mathrm{C\,m^{-3}}} \] \[ \boxed{R=r_N(A/f)^{1/3}\approx1.05A^{1/3}\ \mathrm{fm}.} \]

  • 2. \[ \boxed{ E_b^{(\mathrm{nuc})} =a_V\left[A-\frac{S(A)}2\right] =a_V\left[A-3f^{1/3}A^{2/3} +6f^{2/3}A^{1/3}-4f\right]. } \]

  • 3(a). \[ \boxed{ U_c=\frac{a_CZ(Z-1)}{A^{1/3}} \simeq\frac{a_C}{4}\left(A^{5/3}-2A^{2/3}\right). } \]

  • 3(b). Writing the complete binding energy as \(B(A)\), \[ \boxed{ B(A)=a_V\left[A-3f^{1/3}A^{2/3} +6f^{2/3}A^{1/3}-4f\right] -\frac{a_C}{4}\left(A^{5/3}-2A^{2/3}\right). } \]

  • 4(a). \[ \boxed{ E_{\mathrm{kin}}(d)=2B(A/2)-B(A) -\frac{e^2A^2}{64\pi\varepsilon_0d}. } \]

  • 4(b). At contact, the calculated energies for \(A=100,150,200,250\) are, respectively, \[ \boxed{-29.3,\quad-24.9,\quad-6.77,\quad+23.6\ \mathrm{MeV}.} \] The heavy-nucleus threshold is approximately \(\boxed{A_{\mathrm{crit}}\simeq213}\). Of the four examples, only \(A=250\) permits fission from rest through the contact configuration.

  • 5(a). \[ \boxed{E_{\mathrm{exc}}\approx10.86\ \mathrm{MeV}.} \]

  • 5(b). \[ \boxed{T_{\mathrm{recoil}}\approx1.09\ \mathrm{keV},\qquad E_\gamma^{*}\approx10.86\ \mathrm{MeV},\qquad E_{\gamma,\mathrm{lab}}\approx10.92\ \mathrm{MeV}.} \] Here the star denotes the rest frame of the excited nickel nucleus before emission.

Solution

1(a). Packing factor

In simple cubic packing, neighboring spheres touch along a cube edge, so the lattice spacing is \(2r_N\). A cubic cell contains eight corner nucleons, each shared by eight cells: one complete nucleon per cell.

Consequently, \[ f=\frac{\text{nucleon volume per cell}}{\text{cell volume}} =\frac{\frac43\pi r_N^3}{(2r_N)^3} =\boxed{\frac{\pi}{6}}. \]

1(b). Densities and nuclear radius

The packing-factor definition gives \[ V=\frac{AV_N}{f} =\frac{4\pi r_N^3A}{3f}. \] Equating this to \(4\pi R^3/3\), \[ \boxed{R=r_N\left(\frac Af\right)^{1/3}=r_0A^{1/3}}, \qquad r_0=1.0546\ \mathrm{fm}. \]

At the accuracy of this model, the nuclear mass is \(Am_N\), where \(m_N=1.67\times10^{-27}\ \mathrm{kg}\). Thus \[ \rho_m=\frac{Am_N}{V} =\frac{3fm_N}{4\pi r_N^3}. \] With \(f=\pi/6\), \[ \boxed{ \rho_m=\frac{m_N}{8r_N^3} =3.40\times10^{17}\ \mathrm{kg\,m^{-3}}. } \]

Using \(Z\simeq A/2\), the nuclear charge is approximately \(Ae/2\). The elementary charge consistent with the given electron-volt conversion is \(e=1.602\times10^{-19}\ \mathrm C\). Therefore, \[ \rho_c=\frac{Ae/2}{V} =\frac{3fe}{8\pi r_N^3} =\frac{e}{16r_N^3}, \] so \[ \boxed{\rho_c=1.63\times10^{25}\ \mathrm{C\,m^{-3}}.} \] Both average densities are independent of \(A\).

2. Volume and full surface-layer contributions

The nucleon number density is \[ n=\frac{A}{V}=\frac{f}{V_N}. \] Using the specified surface layer of thickness \(2r_N\), its volume is \[ V_{\mathrm{surface}} =\frac{4\pi}{3}\left[R^3-(R-2r_N)^3\right]. \]

Within the continuum packing model, the number of surface nucleons is therefore \[ \begin{aligned} S(A) &=\frac{f}{r_N^3}\left[R^3-(R-2r_N)^3\right]\\ &=A-f\left[\left(\frac Af\right)^{1/3}-2\right]^3\\ &=\boxed{A-\left(A^{1/3}-2f^{1/3}\right)^3}. \end{aligned} \] This expression applies when \(R\geq2r_N\), as is appropriate to the large nuclei considered here. Expanding without discarding any term, \[ \boxed{ S(A)=6f^{1/3}A^{2/3}-12f^{2/3}A^{1/3}+8f. } \]

There are \(A-S(A)\) interior nucleons contributing \(a_V\) each, and \(S(A)\) surface nucleons contributing \(a_V/2\) each. Hence \[ E_b^{(\mathrm{nuc})} =a_V[A-S(A)]+\frac{a_V}{2}S(A), \] or \[ \boxed{ E_b^{(\mathrm{nuc})} =a_V\left[A-3f^{1/3}A^{2/3} +6f^{2/3}A^{1/3}-4f\right]. } \]

Numerically, \[ \frac{E_b^{(\mathrm{nuc})}}{\mathrm{MeV}} =15.8A-38.2042A^{2/3}+61.5849A^{1/3}-33.0914. \] The \(A^{1/3}\) and constant terms are retained in all subsequent calculations.

3(a). Coulomb self-energy

Substituting \(Q_0=Ze\) into the supplied formula, and then applying the prescribed self-interaction correction, gives \[ U_c=\frac{3e^2Z(Z-1)}{20\pi\varepsilon_0R}. \] Since \(R=r_0A^{1/3}\), define \[ a_C=\frac{3e^2}{20\pi\varepsilon_0r_0}. \] The supplied constants give \[ \frac{e^2}{4\pi\varepsilon_0} =1.44049\ \mathrm{MeV\,fm}, \qquad a_C=0.81955\ \mathrm{MeV}. \] Thus \[ \boxed{U_c=\frac{a_CZ(Z-1)}{A^{1/3}}.} \]

Applying \(Z\simeq A/2\) while retaining the subtraction of one, \[ Z(Z-1)\simeq\frac{A(A-2)}4, \] and therefore \[ \boxed{ U_c(A)=\frac{a_C}{4}\left(A^{5/3}-2A^{2/3}\right). } \]

3(b). Complete binding energy

The positive electrostatic energy reduces the energy required to disassemble the nucleus. Therefore, \[ B(A)=E_b^{(\mathrm{nuc})}(A)-U_c(A), \] which gives \[ \boxed{ B(A)= a_V\left[A-3f^{1/3}A^{2/3} +6f^{2/3}A^{1/3}-4f\right] -\frac{a_C}{4}\left(A^{5/3}-2A^{2/3}\right). } \]

4(a). Kinetic energy at separation \(d\)

Each fragment has \(A/2\) nucleons and charge approximately \(Ae/4\). For two nonoverlapping, spherically symmetric charge distributions, their mutual electrostatic energy equals that of point charges at their centers: \[ U_{12}(d) =\frac{1}{4\pi\varepsilon_0}\frac{(Ae/4)^2}{d} =\frac{e^2A^2}{64\pi\varepsilon_0d}. \] The \(Z(Z-1)\) correction has already been included in each nucleus’s self-energy. It does not apply to this interaction between two distinct sets of protons.

Take the separated free nucleons as the common internal-energy reference. The initial nucleus has energy \(-B(A)\), while the two separated fragments have total energy \[ -2B(A/2)+U_{12}(d)+E_{\mathrm{kin}}(d). \] Energy conservation consequently gives \[ \boxed{ E_{\mathrm{kin}}(d)=2B(A/2)-B(A)-U_{12}(d). } \]

Expanding the binding energies, \[ \boxed{ \begin{aligned} E_{\mathrm{kin}}(d)={}& -3a_Vf^{1/3}(2^{1/3}-1)A^{2/3}\\ &+6a_Vf^{2/3}(2^{2/3}-1)A^{1/3} -4a_Vf\\ &+\frac{a_C}{4} \left[(1-2^{-2/3})A^{5/3} +2(2^{1/3}-1)A^{2/3}\right]\\ &-\frac{e^2A^2}{64\pi\varepsilon_0d}. \end{aligned} } \] The volume contributions cancel because the total number of nucleons is conserved. As the fragments separate, their mutual electrostatic energy decreases and their total kinetic energy increases.

4(b). Contact energies and fission threshold

At contact, \[ d_0=2R(A/2)=2^{2/3}r_0A^{1/3}. \] Using \(e^2/(4\pi\varepsilon_0r_0)=5a_C/3\), \[ U_{12}(d_0)=\frac{5a_C}{48\,2^{2/3}}A^{5/3}. \] Combining this with the expression above, \[ \begin{aligned} E_{\mathrm{kin}}(d_0)={}& \frac{a_C}{4}\left(1-\frac{17}{12}2^{-2/3}\right)A^{5/3}\\ &+(2^{1/3}-1) \left(\frac{a_C}{2}-3a_Vf^{1/3}\right)A^{2/3}\\ &+6a_Vf^{2/3}(2^{2/3}-1)A^{1/3} -4a_Vf. \end{aligned} \] Numerically, \[ \boxed{ \frac{E_{\mathrm{kin}}(d_0)}{\mathrm{MeV}} =0.0220368A^{5/3} -9.82357A^{2/3} +36.1750A^{1/3} -33.0914. } \]

\(A\) Contact separation \(d_0\) (fm) \(E_{\mathrm{kin}}(d_0)\) (MeV)
100 7.77 \(-29.3\)
150 8.89 \(-24.9\)
200 9.79 \(-6.77\)
250 10.55 \(+23.6\)

A negative calculated kinetic energy means that the contact configuration cannot be reached from the initially unexcited nucleus under the assumed energy balance.

For heavy nuclei, solving \(E_{\mathrm{kin}}(d_0)=0\) gives \[ A_{\mathrm{crit}}\simeq213.0. \] Thus the model permits fission through the contact configuration for approximately \[ \boxed{A\gtrsim213.} \] Among the four specified nuclei, only \(A=250\) satisfies this condition. Once contact is accessible, further separation releases additional Coulomb energy.

5(a). Excitation energy in the transfer reaction

Let \(m_1,m_2,m_3,m_4\) denote the tabulated ground-state masses of oxygen, iron, carbon and nickel, respectively. Let \[ m_4^*=m_4+\frac{E_{\mathrm{exc}}}{c^2} \] be the excited nickel mass.

Using the given constants, \[ 1\ \mathrm{a.m.u.}\,c^2 =\frac{(1.6605\times10^{-27})(3\times10^8)^2} {1.602\times10^{-13}}\ \mathrm{MeV} =932.865\ \mathrm{MeV}. \] The ground-state reaction energy is \[ \begin{aligned} Q&=(m_1+m_2-m_3-m_4)c^2\\ &=(15.99491+53.93962-12.00000-57.93535) (932.865)\ \mathrm{MeV}\\ &=-0.764949\ \mathrm{MeV}. \end{aligned} \]

We can retain exact relativistic kinematics. Oxygen and carbon have the same velocity \(v\), hence the same Lorentz factor \[ \gamma=1+\frac{50\ \mathrm{MeV}}{m_1c^2}, \qquad \beta=\frac vc=\sqrt{1-\gamma^{-2}}. \]

Conservation of energy and momentum gives the excited nickel energy and momentum: \[ E_4^{\mathrm{lab}} =\gamma(m_1-m_3)c^2+m_2c^2, \qquad p_4=\gamma(m_1-m_3)v. \] Its invariant rest energy therefore satisfies \[ (m_4^*c^2)^2=(E_4^{\mathrm{lab}})^2-(p_4c)^2. \] Writing \(\Delta m=m_1-m_3\) and using \(\gamma^2(1-\beta^2)=1\), \[ m_4^*=\sqrt{m_2^2+\Delta m^2+2\gamma m_2\Delta m}. \] Consequently, \[ \boxed{ E_{\mathrm{exc}} =c^2\left[ \sqrt{m_2^2+(m_1-m_3)^2+2\gamma m_2(m_1-m_3)} -m_4 \right]. } \]

Numerically, \[ \boxed{E_{\mathrm{exc}}=10.86074\ldots\ \mathrm{MeV} \approx10.86\ \mathrm{MeV}.} \]

For an energy-balance check, the outgoing kinetic energies are \[ T_C=\frac{m_3}{m_1}(50\ \mathrm{MeV}) =37.51193\ \mathrm{MeV}, \] \[ T_{\mathrm{Ni}}=0.86237\ \mathrm{MeV}. \] They satisfy \[ E_{\mathrm{exc}}=50+Q-T_C-T_{\mathrm{Ni}}. \]

The usual nonrelativistic approximation, \[ T_{\mathrm{Ni}}\simeq \frac{(m_1-m_3)^2}{m_1m_4}(50\ \mathrm{MeV}), \] would give \(E_{\mathrm{exc}}\simeq10.862\ \mathrm{MeV}\), the same result at the quoted precision.

For use in part (b), the excited nickel speed is \[ \beta_4=\frac{V}{c} =\frac{p_4c}{E_4^{\mathrm{lab}}} =\frac{\gamma(m_1-m_3)\beta} {m_2+\gamma(m_1-m_3)} =0.00564849, \] so \(V\approx1.69\times10^6\ \mathrm{m\,s^{-1}}\).

5(b). Recoil and photon energies

First work in the rest frame of the excited nickel nucleus. Momentum conservation requires the final nickel nucleus to recoil opposite to the photon, with momentum magnitude \[ p_{\mathrm{recoil}}=\frac{E_\gamma^*}{c}. \]

Set \[ W=m_4c^2,\qquad E=E_{\mathrm{exc}}. \] Energy conservation then gives \[ W+E=E_\gamma^*+\sqrt{W^2+(E_\gamma^*)^2}. \] Solving, \[ \boxed{ E_\gamma^* =\frac{(W+E)^2-W^2}{2(W+E)} =E-\frac{E^2}{2(W+E)}. } \] Thus the recoil kinetic energy is \[ \boxed{ T_{\mathrm{recoil}} =E-E_\gamma^* =\frac{E^2}{2(W+E)} \simeq\frac{E^2}{2W}. } \] The last approximation is valid because \(E\ll W\); it is equivalent to the nonrelativistic recoil expression \(p_{\mathrm{recoil}}^2/(2m_4)\).

Here \[ W=54045.8700\ \mathrm{MeV}. \] Using the excitation energy from part (a), \[ \boxed{ T_{\mathrm{recoil}} =0.00109104\ \mathrm{MeV} \approx1.09\ \mathrm{keV}, } \] and \[ \boxed{ E_\gamma^*=10.85965\ldots\ \mathrm{MeV} \approx10.86\ \mathrm{MeV}. } \] The extra digits display the small recoil subtraction.

For forward emission, Lorentz transformation of the photon energy to the laboratory gives \[ E_{\gamma,\mathrm{lab}} =\Gamma_4\left(E_\gamma^*+Vp_\gamma^*\right) =\Gamma_4E_\gamma^*(1+\beta_4), \] where \(\Gamma_4=(1-\beta_4^2)^{-1/2}\). Hence \[ \boxed{ E_{\gamma,\mathrm{lab}} =E_\gamma^*\sqrt{\frac{1+\beta_4}{1-\beta_4}}. } \] The forward Doppler factor is \(1.00566453\), so \[ \boxed{ E_{\gamma,\mathrm{lab}} =10.92117\ldots\ \mathrm{MeV} \approx10.92\ \mathrm{MeV}. } \]

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.