IPhO 1997 · Theory problem 2

Nuclear Masses and Stability

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All energies in this question are expressed in MeV - millions of electron volts.
One MeV = \(1.6\times10^{-13}\) J, but it is not necessary to know this to solve the problem.

The mass \(M\) of an atomic nucleus with \(Z\) protons and \(N\) neutrons (i.e. the mass number \(A=N+Z\)) is the sum of masses of the free constituent nucleons (protons and neutrons) minus the binding energy \(B/c^2\).

\[ Mc^2=Zm_pc^2+Nm_nc^2-B \]

The graph shown below plots the maximum value of \(B/A\) for a given value of \(A\), vs. \(A\).
The greater the value of \(B/A\), in general, the more stable is the nucleus.

Figure 1

(a) Above a certain mass number \(A_\alpha\), nuclei have binding energies which are always small enough to allow the emission of alpha-particles (\(A=4\)). Use a linear approximation to this curve above \(A=100\) to estimate \(A_\alpha\). (3 marks)

For this model, assume the following:

  • Both initial and final nuclei are represented on this curve.
  • The total binding energy of the alpha-particle is given by \(B_4=25.0\) MeV (this cannot be read off the graph!).

(b) The binding energy of an atomic nucleus with \(Z\) protons and \(N\) neutrons (\(A=N+Z\)) is given by a semi-empirical formula:

\[ B=a_vA-a_sA^{2/3}-a_cZ^2A^{-1/3} -a_a\frac{(N-Z)^2}{A}-\delta \]

The value of \(\delta\) is given by:

\(+a_pA^{-3/4}\) for odd-\(N\)/odd-\(Z\) nuclei

\(0\) for even-\(N\)/odd-\(Z\) or odd-\(N\)/even-\(Z\) nuclei

\(-a_pA^{-3/4}\) for even-\(N\)/even-\(Z\) nuclei

The values of the coefficients are:

\[ a_v=15.8\text{ MeV};\quad a_s=16.8\text{ MeV};\quad a_c=0.72\text{ MeV};\quad a_a=23.5\text{ MeV};\quad a_p=33.5\text{ MeV}. \]

(i) Derive an expression for the proton number \(Z_{\max}\) of the nucleus with the largest binding energy for a given mass number \(A\). Ignore the \(\delta\)-term for this part only. (2 marks)

(ii) What is the value of \(Z\) for the \(A=200\) nucleus with the largest \(B/A\)? Include the effect of the \(\delta\)-term. (2 marks)

(iii) Consider the three nuclei with \(A=128\) listed in the table on the answer sheet. Determine which ones are energetically stable and which ones have sufficient energy to decay by the processes listed below. Determine \(Z_{\max}\) as defined in part (i) and fill out the table on your answer sheet.

In filling out the table, please:

  • Mark processes which are energetically allowed thus: √
  • Mark processes which are NOT energetically allowed thus: 0
  • Consider only transitions between these three nuclei.

Decay processes:

(1) \(\beta^-\) - decay; emission from the nucleus of an electron
(2) \(\beta^+\) - decay; emission from the nucleus of a positron
(3) \(\beta^-\beta^-\) - decay; emission from the nucleus of two electrons simultaneously
(4) Electron capture; capture of an atomic electron by the nucleus.

The rest mass energy of an electron (and positron) is \(m_ec^2=0.51\) MeV; that of a proton is \(m_pc^2=938.27\) MeV; that of a neutron is \(m_nc^2=939.57\) MeV.

(3 marks)

(b) (iii)

Nucleus/Process \(\beta^-\) - decay \(\beta^+\) - decay Electron-capture \(\beta^-\beta^-\) - decay
\({}^{128}_{53}\mathrm{I}\)
\({}^{128}_{54}\mathrm{Xe}\)
\({}^{128}_{55}\mathrm{Cs}\)

Notation : \({}^{A}_{Z}X\)

\(X\) = Chemical Symbol

Official English text of the problem.

Answers

  • (a) \(\boxed{A_\alpha\approx2.1\times10^2}\). The straight-line fit used below gives approximately \(215\).
  • (b)(i) \(\displaystyle \boxed{Z_{\max}=\frac{2a_aA}{4a_a+a_cA^{2/3}}=\frac{A}{2+\frac{a_c}{2a_a}A^{2/3}}}\). This is the continuous maximum; with pairing omitted, choose the nearest integer.
  • (b)(ii) \(\boxed{Z=80}\), with \(N=120\) and \(B/A\approx8.05\) MeV.
  • (b)(iii) The continuous result is \(\boxed{Z_{\max}\approx53.58}\); the integer binding-energy maximum is \(Z=54\). Within the specified transitions, \({}^{128}\mathrm{Xe}\) is stable, while \({}^{128}\mathrm{I}\) and \({}^{128}\mathrm{Cs}\) can decay.
Nucleus/Process \(\beta^-\) - decay \(\beta^+\) - decay Electron-capture \(\beta^-\beta^-\) - decay
\({}^{128}_{53}\mathrm{I}\) √ 0 0 √
\({}^{128}_{54}\mathrm{Xe}\) 0 0 0 0
\({}^{128}_{55}\mathrm{Cs}\) 0 √ √ 0

Entries whose daughters lie outside these three nuclei are marked \(0\) under the stated restriction.

Solution

(a)

Alpha decay conserves the total numbers of protons and neutrons. Consequently, the free-nucleon rest energies cancel when calculating the energy release:

\[ Q_\alpha =\bigl[M(A,Z)-M(A-4,Z-2)-M(4,2)\bigr]c^2 =B(A-4)+B_4-B(A). \]

The decay is energetically allowed when \(Q_\alpha>0\).

Write the straight-line approximation to the plotted quantity as

\[ b(A)\equiv\frac{B(A)}{A}\simeq b_0-kA. \]

Representative readings from the graph are

\[ b(100)\simeq8.85\text{ MeV}, \qquad b(200)\simeq8.05\text{ MeV}. \]

Thus,

\[ k\simeq8.0\times10^{-3}\text{ MeV}, \qquad b_0\simeq9.65\text{ MeV}, \]

where \(A\) is dimensionless. The total binding energy is therefore

\[ B(A)\simeq b_0A-kA^2. \]

Using the finite change of four nucleons,

\[ \begin{aligned} Q_\alpha &=B_4+b_0(A-4)-k(A-4)^2-b_0A+kA^2\\ &=B_4-4b_0+8kA-16k. \end{aligned} \]

At the threshold, \(Q_\alpha=0\), giving

\[ A_\alpha =\frac{b_0-B_4/4}{2k}+2. \]

Numerically,

\[ A_\alpha \simeq\frac{9.65-25.0/4}{0.016}+2 =214.5. \]

Since \(Q_\alpha\) increases with \(A\), alpha emission is energetically possible above this threshold. Allowing for the precision of the graph,

\[ \boxed{A_\alpha\approx2.1\times10^2.} \]

(b)(i)

At fixed \(A\), substitute \(N=A-Z\), so that \(N-Z=A-2Z\). Omitting pairing,

\[ B(Z)=a_vA-a_sA^{2/3}-a_cZ^2A^{-1/3} -\frac{a_a}{A}(A-2Z)^2. \]

Treating \(Z\) initially as continuous, the maximum satisfies

\[ \frac{\partial B}{\partial Z} =-2a_cZA^{-1/3} +\frac{4a_a}{A}(A-2Z)=0. \]

Multiplying by \(A\) and collecting the terms in \(Z\),

\[ (2a_cA^{2/3}+8a_a)Z=4a_aA. \]

Hence,

\[ \boxed{ Z_{\max} =\frac{2a_aA}{4a_a+a_cA^{2/3}} =\frac{A}{2+\frac{a_c}{2a_a}A^{2/3}}. } \]

This stationary point is a maximum because

\[ \frac{\partial^2B}{\partial Z^2} =-2a_cA^{-1/3}-\frac{8a_a}{A}<0. \]

Without pairing, the physical integer maximum is the integer nearest this value.

(b)(ii)

For \(A=200\), the continuous maximum without pairing is

\[ Z_0 =\frac{200}{2+\frac{0.72}{47.0}200^{2/3}} \simeq79.242. \]

Because \(A\) is even, even \(Z\) also means even \(N\), while odd \(Z\) means odd \(N\). The pairing magnitude is

\[ P=a_pA^{-3/4} =33.5(200)^{-3/4} \simeq0.6299\text{ MeV}. \]

Since the formula contains \(-\delta\), pairing adds \(P\) to the binding energy for even \(Z\), and subtracts \(P\) for odd \(Z\).

The unpaired binding energy can be written

\[ B_0(Z)=B_0(Z_0) -\left(a_cA^{-1/3}+\frac{4a_a}{A}\right)(Z-Z_0)^2. \]

Therefore, within either parity class, the best integer is the one nearest \(Z_0\). The best odd choice is \(79\), and the best even choice is \(80\). Direct substitution, including pairing, gives:

\(Z\) \(N\) \(\delta\) (MeV) \(B\) (MeV)
78 122 \(-0.630\) 1609.55
79 121 \(+0.630\) 1609.17
80 120 \(-0.630\) 1610.12

In particular,

\[ B(80)-B(79)\simeq0.954\text{ MeV}>0. \]

Thus,

\[ \boxed{Z=80,\qquad N=120,} \]

with

\[ \frac BA=\frac{1610.12}{200}\simeq8.05\text{ MeV}. \]

(b)(iii)

First, the result from part (i) gives

\[ Z_{\max} =\frac{128}{2+\frac{0.72}{47.0}128^{2/3}} \simeq53.577. \]

Thus the nearest integer is \(54\), which is also favoured by pairing.

For these nuclei,

\[ P=33.5(128)^{-3/4}\simeq0.880313\text{ MeV}. \]

Iodine and caesium have odd \(N\) and odd \(Z\), so \(\delta=+P\). Xenon has even \(N\) and even \(Z\), so \(\delta=-P\).

Decay energetics require nuclear masses as well as binding energies. Writing \(E_Z=M_Zc^2\),

\[ E_Z =128m_nc^2+Z(m_p-m_n)c^2-B_Z =128m_nc^2-(1.30\text{ MeV})Z-B_Z. \]

Substitution gives the following values. Extra digits are retained to resolve the small double-beta energy.

Nucleus \(B_Z\) (MeV) \(E_Z-E_{54}\) (MeV)
\({}^{128}_{53}\mathrm{I}\) 1104.656 3.195830
\({}^{128}_{54}\mathrm{Xe}\) 1106.552 0
\({}^{128}_{55}\mathrm{Cs}\) 1103.172 2.079906

Neglecting atomic electron binding energies and neutrino rest masses, energy conservation gives

\[ \begin{aligned} Q_{\beta^-}(Z)&=E_Z-E_{Z+1}-m_ec^2,\\ Q_{\beta^+}(Z)&=E_Z-E_{Z-1}-m_ec^2,\\ Q_{\mathrm{EC}}(Z)&=E_Z+m_ec^2-E_{Z-1},\\ Q_{\beta^-\beta^-}(Z)&=E_Z-E_{Z+2}-2m_ec^2. \end{aligned} \]

The electron rest energy is subtracted for emission and added for capture. A process is energetically allowed when \(Q>0\).

All possible transitions within the specified set are:

Initial nucleus Process Daughter \(Q\) (MeV) Allowed?
\({}^{128}\mathrm{I}\) \(\beta^-\) \({}^{128}\mathrm{Xe}\) \(+2.686\) √
\({}^{128}\mathrm{Xe}\) \(\beta^-\) \({}^{128}\mathrm{Cs}\) \(-2.590\) 0
\({}^{128}\mathrm{Xe}\) \(\beta^+\) \({}^{128}\mathrm{I}\) \(-3.706\) 0
\({}^{128}\mathrm{Cs}\) \(\beta^+\) \({}^{128}\mathrm{Xe}\) \(+1.570\) √
\({}^{128}\mathrm{Xe}\) Electron capture \({}^{128}\mathrm{I}\) \(-2.686\) 0
\({}^{128}\mathrm{Cs}\) Electron capture \({}^{128}\mathrm{Xe}\) \(+2.590\) √
\({}^{128}\mathrm{I}\) \(\beta^-\beta^-\) \({}^{128}\mathrm{Cs}\) \(+0.096\) √

For the small double-beta energy specifically,

\[ \begin{aligned} Q_{\beta^-\beta^-} &=E_{53}-E_{55}-2m_ec^2\\ &=3.195830-2.079906-1.020\\ &\simeq0.0959\text{ MeV}>0. \end{aligned} \]

Thus simultaneous double-beta decay of iodine to caesium is energetically allowed, even though the second step of the sequential route through xenon is forbidden.

Xenon has no energetically allowed transition within the specified set. Iodine can undergo beta-minus or double-beta-minus decay; caesium can undergo positron emission or electron capture.

The solution was checked against the official solution.