Unit 6 : Worked Problems

Unit 6 : Worked Problems#

Example 6.1.1 : Alpha decay separation energy#

Calculate the separation energy for the alpha decay of \(^{238}\)U.

Solution

For \(\rm ^{238}U \rightarrow ^{234}Th + \alpha\), the separation energy is given by:

\[ Q = m(238,92) - m(234,90) - m(4,2), \]

where m is the nuclear or atomic mass (interchangeable in this calculation since the mass of the electrons cancels out in the atomic mass).

So

\[ Q = ( 238.050788 ~\textnormal{u} - 234.04359 ~\textnormal{u} - 4.002603~\textnormal{u} ) \times 931.5~\textnormal{MeV~u}^{-1} = 4.28~\textnormal{MeV} \]

Example 6.1.2 : Nuclear separation distance in alpha decay#

Using the separation energy calculated in 6.1.1, calculate the nuclear separation distance of the \(\alpha\) and \(^{234}\)Th daughter nucleus.

Solution

Nuclear separation \(\rm R = r_0 (A_{\alpha}^{1/3}+A_d^{1/3})\) where \(A_{\alpha}\) and \(\rm A_d\) are the atomic masses of the alpha and daughter nucleus respectively, and \(\rm r_0 = 1.2\) is a constant of proportionality. Therefore:

\[ R = 1.2(4^{1/3}+234^{1/3}) = 9.30~ \textnormal{fm} \]

Example 6.1.3 : Height of the Coulomb barrier to alpha decay#

Calculate the height of the Coulomb barrier to the alpha decay of \(\rm ^{238}U\) at the nuclear separation distance R calculated in 6.1.2.

Solution

The height of the barrier at nuclear separation distance R is calculated as the Coulomb potential of point charges Z\(_{\alpha}\)e and Z\(_d\)e, where Z\(_\alpha\) and Z\(_d\) are the atomic numbers of the alpha particle and daughter nucleus. The height of the barrier is therefore:

\[\begin{split} \begin{align*} V(R) &= \frac{2Z_dke^2}{R} \\ &= \frac{2(90)1.44~\textnormal{MeV~fm}}{9.30~\textnormal{fm}} \\ &= 27.9~\textnormal{MeV}, \end{align*} \end{split}\]

where \(ke^2=\frac{e^2}{4\pi\epsilon_0}\).

Example 6.1.4 : Outer extent of the Coulomb barrier to alpha decay#

Calculate the distance at which the Coulomb potential drops to the energy of the \(\alpha\) particle and, from this, the width of the Coulomb barrier in the alpha decay of \(^{238}\)U using your answers from earlier steps.

Solution

The energy of the alpha particle is \(T_e \approx Q = 4.28~\textnormal{MeV}\) (from Example 6.1.1). The distance at which the Coulomb potential R\(_Q\) is equal to Q is given by rearranging the equation for the Coulomb potential:

\[ R_Q = \frac{2Z_dke^2}{Q} = \frac{2(90)1.44~\textnormal{MeV~fm}}{4.28~\textnormal{MeV}} = 60.6~\textnormal{fm} \]

The width of the Coulomb barrier is therefore \(w = R_Q - R = 60.6~\textnormal{fm} - 9.30~\textnormal{fm} = 51.3~ \textnormal{fm} \).

Example 6.1.5 : Alpha-Coulomb barrier collision frequency#

Now find the frequency at which the \(\alpha\) particle collides with the Coulomb barrier in the alpha decay of \(^{238}\)U using your earlier answers.

Solution

Since the alpha particle at this energy is non-relativistic, the velocity of the alpha can be calculated from its kinetic energy \(T_e \approx Q = 4.28~\textnormal{MeV}\):

\[\begin{split} \begin{align*} T_e &= \frac{mv^2}{2} \approx Q\\ \textnormal{Therefore}\\ \frac{v}{c} &= sqrt{\frac{2Q}{m}} \\ &= sqrt{\frac{2(4.28)}{4.002603~u (931.5 \textnormal{MeV u}^{-1})}} \\ &= 0.0479~\textnormal{m~s}^{-1}c^{-1}\\ \textnormal{and}\\ v &= 1.44 \times 10^7~\textnormal{m~s}^{-1} \end{align*} \end{split}\]

The frequency of collisions of an alpha particle with this velocity is given by:

\[\begin{split} \begin{align*} f &= \frac{v}{2R} \\ &= \frac{1.88 \times 10^6~\textnormal{m~s}^{-1}}{2(9.30~\textnormal{fm})(10^{-15}~\textnormal{m~fm}^{-1}} \\ &= 0.772 \times 10^{21}~\textnormal{s}{-1} \end{align*} \end{split}\]

Example 6.1.6 : Alpha tunnelling probability#

Now calculate the tunnelling probability of the alpha particle in the decay of \(^{238}\)U by integrating over the Gamow factor.

Solution

The tunnelling probability is given by \(P = e^{-2G}\), where G is the Gamow factor:

\[ G = \sqrt{\frac{2m}{\hbar^2}}\int_R^{R_Q} \sqrt{V(r) - Q}~dr \]

Evaluating this integral gives us:

\[ G = \sqrt{\frac{2m}{\hbar^2Q}}2Z_d ke^2[arccos\sqrt{x}-\sqrt{x(1-x)}] \]

where \(x = R/R_Q = Q/V(R)\).

Putting in the numbers, this gives a tunnelling probability of

\[ P = e^{-2G} = 0.352 \times 10^{-38} \]

Example 6.1.7 : Alpha decay constant and half life#

Assuming that the preformation probability \(P_\alpha\) of the alpha particle is equal to 1, calculate the decay constant, and therefore half-life of \(^{238}\)U.

Solution

The decay constant is given by:

\[ \lambda = fP_\alpha P = f P = 2.72 \times 10^{-18} \]

And so the half-life of \(^{238}\)U is

\[\begin{split} \begin{align*} T_{1/2} &= \frac{ln(2)}{\lambda} \\ &= 2.55 \times 10^{17}~\textnormal{s} \\ &= 8.07 \times 10^9~\textnormal{years} \end{align*} \end{split}\]

Example 6.1.8 : Differences between calcualated and measured half-life#

The measured half-life of \(^{238}\)U is approximately \(4.468 \times 10^9\) years. Comment on the difference between this value and the half-life we have calculated.

Solution

The half-life calculated assumed a spherical nucleus and does not take into consideration:

  1. The non-spherical nature of heavy nuclei.

  2. The effects of nuclear angular momentum.

A 4% change in radius can give a factor of 5 difference in half-life, and in fact the measured half-life can be used to infer information about the radius.

Example 6.2.1 : Beta decay selection rules#

Use the beta decay selection rules to classify the following transitions:

  1. \(^{76}\)Br(1-) to \(^{76}\)Se(0+)

    Solution

    \(\Delta \pi = -\), so L is odd, since \(\Delta \pi = (-1)^L\).

    Since \(\Delta J = L + S = 1\), and L is odd, L = 1 and S = 0 is the only possible combination.

    This transition is therefore a 1st forbidden Fermi decay.

  2. \(^{22}\)Na(3+) to \(^{22}\)Na(0+)

    Solution

    There is no change in parity and \(\Delta \pi = +\), so L is even and can be 0, 2, etc.

    \(\Delta J = L + S = 3\), so L = 2 and S = 1.

    This transition is therefore a 2nd forbidden Gamow-Teller decay.

  3. \(^{40}\)K(4-) to \(^{40}\)Ca(0+)

    Solution

    There is a change in parity and \(\Delta \pi = -\), so L is odd.

    \(\Delta J = L + S = 4\), so L = 3 and S = 1.

    Therefore this is a 3rd forbidden Gamow-Teller decay

Example 6.2.2 : Beta decay endpoint value#

Calculate the endpoint value for the decay of \(^{81}\)Kr by beta minus, beta plus and electron capture, and therefore demonstrate which form of beta decay the isotope undergoes.

Solution

  1. Beta minus decay \( ^{81}_{36}Kr \rightarrow ^{81}_{37}Rb + \beta^- + \bar{\nu}_e\):

\[ Q = M_{Kr} - M_{Rb} - m_e = 80.916593~\textnormal{u} - 80.918994~\textnormal{u} - 5.4858 \times 10^{-4}~\textnormal{u} < 0 \]

  1. Beta plus decay \( ^{81}_{36}Kr \rightarrow ^{81}_{35}Br + \beta^+ + \nu_e\):

\[ Q = M_{Kr} - M_{Br} - 2m_e = 80.916593~\textnormal{u} - 80.916291~\textnormal{u} - 2(5.4858) \times 10^{-4}~\textnormal{u} < 0 \]

  1. Electron capture \( ^{81}_{36}Kr + e^- \rightarrow ^{81}_{35}Br + \nu_e\)

\[ Q = M_{Kr} - M_{Br} = 80.916593~\textnormal{u} - 80.916291~\textnormal{u} = 0.281~\textnormal{MeV} \]

Therefore, electron capture is the only form of beta decay that \(^{81}\)Kr can undergo.