Unit 1 : Worked Problems

Unit 1 : Worked Problems#

Problem - Carbon Mass Calculation

Given the mass of a carbon-12 atom is \(12u\), estimate the binding energy of the carbon nucleus. Comment on the difference between this and C-13 which has a binding energy of \(7.47~\textnormal{MeV/nucleon}\).

Given the mass of a carbon-12 atom is \(12u\), estimate the binding energy of the carbon nucleus. How does this compare to C-13 which has a binding energy of \(7.47~\textnormal{MeV/nucleon}\).

Core Knowledge

First we start with the mass of the proton, neutron, and electrons in the atomic mass units. Since the mass if that of a carbon-12 atom, we include the electrons in our calculation.

\[\begin{split} m_{p} = 1.00728 u \\ m_{n} = 1.00866 u \\ m_{e} = 0.00055 u \\ \end{split}\]

Method

Now we can write the binding energy as

\[\begin{split} B = M_{C12} - 6m_{n} - 6m_{p} - 6m_{e} \\ Bc^{2} = (12 - 1.00729 \times 6 - 1.00866 \times 6 -0.00055 \times 6) u \\ Bc^{2} = 0.099 u = 0.099 u \times 931.5~\textnormal{MeV/u} = 92.22~\textnormal{MeV} \end{split}\]

If we want to get this per nucleon we just need to divide by the total nucleons in a carbon 12 nucleus.

\[\begin{split} B/A = 92.22~\textnormal{MeV} / 12 = 7.68~\textnormal{MeV/nucleon}\\ \end{split}\]

A key thing to remember is that the electron mass does contribute if we are working in atomic mass values, whilst if we were instead given a nuclear mass value it does not. For heavy mass nuclei including the electrons has a smaller effect as the number of protons relative to neutrons drop. Given that the mass of an electron is around \(0.511~\textnormal{MeV/c}^{2}\) we can expect shifts in the total binding energy on the order of several MeV depending on the nuclei.

In this course we will predominantly work in atomic mass values, and tell you when this is not the case.

Application

The total binding per nucleon is higher in C-12 than C-13. It is closer to the line of stability, and we expect C-13 to be less stable. This matches what we expect from the abundance of elements, C-12 makes up 98.9% of all carbon isotopes, whilst C-13 makes up only 1.06%. Similarly C-14 has a lower binding energy than C-12 at 7.47 MeV. This is actually higher than C-13 (we’ll cover why next week), but it’s abundance is lower because is a radio-isotope.

Problem - Nucleon Momentum Estimation

Use the Heisenberg Uncertainty principle to estimate the momentum of nucleons inside the nucleus given that the nuclear radius is on the order of \(5~\textnormal{fm}\). Are the nucleons relativistic?

Use the Heisenberg Uncertainty principle to estimate the momentum of nucleons inside the nucleus given that the nuclear radius is on the order of \(5~\textnormal{fm}\). Are the nucleons relativistic?

Core Knowledge

The Heisenberg principle states that

\[ \Delta x \Delta p \geq \frac{\hbar}{2} \]

Method

In the case of a solid sphere we can measure the nucleons position with an uncertainty similar to the nuclear radius, hence

\[ c\Delta p \approx \frac{\hbar c}{2\Delta x} \]

as usual multiplying both terms by \(c\) to make use of the simple to remember quantity of \(\hbar c \approx 200~\textnormal{MeV}~\textnormal{fm}\). This implies that the momenta of the nucleons will be of the order of

\[ pc \approx \frac{200~\textnormal{MeV}~\textnormal{fm}}{2 \times 5~\textnormal{fm}} = 20~\textnormal{MeV}. \]

Application

This 20 MeV momentum is considerable, however it is much smaller than the nucleons mass, which means that albeit fast, they will not be in the relativistic regime. The momentum can be written as

\[ p = m v \]

and the ratio \(pc/E\) can be expressed as

\[ \frac{pc}{E} = \frac{mvc}{mc^{2}} = \frac{v}{c} = \beta \]

which gives \(\beta \approx 0.02\).

Despite the fact that we consider the nucleus as a solid sphere of protons and neutrons tightly packed together, the quantum mechanical reality is quite different as the nucleons move quite quickly around the allowed space when we consider movement between energy levels.

Problem - Shielding Cross-section Estimate

Consider a \(1~\textnormal{TBq}\) source of Cs-137 that produces an effective dose of \(1.073~\textnormal{Sv/hr}\) at a distance of 30cm (Delecroix Handbook 2002). The decay chain of Cs-137 produces a clear gamma ray peak at \(662~\textnormal{keV}\).

Assuming a cross-section of \(662~\textnormal{keV}\) gamma rays on lead nuclei of \(\sigma=4.3 \times 10^{-23}~cm^{2}\) determine the lead shielding thickness required to reduce the dose rate to \(1~\textnormal{mSv/hr}\).

What would happen if the cross-section was half this?

Consider a \(1~\textnormal{TBq}\) source of Cs-137 that produces an effective dose of \(1.073~\textnormal{Sv/hr}\) at a distance of 30cm (Delecroix Handbook 2002). The decay chain of Cs-137 produces a clear gamma ray peak at \(662~\textnormal{keV}\).

Assuming a cross-section of \(662~\textnormal{keV}\) gamma rays on lead nuclei of \(\sigma=4.3 \times 10^{-23}~cm^{2}\) determine the lead shielding thickness required to reduce the dose rate to \(1~\textnormal{mSv/hr}\).

What would happen if the cross-section was half this?

Core Knowledge

If we consider the intensity of our source passing through the lead shielding we get

\[ N = N_{0}~e^{-x/\lambda} \]

and

\[ \lambda = \frac{1}{n\sigma} \]

Method

To calculate the thickness of lead first we need to know the attenuation rate of our gamma rays in the lead shielding. We can estimate this based on the cross-section and the density of lead itself, \(\rho\).

Given the density, \(\rho=11.36 \textnormal{g/cm}^{3}\), and the mass of a lead atom to be \(m_{Pb} = 207.293u\) we can estimate \(n\) as

\[ n = \frac{11.36 \textnormal{g/cm}^{3}}{207.293u \times 1.66056\times 10^{-24} \textnormal{g/u} } = 3.3 \times 10 ^{22} \textnormal{cm}^{-3} \]

This leads to a mean free path of

\[ \lambda = \frac{1}{n\sigma} = 0.704 \textnormal{cm} \]

If we consider the intensity of our source passing through the lead shielding and take the natural log of our \(N\) equation we get

\[ x = -\lambda ~\textnormal{ln}\left( \frac{N}{N_{0}}\right). \]

Application

Now, since the Sievert is just a scaling factor based on the number of ionising particles passing on the other side of our shielding, we can replace \(N\) with our relative dose rates. If we consider our known attenuation and the drop in rate required we get

\[ x = -0.704 ~\textnormal{ln} \left( \frac{1~\textnormal{mSv/hr}}{1.073~\textnormal{Sv/hr}}\right) = 4.91~\textnormal{cm} \]

Note that this is an estimate based on just the 662keV peak, and neglects the \(r^{2}\) fall-off for point radioactive sources, but it highlights how the cross-section has an effect on the result. If the cross-section’s for gamma radiation were lower, we would expect much thicker shielding requirements.

Problem - Closest Approach Estimate

Considering Polonium-210 as an alpha emitter in Rutherford’s gold foil experiment, estimate the minimum theoretical limit to which alpha scattering may be able to place a limit on the size of the nucleus.

The atomic mass contributions from each of these are taken from the NIST material database as

\[\begin{split}M(\textnormal{Pb}_{210}) = 209.9841u \\ M(\textnormal{Pb}_{206}) = 205.9744u \\ M(\textnormal{He}_{4}) = 4.0026u \\\end{split}\]

Considering Polonium-210 as an alpha emitter in Rutherford’s gold foil experiment, estimate the minimum theoretical limit to which alpha scattering may be able to place a limit on the size of the nucleus.

The atomic mass contributions from each of these are taken from the NIST material database as

\[\begin{split}M(\textnormal{Pb}_{210}) = 209.9841u \\ M(\textnormal{Pb}_{206}) = 205.9744u \\ M(\textnormal{He}_{4}) = 4.0026u \\\end{split}\]

Core Knowledge

Now to determine the energy of the alphe we need to consider the mass deficit of the decay that took place to produce it.

\[ \textnormal{Po}_{210} \rightarrow \textnormal{Pb}_{206} + \alpha. \]

Note that here for the alpha mass contribution we have used the mass of a Helium atom as this encapsulates that the 2 extra electrons have to go somewhere.

This means the mass deficit energy, \(Q\), is given by

\[ Qc^{2} = (209.9828u - 205.9794u - 4.0026u)c^{2} = 0.0058u ~c^{2} \]

Application

This energy will be split between the alpha and daughter nuclei, however if we assume that all the energy goes into the kinetic energy of the alpha given it has much smaller mass this gives the maximum kinetic energy of the alpha in standard units as

\[ K_\alpha = 5.40~\textnormal{MeV} = 8.65175 \times 10^{-13} J \]

To consider the limit of the alpha scattering experiments we need to determine their distance of closest possible approach under the hyperbolic trajectory assumption.

Consider the case when the alpha particle is directed head-on towards the nucleus, so that when it reaches the distance of closest approach it turns around and travels back in the direction it came. At this turning point all of the alpha’s kinetic energy has been transformed into electric potential energy.

The potential some distance, \(r\), away is given by

\[ E_{Q} = \frac{1}{4\pi\epsilon_{0}} \frac{(2e)\cdot(Ze)}{r} \]

where 2e is the charge of the alpha, and Ze is the charge of the gold nucleus.

At the distance of closest approach, \(d\), this potential is equal to the starting kinetic energy of the alpha,

\[ K_{\alpha} = \frac{1}{4\pi\epsilon_{0}} \frac{(2e)\cdot(Ze)}{d}. \]

This means the distance is simply

\[ d = \frac{1}{4\pi\epsilon_{0}} \frac{(2e)\cdot(Ze)}{K_{\alpha}}. \]

Subbing our maximum possible kinetic energy into our equation for the distance gives

\[ d = 4.905 \times 10^{-14}~\textnormal{m} \]

Application

So our alpha particles only manage to get within \(49~\textnormal{fm}\) from the nucleus and that is if they hit the gold nucleus head on.

If we consider that typical \(\alpha\) energies are always between \(0-11 \textnormal{MeV}\) from known natural alpha emitters this gives a minimum distance of closest approach of around \(25~\textnormal{fm}\). This is still much bigger than the actual size of the gold nucleus (\(6.6~\textnormal{fm}\)).