Unit 2 : Equations & Terms

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Unit 2 : Equations & Terms#

Equations#

To aid with exam revision below is a list of all equations that you should learn for this unit in the course.

Form Factor Modification

(1)#\[\begin{equation} \left(\frac{d\sigma}{d\Omega}\right)_{\textnormal{distributed}} = |F({\bf q}^{2})|^{2}\times \left(\frac{d\sigma}{d\Omega}\right)_{\textnormal{point-like}}, \end{equation}\]

Relation between Form Factor and Charge Distribution

(2)#\[\begin{equation} F(q) = \int_{-\infty}^{\infty} e^{i~{\bf q} \cdot {\bf r} / \hbar} ~\rho({\bf r})~d^{3} {\bf r} \end{equation}\]

Saxon Woods Distribution

(3)#\[\begin{equation} \rho(r) = \frac{\rho_0}{1+{\textnormal{exp}}[\frac{r-a}{d}]} \end{equation}\]

  • \(a\) is the “edge” of the nucleus defining the width of the top-hat.

  • \(d\) is the so-called Skin or Surface Thickness.

Typical Values For the charge density distribution

(4)#\[\begin{equation} \rho_{\textnormal{charge}}^{0} = \frac{\rho^0_{\textnormal{charge}}}{1+\textnormal{exp}[\frac{r-a}{d}]} \end{equation}\]

  • \(a = 1.07 A^{1/3} ~\textnormal{fm}~~\textnormal{(radius)}\)

  • \(d = 0.54~\textnormal{fm}~\textnormal{ (skin thickness)}\)

  • \(\rho^{0} = 0.06-0.08~e~\textnormal{fm}^{-3}~\textnormal{ (average charge density)}\)

  • Note: We expect \(\rho^{0}\) to be lower for heavy nuclei as \(Z\) tends away from \(N\).

Typical Values For the mass density distribution

(5)#\[\begin{equation} \rho_{\textnormal{mass}}(r) = \frac{\rho^0_{\textnormal{mass}}}{1+\textnormal{exp}[\frac{r-a}{d}]} \end{equation}\]

  • \(a = 1.2 A^{1/3} ~\textnormal{fm}~~\textnormal{(radius)}\)

  • \(d = 0.75~\textnormal{fm}~\textnormal{ (skin thickness)}\)

  • \(\rho^{0} = 0.17~e~\textnormal{fm}^{-3}~\textnormal{ (average charge density)}\)

Liquid Drop Model Binding Energy Estimate

(6)#\[\begin{equation} B(A,Z) = a_{v}A - a_{s}A^{2/3} - a_{c}Z(Z-1)A^{-1/3} - \left[a_a(A-2Z)^{2}A^{-1}\right]_{\textnormal{assym}}- \delta_{\textnormal{pair}} \end{equation}\]

  • \(a_{v} = 15.5 \textnormal{MeV}\),

  • \(a_{s} = 16.8 \textnormal{MeV}\),

  • \(a_{c} = 0.72 \textnormal{MeV}\),

  • \(a_{a} = 23.0 \textnormal{MeV}\),

  • \(a_{p} = 12.0 \textnormal{MeV}\).

Liquid Drop Model Binding Energy Estimate per Nucleon

(7)#\[\begin{align} B(A,Z)/A = a_{v} - a_{s}A^{-1/3} - a_{c} Z(Z-1) A^{-4/3} - a_{a}(A-2Z)^{2}A^{-2} + \delta A^{-1} \end{align}\]

  • \(a_{v} = 15.5 \textnormal{MeV}\),

  • \(a_{s} = 16.8 \textnormal{MeV}\),

  • \(a_{c} = 0.72 \textnormal{MeV}\),

  • \(a_{a} = 23.0 \textnormal{MeV}\),

  • \(a_{p} = 12.0 \textnormal{MeV}\).

Coulomb Energy Estimate

(8)#\[\begin{equation} E_{e} = \frac{3}{5} \frac{Z^{2}e^{2}}{4\pi \epsilon_{0}R} \end{equation}\]

Coulomb Constant assuming spherical integration

(9)#\[\begin{equation} a_{c} = \frac{3}{5} \times \frac{e^{2}}{4\pi \epsilon_{0} r_{0}} \approx 0.7~\textnormal{MeV} \end{equation}\]

Coulomb Final Term

(10)#\[\begin{equation} B_{C} = -a_{C} \frac{Z(Z-1)}{A^{1/3}} \end{equation}\]

Nucleon Pairing Term

(11)#\[\begin{split}\begin{align} \delta_{\textnormal{pair}} &= +a_{p}A^{-1/2} ~~~~\textnormal{for N odd AND Z odd}\\ \delta_{\textnormal{pair}} &= 0 ~~~~~~~~~~~~~~~~~~~\textnormal{for N even OR Z even} \\ \delta_{\textnormal{pair}} &= -a_{p}A^{-1/2} ~~~~\textnormal{for N even AND Z even}\\ \end{align}\end{split}\]

Nucleon Asymmetry Term

(12)#\[\begin{align} B_{A} = -a_{a} (A-2Z)^{2}A^{-1} \end{align}\]
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