gamma decay | Nuclear Physics 101

GEANT4 simulation of HPGe Clover detector

February 8, 2021

I recently go back to GEANT4 simulation. To simulate the gamma spectrum of the high-purity Germanium (HPGe) Clover detector.

The code is in here: https://github.com/goluckyryan/HPGeClover

Here are some visualizations

The simulated setup for the 16N isomer ration determination. The simulation generate all possible gamma ray energy. We can see a lot of 120 keV gamma being stopped by the vacuum pipe.

The simulation for the 16N isomer decay at the center is like

the simulation caught all feature, but the intensity of the double escape peak from the 6130 keV peak is smaller than the experiment. And the peak to Compton scattering background is different.

Since the strength of the escape peaks are very sensitive to the position and geometry of the crystal, the simulation condition need to be adjusted.

Here is the program structure. This is also a generic program structure.

comments or to do for the code

is there a way to get number of clover at EventAction ? I tried to use the G4LogicalVolumeStore, but it seems that the detector is constructed after EventAction.
is there any way to enable or disable geometry from command ?
change clover position from command?
set list of gamma energy and distribution from command
read file for energy and distribution

Gamma Transition

February 28, 2016

The gamma decay brings a excited nucleus to a lower energy state by emitting a photon. Photon carry angular momentum with intrinsic spin of 1. Its parity is positive. Thus, we have three constrain from conversation laws immediately.

$E_f = E_i + \hbar \omega$

$J_f^{\pi_f} = J_i^{\pi_i} + L$

where $E$ is the energy of the nucleus, $\hbar \omega$ and $L$ are the photon energy and angular momentum respectively. We also have to consider the parities of electric and magnetic transition are different.

To calculate the transition rate, we can start from a classical equation of power emission from an antenna, since the photon energy is quantized, the transition rate [number of photon emitted per time] is the power divided by a photon energy.

$T(qL) = \frac{2(2L+1)}{\epsilon_0 L [(2L+1)!!]^2 \hbar} (\frac{\omega}{c})^{2L+1} B(qL)$

where $qL$ is the electromagnetic multipole with angular momentum $L$ , and $B(qL)$ the the reduced transition probability, it is equal to the square of the magnitude of the transition matrix element $M_{fi}(qL)$ .

In the electric transition, the multipole is

$qL = e r^L$

we assume the transition is conduced by a single nucleon and the rest of the nucleus is unaffected. The transition matrix element than can be written as

$M_{fi}(qL) = \left<j_f m_f|e r^L Y_{LM}(\Omega)| j_i m_i\right>$

The single particle wave function can be written as

$\left|j m\right>= R_{nl}(r) [Y_l \times \chi_{1/2}]_{jm}$

The matrix elements becomes,

$M_{fi}(qL) = e \int_{0}^{\infty} R_{n_f l_f}^* r^L R_{n_i l_i} r^2 dr \times \left<Y_{LM}\right>$

To evaluate the radial integral, we make another assumption that the nucleus is a sphere of uniform density with radius $R=r_0 A^{1/3}$ ,

$R_{nl}(r) = \frac{\sqrt{3}}{R^{3/2}}$ , for $r<R$ , so that $\int_{0}^{R} |R_{nl}(r)|^2 r^2 dr = 1$

Then the radial integral is

$\left<r^L \right>=\frac{3}{R}\int_{0}^{R} r^{L+2} dr = \frac{3}{L+3} r_0^L A^{L/3}$

The reduced transitional probability

$B_{sp}(qL)=\sum \limits_{M m_f} |\left<j_f m_f|e r^L Y_{LM}|j_i m_i\right>|^2$

$= e^{2} \left< r^{L} \right> ^{2} \sum \limits_{M m_f} \left<Y_{LM}\right>$

the angular part could be assumed as $1/4\pi$ as the total solid angle is $4\pi$ . Thus, with these three assumptions, we have the Weisskopf single particle estimation for the L-pole reduced electric transition probability