Neutron energy resolution in NaI detector: Difference between revisions
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<math> E_n = \frac{m}{2}\left( \frac{d}{t_n} \right)^2 </math> | <math> E_n = \frac{m}{2}\left( \frac{d}{t_n} \right)^2 </math> | ||
<math> \delta E_n = -2\ E_n \frac{\delta t_n}{t_n} = -\frac{2 \sqrt{2}}{\sqrt{m} d}\ E_n^{3/2} \delta t_n </math> | <math> \delta E_n = -2\ E_n \frac{\delta t_n}{t_n} = -\frac{2 \sqrt{2}}{\sqrt{m} d}\ E_n^{3/2} \delta t_n </math> | ||
* <math> \delta t_n = 15\ ns</math> | * <math> \delta t_n = 15\ ns</math> | ||
* <math> mc^2 = 938\ MeV </math> | * <math> mc^2 = 938\ MeV </math> | ||
<math> d = 0.5 m: \ \ \ \delta E_n = 0.83\ E_n^{3/2}\ MeV </math> | <math> d = 0.5 m: \ \ \ \delta E_n = 0.83\ E_n^{3/2}\ MeV </math> | ||
Revision as of 03:37, 28 April 2012
[math]\displaystyle{ E_n = \frac{m}{2}\left( \frac{d}{t_n} \right)^2 }[/math]
[math]\displaystyle{ \delta E_n = -2\ E_n \frac{\delta t_n}{t_n} = -\frac{2 \sqrt{2}}{\sqrt{m} d}\ E_n^{3/2} \delta t_n }[/math]
- [math]\displaystyle{ \delta t_n = 15\ ns }[/math]
- [math]\displaystyle{ mc^2 = 938\ MeV }[/math]
[math]\displaystyle{ d = 0.5 m: \ \ \ \delta E_n = 0.83\ E_n^{3/2}\ MeV }[/math]
[math]\displaystyle{ d = 1.0 m: \ \ \ \delta E_n = 0.42\ E_n^{3/2}\ MeV }[/math]
[math]\displaystyle{ d = 1.5 m: \ \ \ \delta E_n = 0.28\ E_n^{3/2}\ MeV }[/math]
[math]\displaystyle{ d = 2.0 m: \ \ \ \delta E_n = 0.21\ E_n^{3/2}\ MeV }[/math]