Neutron energy resolution in plastic scintillation detector with start signal from NaI detector

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[math] E_n = \frac{m}{2}\left( \frac{d}{t_n} \right)^2 [/math]

[math] \delta E_n = -\frac{2 \sqrt{2}}{\sqrt{m} d}\ E_n^{3/2} \delta t_n [/math]

• [math] \delta t_n = 15\ ns[/math]
• [math] mc^2 = 938\ MeV [/math]

[math] d = 0.5\ m: \ \ \   ( \delta E_n/E_n ) = 0.83\ E_n^{1/2}\ MeV [/math]
[math] d = 0.7\ m: \ \ \   ( \delta E_n/E_n ) = 0.59\ E_n^{1/2}\ MeV [/math]
[math] d = 1.0\ m: \ \ \   ( \delta E_n/E_n ) = 0.42\ E_n^{1/2}\ MeV [/math]
[math] d = 1.5\ m: \ \ \   ( \delta E_n/E_n ) = 0.28\ E_n^{1/2}\ MeV [/math]
[math] d = 2.0\ m: \ \ \   ( \delta E_n/E_n ) = 0.21\ E_n^{1/2}\ MeV [/math]