Determining the uncertainty of Eγ

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To determine the uncertainty in Eγ we pick an angle for the neutron within [[math]\theta_n[/math], [math]\theta_n[/math] + Δ [math]\theta_n[/math]] and a momentum of the neutron between [[math]P_n[/math], [math]P_n[/math] + Δ [math]P_n[/math]].

What are reasonable Δ[math]\theta_n[/math] and Δ [math]P_n[/math]?

[math]P_n[/math] is determined by time of flight.

Knowns:

[math]m_n[/math] = 939.565 ± 0.00028 [math]MeV/c^2[/math]

d = 3 ± 0.005 m

t = 50 ± 1 ns

Fractional Uncertainties

[math]\frac{\delta_m}{m}=\frac{0.00028}{939.565}=0.00003%[/math]

[math]\frac{\delta_d}{d}=\frac{0.005}{3}=0.2%[/math]

[math]\frac{\delta_t}{t}=\frac{1}{50}=2%[/math]

[math]v=\frac{d}{t}=\frac{3\pm0.2%}{50\pm 2%}[/math] = 0.2c ± 2.2%

[math]P_n=m_nv[/math] = 188MeV/c ± 2.2%

Δ[math]P_n=4MeV/c[/math]

Δ[math]\theta_n[/math] can be determined knowing that the detector is 3 meters away and the dimensions of the detector are 5cm wide by 5cm tall.

Δ[math]\theta_n=tan^{-1}(\frac{5}{300})=0.0167rads=0.95degrees[/math]

Applying the consevation of energy and momentun to the system we come up with three equations:

[math]1:\ E_{\gamma}-1877.9-\sqrt{m_n^2+P_n^2}-\sqrt{m_p^2+P_p^2}[/math]

[math]2:\ E_{\gamma}-P_ncos(\theta_n)-P_pcos(\theta_p)[/math]

[math]3:\ P_nsin(\theta_n-P_psin(\theta_p)[/math]

Knowing the uncertainty of the momentum and angle of the neutron, the uncertainty of the energy can be calculated using these three equations. The resulting uncertainty of the energy is found to be 2.8 MeV.

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