Determining the uncertainty of Eγ - Revision history

Oborn at 15:49, 8 June 2010

2010-06-08T15:49:07Z

2009-02-05T06:27:20Z

2008-06-12T10:38:11Z

2008-06-12T10:02:31Z

2008-06-12T09:56:49Z

2008-06-12T09:56:02Z

2008-06-12T09:54:49Z

2008-06-12T09:54:34Z

2008-06-12T09:52:44Z

2008-06-12T09:52:24Z

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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>].
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 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>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>2:\ E_{\gamma}-P_ncos(\theta_n)-P_pcos(\theta_p)</math>
-<math>3: P_nsin(\theta_n-P_psin(\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.
 [http://wiki.iac.isu.edu/index.php/PhotoFission_with_Polarized_Photons_from_HRRL Go Back]

← Older revision		Revision as of 06:27, 5 February 2009
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	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.		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.
		+
		+	[http://wiki.iac.isu.edu/index.php/PhotoFission_with_Polarized_Photons_from_HRRL Go Back]

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 <math>\frac{\delta_t}{t}=\frac{1}{50}=2%</math>
-<math>v=\frac{d}{t}=\frac{3 +/- 0.2%}{50 +/- 2%}</math> = 0.2c ± 2.2%
+<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%

← Older revision		Revision as of 10:02, 12 June 2008
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	<math>3: P_nsin(\theta_n-P_psin(\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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 Applying the consevation of energy and momentun to the system we come up with three equations:
-<math>E_{\gamma}-1877.9-\sqrt{m_n^2+P_n^2}-\sqrt{m_p^2+P_p^2}</math>
+<math>1: E_{\gamma}-1877.9-\sqrt{m_n^2+P_n^2}-\sqrt{m_p^2+P_p^2}</math>
-<math>E_{\gamma}-P_ncos(\theta_n)-P_pcos(\theta_p)</math>
+<math>2: E_{\gamma}-P_ncos(\theta_n)-P_pcos(\theta_p)</math>
-<math>P_nsin(\theta_n-P_psin(\theta_p)</math>
+<math>3: P_nsin(\theta_n-P_psin(\theta_p)</math>

← Older revision		Revision as of 09:56, 12 June 2008
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	<math>E_{\gamma}-P_ncos(\theta_n)-P_pcos(\theta_p)</math>		<math>E_{\gamma}-P_ncos(\theta_n)-P_pcos(\theta_p)</math>
		+
		+	<math>P_nsin(\theta_n-P_psin(\theta_p)</math>

← Older revision		Revision as of 09:54, 12 June 2008
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	<math>E_{\gamma}-1877.9-\sqrt{m_n^2+P_n^2}-\sqrt{m_p^2+P_p^2}</math>		<math>E_{\gamma}-1877.9-\sqrt{m_n^2+P_n^2}-\sqrt{m_p^2+P_p^2}</math>

−	<math>E_{\gamma}-P_ncos(\~~thata_n~~)-P_pcos(\theta_p)</math>	+	<math>E_{\gamma}-P_ncos(\theta_n)-P_pcos(\theta_p)</math>

← Older revision		Revision as of 09:54, 12 June 2008
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	<math>E_{\gamma}-1877.9-\sqrt{m_n^2+P_n^2}-\sqrt{m_p^2+P_p^2}</math>		<math>E_{\gamma}-1877.9-\sqrt{m_n^2+P_n^2}-\sqrt{m_p^2+P_p^2}</math>
		+
		+	<math>E_{\gamma}-P_ncos(\thata_n)-P_pcos(\theta_p)</math>

← Older revision		Revision as of 09:52, 12 June 2008
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	Applying the consevation of energy and momentun to the system we come up with three equations:		Applying the consevation of energy and momentun to the system we come up with three equations:

−	<math>E_{\gamma}-1877.9-\sqrt{m_n^2-P_n^2}-\sqrt{m_p^2+P_p^2}</math>	+	<math>E_{\gamma}-1877.9-\sqrt{m_n^2+P_n^2}-\sqrt{m_p^2+P_p^2}</math>