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Didn't really have any idea when I started to write the problem, but I think I've managed to work it out now posting for clarification.
Estimate the time it will take to produce a 100 μCi source of 36Cl by irradiating 1g of natural nickel chloride (molecular weight 129.6) in a neutron flux of 1014 cm-2s-1. The cross section for neutron capture on 35Cl is 43 b and the half-life of 36Cl is 3 x 105 years. 75.8% of natural chlorine consists of 35Cl.
Number of 35-Cl atoms in 1g of NaCl2 is
[tex] N = 2x0.758xN_A/molecular weight = 7.04x10^{21} [/tex]
[tex]A(t) = \lambda N = P(1 - e^{-\lambda t} \approx P\lambda t[/tex]
[tex]t = \frac{A(t)}{P\lambda} = \frac{A(t)t_{1/2}}{ln 2 x \Omega x F x N} [/tex]
[tex]F = incident flux, A(t) = activity, \Omega = cross section for neutron capture[/tex]
[tex] F = 10^{14} cm^{-2}s^{-1} [/tex]
[tex]t_{1/2} = 3.16x10^7 x 3x10^5 = 9.48x10^12 s [/tex]
[tex]\Omega = 43x10^-24 cm [/tex]
[tex]A(t) = 100\mu Ci = 3.7x10^6 Bq [/tex]
Putting in the numbers gives 1.67164x10^6 seconds = 19.35 days
Didn't really have any idea when I started to write the problem, but I think I've managed to work it out now posting for clarification.
Homework Statement
Estimate the time it will take to produce a 100 μCi source of 36Cl by irradiating 1g of natural nickel chloride (molecular weight 129.6) in a neutron flux of 1014 cm-2s-1. The cross section for neutron capture on 35Cl is 43 b and the half-life of 36Cl is 3 x 105 years. 75.8% of natural chlorine consists of 35Cl.
Homework Equations
Number of 35-Cl atoms in 1g of NaCl2 is
[tex] N = 2x0.758xN_A/molecular weight = 7.04x10^{21} [/tex]
[tex]A(t) = \lambda N = P(1 - e^{-\lambda t} \approx P\lambda t[/tex]
[tex]t = \frac{A(t)}{P\lambda} = \frac{A(t)t_{1/2}}{ln 2 x \Omega x F x N} [/tex]
[tex]F = incident flux, A(t) = activity, \Omega = cross section for neutron capture[/tex]
The Attempt at a Solution
[tex] F = 10^{14} cm^{-2}s^{-1} [/tex]
[tex]t_{1/2} = 3.16x10^7 x 3x10^5 = 9.48x10^12 s [/tex]
[tex]\Omega = 43x10^-24 cm [/tex]
[tex]A(t) = 100\mu Ci = 3.7x10^6 Bq [/tex]
Putting in the numbers gives 1.67164x10^6 seconds = 19.35 days
Didn't really have any idea when I started to write the problem, but I think I've managed to work it out now posting for clarification.