Example of Fast NAA:

Example: 100 mg piece of a gold ornament containing 10% copper as impurity was irradiated with thermal neutrons at 2.1×10¹¹ n/cm²/s for 2 days producing ¹⁹⁸Au (t_½ = 2.7 d) and ⁶⁴Cu (t_½ = 12.7 h). Calculate the activity of these nuclides in terms of counts/s at the end of irradiation and after a delay of 2 d and 4 d. How long will it take for the decay of activity due to ⁶⁴Cu to 0.1% of its original activity? ¹⁹⁷Au is monoisotopic and its thermal neutron absorption cross section (σ) is 99 b whereas ⁶³Cu is 30.9% abundant and its σ value is 4.4 b.

Solution        100 mg piece of gold ornament will contain 90 mg gold or ¹⁹⁷Au and 10 mg of copper or 0.309×10 = 3.09 mg ⁶⁴Cu

Using the Eq. (13.8) Activity due to ¹⁹⁸Au at the end of irradiation

=   90×10^-3 × 6.02×10²³ ×2.1×10¹¹(1- 0.5985)/197

=   22.96×10⁸ cps

Activity after a delay of 2 d   = 22.96×10⁸×e(-0.693×2/2.7)

=          13.74×10⁸ cps

Activity after a delay of 4 d   =      22.96×10⁸×e(-0.693×4/2.7)

=          8.22×10⁸ cps

Activity due to ⁶⁴Cu at the end of irradiation = 3.09×10^-3×6.02×10²³×4.4×10^-24\(1-0.073)

= 2.53×10⁷ cps

Activity after a delay of 2 d            = 2.53×10⁷×e(0.693×48/12.7)

= 18.4×10⁵ cps

Activity after a delay of 4 d            = 2.53×10⁷×e (0.693×96/12.7)

= 13.4×10⁴

Using the formula    t = 1/λ lnN_o/N

After substitution, time interval = 12.7/0.693 ln 2.53×10⁷×0.1/100

= 185.8 h = 7.74 d