¿A shaker of salt substitute contains 1.6 oz of K. What is the activity, in milliCuries, of the potassium in the shaker? The activity is 7 microcuries (µCi)
We know K-40 (potassium having atomic mass 40 g) is radioactive and its natural abundance is 0.012% So for 1 mol of potassium contains 0.00012 mol of K-40 Now 1.6 oz of K = 45.36 g of K average atomic weight of K = 39.1 g so 45.36 g of K contains: (45.36 / 39.1) * 0.00012 * 6.022 x 10²³ (atoms of K-40) = 8.4 x 10¹⁹ atoms of K-40 We know, activity A is: A = 0.693 / t1/2 N₀ [t1/2 : half life time and N₀ : initial number of atoms] t1/2 of K-40 = 1.28 x 10⁹ years = 4.04 x 10¹⁶ seconds So A = (0.693 / 4.04 x 10¹⁶) * (8.4 x 10¹⁹ ) = 1441 cps A = 141 x (1/3.7 x 10¹⁰) because 3.7 x 10¹⁰ cps = 1 Ci A = 3.9 x 10⁻⁸ Ci = 3.9 x 10⁻⁵ millicurie
