The half-life of Au-198 is 2.7 days. What will be the activity of 1 mg of Au-198?

Explanation

half life T= 2.7 dayshalf life T= 2.7 days 24hour3600secondshalf life T= 233280 smass m= 1 mg= 10-3gAtomic mass= 198 g/molactivity A =?activity A =λNWhere λ is decay constantIt is found by given formula The relationship between the half-life, T1/2, and the decay constant is given byT1/2 = 0.693/λ.λ = 0.693T.λ = 0.693233280sλ =2.9 10-6s-1N is number of atoms given by;N=[ mass/atomic mass ]Avogadros number`N=[ 10-3g/198g/mol ]6.022 x 10^(23)``N= 3.03 x 10^18Now, we will find the activity Aactivity A =λNactivity A =[2.9 10-6s-1][3.03 x 10^18]activity A = 8.8x 10^12 disintegrations/second or 8.8x 10^12 BqFor Curie, 1 Ci= 3.7x 10^10 BqHence,A= 8.8x 10^12 / 3.7x 10^10 A= 240 Ci