Question

One of the many isotopes used in cancer treatment is 79198Au, with a half-life of 2.70 d. Determine the mass of this isotope that is required to give an activity of 255 Ci. (answer in mg)

Answer 1

given T1/2 = 2.7 days = 2.7*24*60*60 = 233280 s

A = 255 Ci

= 255*3.7*10^10 decay/second

decay constant, lamda = 0.693/T1/2

= 0.693/233280

= 2.97*10^-6 s^-1

let N is the no of molecules present.

we know, A = lamda*N

==> N = A/lamda

= 255*3.7*10^10/(2.97*10^-6)

= 3.177*10^18

no of moles, n = N/Na

= 3.177*10^18/(6.023*10^23)

= 5.27*10^-6 mole

we know, n = mass/molar mass

mass = n*molar mass

= 5.27*10^-6*198

= 1.04*10^-3 grams

= 1.04 mg <<<<<<<------------------Answer