Question

One of the many isotopes used in cancer treatment is 198 79 Au , with a half-life of 2.70 d. Determine the mass of this isotope that is required to give an activity of 285 Ci.

Answer in mg please

Answer 1

Radio Active decay formula = y(t) = Ae^(-?t) ---> 1 Eq

where y = Number of isotopes

A = Initial number of isotopes

? = decay constant

From the equation 1 , rate of decay equation as shown below.

r(t) = -A?e^(-?t) --- 2 Eq

The half-life T(1/2) of an isotope is the time taken for the nuclei of that isotope to decay to half of its original number.

ln(2) / ? = T(1/2) ---> 3 Eq

1 Ci is 3.7x10^10 decays per second

1 mass unit (u) = 1.66x10^-27 kg

From the eq 3 , calucate the ? value.

2.70 days to convert into seconds then T (1/2) =233300 sec

? = ln(2) / T(1/2) = 2.97X10^-6

now , you want to cacluate A value for finding number of isotropes.

285 Ci = 1.054 x 10^13

-1.054x10^13 = -A?e^(-?*0)

From the above equation you can easily find the A value.

1.054x10^13 / 2.97X10^-6 = A

3.548X10^18 = A

For mass cacluation in isotrope = (3.548x10^18)*(198u)

1 mass unit (u) = 1.66x10^-27 kg

= 3.548x10^18x198u

= 1.166 mg