Obtain the amount of `( )_(27)^(60) (Co)` (in `mu (g)` ) necessary to provide a radioactive source of `8.0 (~m) (Ci)` strength. The half life of `( )_(27)^(60) (Co)` is `5.3` years. (Give answer in integer value)

Obtain the amount of ^60Co necessary to provide a radioactive source of 8.0 Ci strength. The half-life of ^60Co is 5.3 years?

What is the amount of "^(60)CO_(27) necessary to provide a radioactive source of 8.0 m Ci strength? The half-life of ^(60)CO_(27) is 5.3 years.

Determine the amount of ._84Po^(210) necessary to provide a source of alpha particles of 5 millicurie strength. The half life of Po is 138 days.

Determine the amount of ""_(84)Po^(210) (polonium ) necessary to provide a source of alpha -particles of 5 millicurice strength , if life of polonium is 138 days (given ,1 curie =3.7xx10^(10) disintegrations /s )

What would be the mean life of ._(27)Co^(60) , if half-life period is 5.3 years?

The radioactive isotope ._(27)^(60)Co which has now replaced radium in the treatment of cancer can be made by a(n,p) or (n, gamma) reaction. For each reactio, indicate the apporipriate target nucleus. If the half life of ._(27)^(60)Co is 7 year evaluate the decay constant in s^(-1) .

Radioactive decay follows first-order kinetic. The mean life and half-life of nuclear decay process are tau = 1// lambda and t_(1//2) = 0.693//lambda . Therefore are a number of radioactive elements in nature, their abundance is directly proportional to half life. The amount remaining after n half lives of radioactive elements can be calculated using the relation: N = N_(0) ((1)/(2))^(n) Half life of .^(60)Co is 5.3 years, the time taken for 99.9% decay will be