Question

BIOPHYSICS

1. Cobalt-60 (60Co), is a synthetic radioactive isotope of cobalt with a half-life of 5.2713 years. It is an isotope that emits gamma rays essential to the medical community for cancer treatments, as well as sterilization of medical devices. Cobalt-60 is used as a radiation source for medical radiotherapy where it is used in cancer treatment to control or kill malignant cells. Cobalt-60 is used as the radiation source in Gamma Knife equipment that enables non-surgical treatment of brain tumours.

Suppose you purchased a high activity cobalt source with 10⁶ Curie.

a) Please find how many kg of ⁶⁰Co are needed to achieve this activity?   (3 points)

b) After how many years will the activity of this source be reduced to 10% of its initial value? (3 points)

Hints: Remember unit conversions (years to second, Ci to Bq). First find number of nuclides that creates this much activity. Use Avogadro’s number to find kg.

Answer 1

Solution:

Activity Of Cobalt-60 is 10^6 Curie. = 10^6 * (3.7*10^10) = 3.7*10^16 disint/s

Half life = 5.2173 years = 5.2173*(3.16*10^7) = 16.5*10^7 s

Decay Constant = 0.693 / (16.5*10^7) = 4.2*10^-9 s^-1

Activity , A = *N

Or, N = A / = 8.8*10^24

Therefore, Cobalt are needed to achieve this activity = (8.8*10^24)*60 / ( 6.023*10^23) = 876.64 g =0.876 kg [Answer]

PartB : Suppose after t years the activity of this Source be reduced to 10% of its initial Value.

We Know ,

A = Ao exp[-t] ----(i)

= 0.693/5.2173 = 0.1328 year^-1

A = 0.1Ao ---- Substitute this values in equation(i) we get

t = 17.3 years . [ Answer]