Question

I know how to solve normal 1/2 life equations but this one is confusing me! Please help me get both parts.

I know the standard equations of

t(1/2) = ln(2) / k & At= A0 e ^ (-kt)

but those didn't work when i tried. i got stuck! thanks! :)

This question has multiple parts. Work all the parts to get the most points. Radioactive cobalt-60 is used extensively in nuclear medicine as a γ-ray source. It is made by a neutron capture reaction from cobalt-59 and is a β emitter, β emission is accompanied by strong γ radiation. The half-life of cobalt-60 is 5.27 years. How long will it take for a cobalt-60 source to decrease to one thirty-second of its original activity? years b The half-life of obalt-60 is 5.27 year The half-life of cobalt-60 is 5.27 years. What fraction of the activity of a cobalt-60 source remains after 2.0 years? Submit

Answer 1

we know that

decay constant (k) = 0.693 / half life

so

k = 0.693 / 5.27

k = 0.1315

now

we know that

ln A = ln Ao - kt

ln A - lnAo = -kt

ln (A/Ao) = -kt

given

A / Ao = 1/3

so

ln (1/3) =- 0.1315 x t

t = 8.3545

so

the time taken is 8.3545 years

b)

now

ln (A/Ao) = -kt

given

t = 2years

so

ln (A/Ao) = -0.1315 x 2

(A/Ao) = 0.7687

0.7687 fraction of the activty remains after 2 years

A /Ao = 0.7687