Question

Researchers have obtained promising results using Cu-67 in the radiotherapy of a number of cancers including non-Hodgkin's lymphoma. If the half-life of this isotope is 2.58 days, by what factor will its activity have decreased after 6.0 days?

Answer 1

find k, the decay constant:

\[t _{1/2}=\frac{ \ln2 }{ k }\]

t1/2= half life
k=decay constant

then find At,

\[A _{t}=A _{0}*e ^{-kt}\]

At= amount after time elapsed
Ao=initial amount
k=decay constant (from first equation)
t= time elapsed

then make a ratio At/Ao to find the factor by which is remains

to find he factor by which is decreased subtract that from 1 i.e. 1-At/Ao

Answer 2

thanks your the best!

Answer 3

no prob dude