Find the half life of a radioactive element, if its activity decreases for 1 month by 10%.
Activity of an isotope is measured by the number of nuclei decaying for a time unit.
Suppose that dNd nuclei decay for a short period of time dt.
Then the isotope activity A is expressed by the formula A=dNddt
It follows from the radioactive decay law that
N(t)=N0e−λt where N(t) is the quantity of the remaining nuclei.
Therefore,
Nd(t)=N0−N(t)=N0−N0e−λt=N0(1−e−λt)
By differentiating with respect to t, we find the expression for activity:
A(t)=dNddt=N0λe−λt
The initial isotope activity is equal to
A(t=0)=A0=N0λ
Hence, A(t)=A0e−λt
As it can be seen, the activity decreases over time by the same law as the amount of undecayed material. Substituting the expression for the half life T=ln2λ in the last formula, we can write:
A(t)=A0e−tln2T
The value of T in the last expression can be found by
e−tln2T=AA0
⇒−tln2T=lnAA0
⇒tln2T=lnA0A
⇒T=tln2lnA0A
In our case, the half life period of the given isotope is
T=tln2lnA0A=30ln2ln10090≈30⋅0.93ln1.11≈197.3days