Answer
See explanation.
Work Step by Step
Fnd the initial activity rate.
First, from the half-life of 17 days, calculate the decay constant.
$ \lambda =\frac{ln2}{ T_{1/2}}=4.72\times10^{-7}s^{-1}$
The mass of a single nucleus is $103m_p=1.72\times10^{-25}kg$.
Find the number of atoms, N, in 0.250g = 0.000250kg.
$$N=\frac{0.000250kg }{1.72\times10^{-25}kg }=1.45\times10^{21}$$
Now find the initial rate of decays.
$|\frac{dN}{dt}|=(1.45\times10^{21})(4.72\times10^{-7}s^{-1})=6.86\times10^{14}Bq$.
b. 68 days later is exactly 4 half-lives.
Calculate the activity.
$$(6.86\times10^{14}Bq)2^{-17d/68d}=4.29\times10^{13}Bq $$
As expected, at the end of 4 half-lives, the activity rate is one-sixteenth of its initial rate.
