Answer
(a) $9.97$mg
(b) $1.39\times10^5$ years
Work Step by Step
(a) Given $h=2.7\times10^5,m_0=10,t=1000$, use the radioactive decay model $m(t)=m_0e^{-rt}$ where
$r=\frac{ln2}{h}=\frac{\ln2}{2.7\times10^5}$ thus $m(1000)=10e^{-1000\times\frac{\ln2}{2.7\times10^5}}\approx9.97$mg
(b) Let $m(t)=7$, we have $10e^{-\frac{t\ln2}{2.7\times10^5}}=7$,
which gives $t=-\frac{2.7\times10^5}{\ln2}\times ln(\frac{7}{10})\approx1.39\times10^5$ years