So \({T_{1/2}} = 4\,\min \).
Again \(M = {M_0}{\left( {\frac{1}{2}} \right)^{t/{T_{1/2}}}}\)
==> \(10 = 80\,{\left( {\frac{1}{2}} \right)^{t/4}}\)
==> \(\frac{1}{8} = {\left( {\frac{1}{2}} \right)^3} = {\left( {\frac{1}{2}} \right)^{t/4}}\)
==> \(t = 12 min.\)
$\left( M _{\text {prot. }} \cong M _{\text {neut. }}=1.67 \times 10^{-27} kg \right)$